Math 324: Multivariable Calculus

Transcription

1 Math 324: Multivariable alculus Matt Robinson February 26, 218 1

2 ONTENTS ONTENTS ontents 1 ouble Integrals over Regions ouble Integrals and Riemann Sums Integration in Polar oordinates Applications of ouble Integrals Surface Area Triple Integrals Applications of Triple Integrals oordinate Systems ylindrical oordinates Spherical oordinates hange of Variables erivatives irectional erivatives The Gradient Vector fields 51 6 Integration over urves Integrating Vector Fields over urves Green s Theorem Operations on Vector Fields The el Operator Integration over Surfaces Parametric Surfaces Surface Integrals Stokes Theorem ivergence Theorem The Fundamental Theorems of Vector alculus

3 1 OUBLE INTEGRALS OVER REGIONS 1 ouble Integrals over Regions In this section we build off what was covered in Math 126, namely with a review of double integrals, iterated integrals, integration in polar coordinates, and applications of double integration. 1.1 ouble Integrals and Riemann Sums Recall that if y f(x), then b a f(x)dx lim n n f(x j ) x j gives the signed area under the curve f(x) on the interval [a, b]. j1 Now consider a function z f(x, y) and a rectangle R in the xy-plane. Then f(x, y)da represents the signed volume under f(x, y) on R. R 3

4 1.1 ouble Integrals and Riemann Sums 1 OUBLE INTEGRALS OVER REGIONS In terms of Riemannan sums, we then have R f(x, y)da lim m,n m,n j,k1 f(x j, y k ) x j y k, where as m, n, the area x j y k for each j m, k n. We will not do Riemann sums in class, but let s recall how to compute double integrals in practice. Recall that an intergral of the form d b c a f(x, y)dxdy, is called an iterated integral and is computed as follows d b d [ b ] f(x, y)dxdy f(x, y)dx dy, c a where the piece in the inner bracket is single-variable function in y. If R [a, b] [c, d] is a rectangular region in the plane, then we need to a way to convert the double integral f(x, y)da into an iterated integral for which we can explicitly compute. R c a 4

5 1.1 ouble Integrals and Riemann Sums 1 OUBLE INTEGRALS OVER REGIONS Theorem 1 (Fubini s Theorem): Suppose f is continuous on the rectangle R [a, b] [c, d]. Then R f(x, y)da d b c a f(x, y)dxdy b d a c f(x, y)dydx. Example 2: ompute the volume of the solid whose base is the rectangle R [1, 5] [2, 7] and whose top lies in the plane z x + 2y + 1. In this example, we have that z f(x, y) x + 2y + 1. So V f(x, y)da R ( x + 2y + 1)dxdy [ 1 2 x2 + 2xy + 1x [( y (8y + 28)dy [4y y] y7 y2 ( ) ( ) 32. Example 3: z x sin xy over R [, 1] [, π]. ] x5 ) dy x1 ( y + 1 )] dy 5

6 1.1 ouble Integrals and Riemann Sums 1 OUBLE INTEGRALS OVER REGIONS R x sin xyda π 1 1 π 1 1 x sin xydxdy x sin xydydx cos xy yπ y dx ( cos πx + 1)dx [ 1π ] x1 sin πx + x 1. x requires by parts Example 4: z xex, R [, 1] [1, 2]. y ( xe x 1 y dydx 1 ) ( 2 xe x dx ln 2 xe x dx 1 ln 2 [xe x 1 ln 2 [ e e x 1 ] ln 2 [e e + 1] ln 2. 1 ) 1 y dy ] e x dx So far, we ve been dealing with the relatively straight-forward case of regions that are rectangles. Let s turn our attention to integrating over a more general region. Suppose we wanted to integrate over a region that looked like 6

7 1.1 ouble Integrals and Riemann Sums 1 OUBLE INTEGRALS OVER REGIONS then we have the integral f(x, y)da b g2 (x) a g 1 (x) which is the signed volume under f(x, y) on. Example 5: Evaluate (2 3x + xy)da f(x, y)dydx, if is the region bounded by the x-axis, the line x 1 and the line y 3x. Always draw a picture of! We have that x 1 and y 3x. Then 1 3x (2 3x + xy)da (2 3x + xy)dydx 1 1 [2y 3xy + 12 ] y3x xy2 dx y (6x 9x x3 ) dx [3x 2 3x ] x1 x x 7

8 1.1 ouble Integrals and Riemann Sums 1 OUBLE INTEGRALS OVER REGIONS Note that since we re not integrating over a rectangle, Fubini s Theorem does not apply. We can t say that the above integral is equal to 3x 1 (2 3x + xy)dxdy. This integral doesn t even make sense. Now, suppose we wanted to integrate over a region that looked like then we have the integral f(x, y)da d h2 (y) Let s look at the previous example again. Example 6: Evaluate c h 1 (y) (2 3x + xy)da f(x, y)dxdy. if is the region bounded by the x-axis, the line x 1 and the line y 3x. 8

9 1.1 ouble Integrals and Riemann Sums 1 OUBLE INTEGRALS OVER REGIONS From the same picture, we have that y 3 and y 3 (2 3x + xy)da y/3 (2 3x + xy)da [ 2x 3 2 x x2 y x 1. Then ] x1 x y 3 dy [ y 23 y + 16 y2 16 y3 ] dy [ y y2 1 ] 6 y3 dy [ 1 2 y 1 12 y y y4 Note that in these previous two examples, if f(x, y) is any continuous function, then 1 2x 3 1 f(x, y)dydx f(x, y)dxdy, where the only way to make sense of the bounds and equality is with a picture of. Example 7: Evaluate y 2 e xy da, First, let s draw the picture of. Then y/3 ] y3 y {(x, y) : y 3, x y}. 9

10 1.2 Integration in Polar oordinates 1 OUBLE INTEGRALS OVER REGIONS y 2 e xy da 3 y 3 3 y 2 e xy dxdy ye xy xy x dy (ye y2 y)dy [ 1 2 ey2 1 2 y2 ] y3 y 1 2 (e ) 1 2 e Integration in Polar oordinates Let s first recall what polar coordinates are. Instead of using the ordered pair (x, y) to describe a point in the plane, we use the ordered pair (r, θ), where r is the radius (or distance from the origin) and θ denotes the angle of rotation from the positive x-axis. Furthermore, recall our basic polar identities x 2 + y 2 r 2, x r cos θ, y r sin θ. We know artesian rectangles look like a normal rectangle but a polar rectangle looks like R [a, b] [c, d] {(x, y) : a x b, c y d}, R [a, b] [α, β] {(r, θ) : a r b, α θ β}, a piece of a pie-shaped wedge! So to integrate a function z f(x, y) on a polar rectangle (perhaps to find a volume), say of the form R [a, b] [α, β] {(r, θ) : a r b, α θ β}, we make the conversion x r cos θ, y r sin θ, da rdrdθ and define β b f(x, y)da f(r cos θ, r sin θ)rdrdθ. R α a 1

11 1.2 Integration in Polar oordinates 1 OUBLE INTEGRALS OVER REGIONS on t forget the extra r term! Even though we know the extra r term comes from the area of a piece of a wedge, it also has geometric significance that we ll revisit later in the course. Similarly to artesian coordinates, we could have more general regions. Suppose we had a region given by Then {(r, θ) : α θ β, g 1 (θ) r g 2 (θ)}. f(x, y)da β g2 (θ) α g 1 (θ) f(r cos θ, r sin θ)rdrdθ. Example 8: ompute xyda if is the region in the first quadrant that lies between the circles x 2 + y 2 4 and x 2 + y 2 2x ((x 1) 2 + y 2 1). onverting both circles to polar coordinates we have r 2 for the first and r 2 cos θ for the second. Then our bounds are θ π and 2 cos θ r 2. 2 Hence π/2 2 xyda (r cos θ)(r sin θ)rdrdθ 4 π/2 π/2 π/ π/ cos θ cos θ sin θ 2 2 cos θ cos θ sin θ 1 4 r4 r2 r 3 drdθ r2 cos θ dθ cos θ sin θ(4 4 cos 4 θ)dθ (cos θ cos 5 θ) sin θdθ (u u 5 )du (u u 5 )du [ 1 2 u2 1 6 u6 ] u1 Example 9: Find the volume of the solid bounded by the plane z and the parabaloid z 1 x 2 y 2. Set up the intergral in both artesian and polar coordinates. u 11

12 1.2 Integration in Polar oordinates 1 OUBLE INTEGRALS OVER REGIONS and so After drawing the picture, we see the artesian domain is given by {(x, y) : 1 x 1, 1 x 2 y 1 x 2 } V (1 x 2 y 2 )da 1 1 x x 2 (1 x 2 y 2 )dydx. Note how difficult this would be to evaluate. Where as, in polar coordinates, we have that {(r, θ) : r 1, θ 2π}, and V 1 4 (1 x 2 y 2 )da 2π 1 2π 1 2π π 2. 2π (1 r 2 cos 2 θ r 2 sin 2 θ)rdrdθ (1 r 2 )rdrdθ [ 14 (1 r2 ) 2 ] r1 dθ dθ r Example 1: Find the volume of the solid that lies under the parabaloid z x 2 + y 2 above the region in the plane bounded by the circle (x 1) 2 + y 2 1. First noting that (x 1) 2 + y 2 1 is equivalent to x 2 + y 2 2x, after converting to polar coordinates, we see that or rather r 2 2r cos θ, r 2 cos θ, and the circle makes a complete revolution as θ ranges from π to π. Therefore, in 2 2 polar coordinates our region is given by {(r, θ) : r 2 cos θ, π 2 θ π 2 }. 12

13 1.3 Applications of ouble Integrals 1 OUBLE INTEGRALS OVER REGIONS Thus, the volume is given by V 1 4 (x 2 + y 2 )da π 2 2 cos θ π 2 3π 2. π 2 π 2 π π 2 2 π π 2 2 r 2 rdrdθ π π cos 4 θdθ 2 π 2 (2 cos 2 θ) 2 dθ π 2 (1 + cos 2θ) 2 dθ (1 + 2 cos 2θ + cos 2 2θ)dθ ( cos 2θ + 1 cos 4θ ) dθ 1.3 Applications of ouble Integrals The volume of a uniform object (same height) is given by Volume (area of )h so if the height is identically one, then 1dA (area of )1 area of. Example 11: ompute the area of. 13

14 1.3 Applications of ouble Integrals 1 OUBLE INTEGRALS OVER REGIONS Area() 1dA 1 ex dydx x 2 1 (e x x 2 )dx [e x 13 ] x1 x3 e e 4 3. an we switch the order of integration here? Yes, but it will be the sum of two x 14

15 1.3 Applications of ouble Integrals 1 OUBLE INTEGRALS OVER REGIONS integrals. Then we have that for 1 that y 1 and x y and for 2 that 1 y e and ln x x 1. So da da + 1 da 2. 1 y e 4 3. dxdy + e 1 1 ln x dxdy We now turn our attention to an application of double integrals in the form of finding the mass of flat plates (negligible height) which are called lamina. Suppose is a region in the xy-plane and we have a lamina in the shape of. The density of the plate varies, in the particular, at the point (x, y), the density is given by the function ρ(x, y), measured in units of mass per unit area. 15

16 1.3 Applications of ouble Integrals 1 OUBLE INTEGRALS OVER REGIONS Example 12: Find the mass of the lamina in the shape of a triangle with vertices (2, 3), (8, 3) and (6, 5), if the density is given by ρ(x, y) x + y. We have that 3 y 5, 2y 4 x 11 y. So the mass is given by m y y 4 [ x yx (x + y)dxdy ] x11 y dy x2y 4 Note that this can analagously be used to find the total charge of an object by replacing density with charge density. Recall that x is proportional to y if there is a constant k such that y kx. x is inversely proportional to y if there is a constant k such that y k x. Example 13: A lamina occupies the portion of the first quadrant enclosed by the polar curve r cos 3θ. The density of the lamina at each point is inversely proportional to its distance from the origin. Let k be the constant of proportionality and find the mass of the lamina. Since dist((x, y), (, )) x 2 + y 2, we have that Our region is described by ρ(x, y) k x2 + y 2. 16

17 1.3 Applications of ouble Integrals 1 OUBLE INTEGRALS OVER REGIONS and so the mass m is given by m k k x2 + y 2 da π/6 cos 3θ π/6 cos 3θ π/6 cos 3θdθ k 3 sin 3θ θπ/6 k 3. θ k r 2 rdrdθ kdrdθ The previous application was dealing with finding masses of lamina, so now we ll turn our attention to the closely related topic of finding the moments and the center of mass of a lamina. efinition 14: The moment about the x-axis of a lamina (which is a measure of the tendency 17

18 1.3 Applications of ouble Integrals 1 OUBLE INTEGRALS OVER REGIONS of the lamina to spin about the x-axis) is given by the equation M x yρ(x, y)da and the moment about the y-axis is given by M y xρ(x, y)da. The center of mass of the lamina is the point (x, y), where x M y m, y M x m. Note that it is possible for the center of mass to lie outside of the lamina: Example 15: A lamina occupies the region in the first quadrant bounded by the circle x 2 +y 2 a 2. Its density function is ρ(x, y) xy 2. Find the center of mass. 18

19 1.3 Applications of ouble Integrals 1 OUBLE INTEGRALS OVER REGIONS Our region is the polar rectangle [, a] [, π/2]. Let s first compute the mass. m xy 2 da π/2 a π/2 a5 5 a5 π/2 15 sin3 θ a5 15. r cos θr 2 sin 2 θrdrdθ cos θ sin 2 θdθ a sin 2 θ cos θdθ θπ/2 θ Then the moment about the x-axis is M x yρ(x, y)da π/2 a π/2 sin 3 θ cos θdθ ( ) ( ) 1 a a6 24 r 4 dr r sin θr cos θr 2 sin 2 θrdrdθ a r 5 dr 19

20 1.3 Applications of ouble Integrals 1 OUBLE INTEGRALS OVER REGIONS and the moment about the y-axis is M y xρ(x, y)da π/2 a π/2 a6 6 a6 6 a6 24 a6 24 a6 48 π/2 π/2 r cos θr cos θr 2 sin 2 θrdrdθ cos 2 θ sin 2 θdθ π/2 π/2 π/2 a6 π 96. Then the center of mass is given by a r 5 dr cos 2 θ sin 2 θdθ ( ) ( 1 + cos 2θ 1 cos 2θ 2 (1 cos 2 2θ)dθ (1 12 (1 + cos 4θ) ) dθ (1 cos 4θ)dθ x M y m a6 π a 5 5πa 32, y M x m a6 24 5a a 5 So the center of mass is given by the coordinates ( 5πa 32, 5a ). 8 2 ) dθ 2

21 1.4 Surface Area 1 OUBLE INTEGRALS OVER REGIONS One final application deals with finding the moment of inertia which gives information on the resistance to change of rotation. That is, if we increase mass or increase how far the mass is from the axis, we increase moment of inertia (and make it harder to rotate). efinition 16: The moment of inertia for a single particle of mass m a distance of r units from the axis is defined to be mr 2. For a lamina with density ρ(x, y), the moments of inertia with respect to the axes is defined as follows: I y x 2 ρ(x, y)da moment of inertia from the y-axis, I o I x 1.4 Surface Area y 2 ρ(x, y)da moment of inertia from the x-axis, (x 2 + y 2 )ρ(x, y)da I x + I y moment of inertia about the origin. Recall that the area of of the parallelogram determined by two vectors u and v is given by u v. Now, suppose we have some surface in R 3 given by the function z f(x, y) over the region R, and we wish to compute the surface area of z f(x, y) over R. Now, let s assume a very fine partition of R into subrectangles R ij. Then as usual, the area of this subrectangle is given by A ij x i y j. At a point (x i, y j ) in R ij, we have the point P ij (x i, y j, f(x i, y j )) on the surface z f(x, y). The surface area of z f(x, y) over the subrectangle R ij is approximately equal to the area of the tangent plane at the point P ij. That is, S ij T ij, where S ij is the actual surface area over R ij and T ij the area of the parallelogram of the tangent plane at P ij above R ij. Hence the total surface area if given by the sum m,n i,j1 S ij m,n i,j1 T ij. raw pic of cente 21

22 1.4 Surface Area 1 OUBLE INTEGRALS OVER REGIONS Thus, we need a way to nicely characterize the area of the piece of the tangent plane T ij. Recall from Math that the two vectors, 1, y f(x i, y j ) and 1,, x f(x i, y j ) lie in the tangent plane at P ij and so scalling by the subdivision, we have that u x i,, x f(x i, y j ) x i and v, y j, y f(x i, y j ) y j lie entirely in the tangent plane above R ij. Thus the area T ij is given by T ij u v x f(x i, y j ) x i y j, y f(x i, y j ) x i y j, x i y j x f(x i, y j ), y f(x i, y j ), 1 A ij ( x f(x i, y j )) 2 + ( y f(x i, y j )) A ij. Summing these up, and taking the partition finer and finer as in Riemann sums, we get the following definition efinition 17: If f and its first partial derivatives are continuous on the region in the xy-plane, the surface area of S described by z f(x, y) over is given by Surface Area ds 1 + [ x f(x, y)] 2 + [ y f(x, y)] 2 da. Note that ds is called our surface area differential and is given by ds 1 + [ x f(x, y)] 2 + [ y f(x, y)] 2 da. Example 18: Find the area of the part of the paraboloid z x 2 + y 2 that lies under the plane z 9. Let s first compute some partial derivatives of f(x, y) x 2 + y 2. We have that Thus f x 2x, Surface Area ds 1 See my 126 notes, Section 5.2 on Tangent Planes f y 2y. 1 + (2x2 ) + (2y) 2 da 1 + 4(x2 + y 2 )da. 22

23 1.4 Surface Area 1 OUBLE INTEGRALS OVER REGIONS Projecting this surface down to the xy-plane, we see that our region is the disk of radius 3, witch is the polar rectangle [, 3] [, 2π]. Turning the above into an iterated integral, we see that Surface Area 1 + 4(x2 + y 2 )da 2π r2 rdrdθ 2π (1 + 4r2 ) 3/2 r3 r π 6 ( ). 23

24 2 TRIPLE INTEGRALS 2 Triple Integrals Thinking about the applications we ve just covered (finding mass and center of mass), it makes sense to now extend our notion of integration to higher dimensions, namely triple integrals. onsider the case where R is a rectangular prism in R 3 (3-dimensional Euclidean space). That is, R [a, b] [c, d] [r, s] {(x, y, z) : a x b, c y d, r z s}. And we wish to know the mass of the object R, where the denisty is then a function of three variables, ρ(x, y, z). Then the mass is given by the triple integral ρ(x, y, z)dv, where dv represents our infinitesimal volume R dv V x y z, taken in the form of a Riemann sum. This means our triple intgral is defined in the usual way via Riemann sums, i.e., R ρ(x, y, z, )dv lim l,m,n l,m,n i,j,k1 ρ(x i, y j, z k ) x i y j z k. To actually compute these triple integrals, we use similar methods as in the twodimensional case. Theorem 19 (Fubini s Theorem): If f is continuous on the rectangular box R [a, b] [c, d] [r, s], then R f(x, y, z)dv s c b r b a f(x, y, z)dxdydz. Note: that just in the two-dimensional case, you can switch the order of integration here, however, there will be six different permutations all yielding the same result. Example 2: Evaluate the triple integral R xyz2 dv, where R [, 1] [ 1, 2] [, 3]. 24

25 2 TRIPLE INTEGRALS Let s integrate as follows: R xyz 2 dv xdx xyz 2 dzdydx 9xydydx We can also do triple integrals over more genreal regions besides boxes, however, since we re in three-dimensions now, we ll have three general tpyes. efinition 21: A solid E is of type I if it lies above a region in the xy-plane between two continuous functions of only x and y, that is, E {(x, y, z) : (x, y), u 1 (x, y) z u 2 (x, y)}. If f(x, y, z) is any continuous function on E, then f(x, y, z)dv [ u2 (x,y) E u 1 (x,y) f(x, y, z)dz ] da. 25

26 2 TRIPLE INTEGRALS This is interpreted as we first integrate with respect to z (holding x and y constant), leaving just to compute a double integral over in the variables of x and y (as we ve covered in the previous section). efinition 22: A solid E is of type II if it lies above a region in the yz-plane between two continuous functions of only y and z, that is, E {(x, y, z) : (y, z), u 1 (y, z) x u 2 (y, z)}. If f(x, y, z) is any continuous function on E, then E f(x, y, z)dv [ u2 (y,z) u 1 (y,z) f(x, y, z)dx efinition 23: A solid E is of type III if it lies above a region in the xz-plane between two continuous functions of only x and z, that is, ] da. E {(x, y, z) : (x, z), u 1 (x, z) y u 2 (x, z)}. 26

27 2 TRIPLE INTEGRALS If f(x, y, z) is any continuous function E, then E f(x, y, z)dv [ u2 (x,z) u 1 (x,z) f(x, y, z)dy Example 24: Evaluate zdv, where E is the tetrahedron bounded by the four planes x, E y, z, and x + y + z 1. Note that this tetrahedron has vertices (,, ), (1,, ), (, 1, ), and (,, 1). Thus if we project on the yz-plane, we see that y 1 and z 1 y and ] da. 27

28 2 TRIPLE INTEGRALS we re of type II between the two surfaces x and x 1 y z. Then [ 1 y z ] zdv zdx da E y 1 y z 1 1 y 1 ( 1 1 zdxdzdy ((1 y)z z 2 )dzdy 2 (1 y)3 1 ) (1 y)3 dy 3 (1 y) 3 dy 1 24 (1 y)4 y1 y Example 25: Evaluate E x2 + z 2 dv where E is the region bounded by the surfaces y 4 and y x 2 + z 2. Let s first set this up as a type I integral, that is projecting E onto the xy-plane, we see that 2 x 2 and x 2 y 4, and we get the triple integral 2 4 y x 2 x2 + z 2 dv x2 + z E 2 x 2 y x 2 dzdydx. 2 This integral is very difficult to evaluate, so let s try setting this up as a type III integral instead. That is, E x2 + z 2 dv 4 x2 + z 2 dyda x 2 +z 2 (4 x 2 z 2 ) x 2 + z 2 da, where is the disk of radius 2 in the xz-plane. Thus, let s switch to polar coordinates 28

29 2 TRIPLE INTEGRALS in the xz-plane, where r 2 x 2 + z 2, r 2 and θ 2π. Hence x2 + z 2 dv (4 x 2 z 2 ) x 2 + z 2 da E 2π 2 (4r 2 r 4 )drdθ ( 4 2π ) π π 15. Example 26: Set up but do not solve the integral f(x, y, z)dv as an iterated integral in six E different ways, where E is the solid bounded by y 2 + z 2 9, x 2, and x 2. Note that this is a cylinder of height 4 in along the x-axis of radius 3. In the type I case, projecting to the xy-plane, we have a projection to the rectangle [ 2, 2] [ 3, 3] with 9 y 2 z 9 y 2 and so y 2 f(x, y, z)dv f(x, y, z)dzdxdy E y y f(x, y, z)dzdydx. 9 y 2 In the type II case, projecting to the yz-plane, we have a projection of a circle of radius 3 with 2 x 2. Thus 3 9 z 2 2 f(x, y, z)dv f(x, y, z)dxdydz E 3 9 z y y 2 2 f(x, y, z)dxdzdy. In the type III case, projecting to the xz-plane, we again have a prokection to a rectangle of the form [ 2, 2] [ 3, 3] with 9 z 2 y 9 z 2 and hence z 2 f(x, y, z)dv f(x, y, z)dydzdx E z z z 2 dydxdz 29

30 2.1 Applications of Triple Integrals 2 TRIPLE INTEGRALS 2.1 Applications of Triple Integrals Similar to case of calculating area via double integrals with A() da, we can also define the volume of solids V (E) to be V (E) dv. We also have the various definitions for computing mass, center of mass, and moments: efinition 27: If the solid E has density function ρ(x, y, z), then the mass of E is given by m ρ(x, y, z)dv. The moments of E about the three coordinates planes are given by M xy zρ(x, y, z)dv, M xz yρ(x, y, z)dv, E E M yz E E xρ(x, y, z)dv. The center of mass of the solid E is the point (x, y, z), where x M yz m, y M xz m, E z M xy m. The moments of inertia about the three corrdinates axes are given by I x (y 2 + z 2 )ρ(x, y, z)dv E I y (x 2 + z 2 )ρ(x, y, z)dv E I z (x 2 + y 2 )ρ(x, y, z)dv. E 3

31 3 OORINATE SYSTEMS 3 oordinate Systems When computing double integrals, we ve seen many examples of when the domain is circular in nature, it makes sense to convert to polar coordinates to carryout the integration. What is the natural generalization of this to triple integrals? 3.1 ylindrical oordinates The first natural notion of extending polar coordinates to R 3, is keeping the same definition for (r, θ) as in polar coordinates and then let z stay the same. That is, efinition 28: The cylindrical coordinate system on R 3 represents a point P as an ordered triple, P (r, θ, z), where r is the distance from P to the z-axis, θ is the angle of roation from the positive x-axis, and z the signed distance from the xy-plane. 31

32 3.1 ylindrical oordinates 3 OORINATE SYSTEMS The conversion between artesian coordinates and cylindrical coordinates works simillar as in polar coordinates, that is, (x, y, z) (r, θ, z), is given by and conversely, x r cos θ, y r sin θ, z z, r 2 x 2 + y 2, tan θ y x, z z. Note, that you can rotate cylindrical coordinates to be along any axis, depending on the desired situation. E.g., for cylindrical coordinates along the y-axis, we have that x r cos θ, z r sin θ, y y. Now, let s see how to integrate using cylindrical coordinates: Suppose we have a type I solid E. That is, E {(x, y, z) : (x, y), u 1 (x, y) z u 2 (x, y)}, and if f(x, y, z) is any continuous function on E, we have that [ ] u2 (x,y) f(x, y, z)dv f(x, y, z)dz da. E u 1 (x,y) Now, suppose we can write in polar coordinates, that is, then from the above, we get E f(x, y, z)dv {(r, θ) : α θ β, g 1 (θ) r g 2 (θ)}, β g2 (θ) u2 (r cos θ,r sin θ) α g 1 (θ) u 1 (r cos θ,r sin θ) f(r cos θ, r sin θ, z)rdzdrdθ. Example 29: Find the volume of the solid E that lies inside the cylinder x 2 + y 2 1, outside the parabaloid z 1 x 2 y 2, and below the plane z 2. 32

33 3.1 ylindrical oordinates 3 OORINATE SYSTEMS In cylindrical coordinates, we see that E is the region of θ 2π, r 1, and 1 r 2 z 2. omputing the volume, V dv Example 3: Evaluate E 2π 1 2 2π 1 1 r 2 rdzdrdθ (1 + r 2 )rdrdθ 2π 2 (1 + r2 ) r1 r π. 2 4 y y 2 x 2 +y 2 xzdzdxdy. We see that E is the cylinderical region of θ 2π, r 2, r z 2. 33

34 3.2 Spherical oordinates 3 OORINATE SYSTEMS Hence 2 4 y y 2 x 2 +y 2 xzdzdxdy 2π 2 2 2π. r cos θdθ r 2 z cos θdzdrdθ 2 2 r r 2 zdzdr ylindrical coordinates are a good first notion of generalizing polar coordinates, however, there are other geometric generalization for which cylindrical coordinates aren t helpful. This leads us to our next natural generalization. 3.2 Spherical oordinates In contrast to cylindrical coordinates, where we just kept polar coordinates and added the z-component, we re redefining a couple of the components here. efinition 31: The spherical coordinate system on R 3 represents a point P as the ordered triple, P (ρ, θ, φ), where ρ is the distance from P to the origin, θ is the angle of rotation from the positive x-axis (the same as in polar coordinates), and φ is the angle of rotation down from the positive z-axis. Note that by convention ρ and φ π; θ is called the polar angle and φ is called the azimuthal angle. 34

35 3.2 Spherical oordinates 3 OORINATE SYSTEMS Let s use some simple geometry to derive the conversion betrween artesian and sphereical coordinares. From the above picture, we can see that and as usual Putting it all together, is given by and conversely, ρ cos φ z, ρ sin φ r, x r cos θ, y r sin θ. (x, y, z) (ρ, θ, φ), x ρ cos θ sin φ, y ρ sin θ sin φ, z ρ cos φ, ρ 2 x 2 + y 2 + z 2, tan θ y x, tan φ r z, where r is given as the usual radial coordinate. 35

36 3.2 Spherical oordinates 3 OORINATE SYSTEMS Now, let s see how to compute a triple integral using spherical coordinates. Suppose we wish to evaluate f(x, y, z)dv, E in spherical coordinates. We know how to sitch from artesian coordiantes to spherical coordinates using the formulas x rρ cos θ sin φ, y r sin θ sin φ, z ρ cos φ, and we can describe the region E in terms of spherical inequalities. Thus we so far have f(x, y, z)dv f(ρ cos θ sin φ, ρ sin θ sin φ, ρ cos φ)dv. E E However, unlike the cylindrical case, where our cylindrical-volume differential was easily obtained from our polar-area differential to be dv rdrdθdz, we don t yet have formula to define our sphirical-volume differential. First, let s consider a nearly infinitisimal small sphirical prism, that is, E [ρ, ρ + ρ] [θ, θ + θ] [φ, φ + φ]. If we recall the length of an arc is given by we then have the following side lengths of Rϕ, ρ, ρ φ, ρ sin φ θ. 36

37 3.2 Spherical oordinates 3 OORINATE SYSTEMS Now, since we re dealing with an infinitesimal rectangles, the volume is approximately calculated in the same way as artersian rectangles, that is, V ( ρ)(ρ φ)(ρ sin φ θ), and hence taking limits, we see that our spherical-volume differential is given by dv ρ 2 sin φdρdθdφ. You can make this more precise, as in the case of polar coordinates, by using the volume of a spherical cones. Finally, if E [a, b] [α, β] [γ, δ] is a sphirical prism γ α b f(x, y, z)dv f(ρ cos θ sin φ, ρ sin θ sin φ, ρ cos φ)ρ 2 sin φdρdθdφ E δ β a 37

38 3.3 hange of Variables 3 OORINATE SYSTEMS Example 32: Use spherical coordinates to compute the volume of the ball of radius r. Since a ball of radius R is given by E [, R] [, 2π] [, π], we have V (E) dv E π 2π R R3 3 π 2π ρ 2 sin φdρdθdφ sin φdθdφ 2 π 3 πr3 sin φdφ 4 3 πr hange of Variables We ve seen a few examples in this course of changing variables already. By a change of variable, we mean a different coordinate system. In R 2, we ve seen polar coordinates, where (x, y) (r, θ), and in R 3, we ve seen cylindrical coordinates and spherical coordinates (x, y, z) (r, θ, z) (x, y, z) (ρ, θ, φ). These were all examples of change of variables, and the main goal of introducing these was to allow us to describe regions in a new way as to facilitate integration. For integration, however, we had to use very example-specific methods to determine the area differential for polar coordinates or the volume differentials for cylindrical and sphereical coordinates. This leads to the question: Is there a more general method of determining these differentials? To answer this, let s recall from Math 124, u-substitution: b a f(x)dx d c f(g(u))g (u)du, 38

39 3.3 hange of Variables 3 OORINATE SYSTEMS where g(c) a and g(d) b. This is actually an example of change of variables with given by the conversion (x) (u), x g(u). And the length differential is represented by g (u)du in complete generality for 1- dimension. We can generalize this type of change of variable to R 2. Indeed, let s take with conversion x g(u, v), (x, y) (u, v), y h(u, v). This is sometimes called a transformation and denoted T (u, v) (x, y). It transforms a region in the uv-plane to a region in the xy-plane. Example 33: What is the image under T of the region in the uv-plane bounded by u, u 1, v, v π, where T is the transformation given by x u cos v, y u sin v. 2 This is just polar coordinates! 39

40 3.3 hange of Variables 3 OORINATE SYSTEMS Example 34: What is the image of the square R {(u, v) : u 1, v 1} under the transformation x u 2 v 2, y 2uv? We see the line u corresponds x v 2, y, so we lie on the negative x-axix moving to as v goes to. Similiarly when v, y and x u 2, so it loes on the positive x-axis moving to as u goes to. When v 1, x u 2 1 and y 2u and so substituting, x y2 4 1, and similar yor u 1, x 1 y2 4. Now, let s return to using change of variables for integration. Suppose we have a transformation T (u, v) (x, y) with x g(u, v), y h(u, v) which takes some region R in the uv-plane to the region in the xy-plane. Then, in similiarity to u-substitution, for double integrals, f(x, y)dxdy f(g(u, v), h(u, v))j(u, v)dudv. R What is this function J(u, v) replacing the single derivative in the u-substitution case? Notational Aside: As a function T : R 2 R 2 with T (u, v) (g(u, v), h(u, v)). We will also write x x(u, v) and y y(u, v) supressing the g and h functions. This is in similarity to single-variable calculus, where in u-substitution, we write u u(x) and du u (x)dx. 4

41 3.3 hange of Variables 3 OORINATE SYSTEMS efinition 35: The Jacobian of the transformation T (u, v) (x, y), denoted (x,y) (u,v) (x, y) (u, v) x x u v y y. u v is given by Its derivation is similar in nature to the surface area derivation, but it s not particularly enlightening to go through that again. See the textbook for further details. Theorem 36: Suppose that T is a transformation with x g(u, v) and y h(u, v) such that g and h are continuous and have continuous first partial derivatives. If the Jacobian is not identically zero, and T (R), then f(x, y)da f(g(u, v), h(u, v)) (x, y) (u, v) dudv. Example 37: ompute the Jacobian for polar coordinates. R (x, y) (r, θ) x r y r x θ y θ r cos 2 θ + r sin 2 θ r and so J r, which we already know since the polar-area differential is given by da rdrdθ. Example 38: Use the change of variables x u 2 v 2 and y 2uv to compute the integral yda, where is the region bounded by the lines y, x 1 y 2 /4 and x y 2 /4 1. From Example 34, we know the transformation T takes the unit sqaure R onto. Thus, we need to only compute the Jacobian (x, y) (u, v) 2u 2v 2v 2u 4u 2 + 4v 2. 41

42 3.3 hange of Variables 3 OORINATE SYSTEMS And so yda (2uv)(4u 2 + 4v 2 )dudv R (u 2 + v 2 )2uvdudv (u 2 + v 2 ) 2 u1 u vdv So far, we ve introduced change-of-variables in R 2, but an identical process works for change-of-variables in R 3. efinition 39: If T is a transformation which maps a region R in uvw-space to a region E in xyzspace, by the equations x g(u, v, w), y h(u, v, w), z k(u, v, w), then the Jacobian of T, denoted (x,y,z) (u,v,w) (x, y, z) (u, v, w) is computed as x x x u v w y y y u v w z z. Also, in similiarty to before, we have the equality f(x, y, z)dv f(g(u, v, w), h(u, v, w), k(u, v, w)) (x, y, z) (u, v, w) dudvdw. E Example 4: ompute the Jacobian for spherical coordinates. R ρ cos θ sin φ ρ cos θ sin φ ρ cos θ sin φ (x, y, z) ρ θ φ (ρ, θ, φ) ρ sin θ sin φ ρ sin θ sin φ ρ sin θ sin φ ρ θ φ ρ cos φ ρ cos φ ρ cos φ ρ θ φ cos θ sin φ ρ sin θ sin φ ρ cos θ cos φ sin θ sin φ ρ cos θ sin φ ρ sin θ cos φ cos φ ρ sin φ u z v cos φ(ρ 2 cos 2 θ cos φ sin φ + ρ 2 sin 2 θ cos φ sin φ) w ρ sin φ( ρ cos 2 θ sin 2 φ ρ sin 2 θ sin 2 φ) ρ 2 sin φ cos 2 φ + ρ 2 sin φ sin 2 φ ρ 2 sin φ 42

43 3.3 hange of Variables 3 OORINATE SYSTEMS In summary, this might ve seemed like a fairly convoluted process, however, when doing these types of problems, it s best to keep in mind the three change-of-variables we re familiar with (polar, cylindrical, spherical), and think about the order in which we compute things with them. Let s try and outline the methodology of using arbitrary coordinates: 1. hoose a change-of-variable x g(u, v), y h(u, v) that will simplify the problem (this is not unique). 2. etermine the bounds on u and v which correspond to the desired bounds on x and y. This depends on your choice in Step ompute the Jacobian of the transformation chosen. 4. Replace the x s and y s with u and v in the integration, along with dxdy J(u, v)dudv. 5. Integrate! 43

44 4 ERIVATIVES 4 erivatives We now turn our attention to revisiting derivatives, which we will then relate to integration later in the course. We know how to take partial derivatives from Math 126 2, but we need another technique. Recall the single-variable chain rule: If y f(x) and x g(t), then dy dt dy dx dx dt. efinition 41: Suppose z f(x, y) with x g(t) and y h(t), then the multivariable chain rule is dz dt z dx x dt + z dy y dt. Example 42: If z x 3 + xy 2 + y, x e 2t and y sin t, compute dz. dt By the chain rule, we have that dz dt (3x2 + y 2 )(2e 2t ) + (2xy + 1)(cos t) (3e 4t + sin 2 t)(2e 2t ) + (2e 2t sin t + 1) cos t. Example 43: If z sin(xy) and x 3t, y t 2, find dz. dt By the chain rule, dz dt (y cos(xy))(3) + (x cos(xy))(2t) (3t 2 + 6t 2 ) cos(3t 3 ) 9t 2 cos(3t 3 ). Proposition 44: Suppose z f(x, y) and x g(s, t), y h(s, t), then z s z x x s + z y y s, z t z x x t + z y y t. Example 45: Suppose z e 2x cos y and x s 2 + st, y st 2, then compute z s 2 See my 126 notes, Section 5.1 on Partial erivatives and z t. 44

45 4 ERIVATIVES and z s (2e2x cos y)(2s + t) + ( e 2x sin y)(t 2 ) 2e 2(s2 +st) cos(st 2 )(2s + t) e 2(s2 +st) sin(st 2 )t 2 z t 2e2(s2 +st) cos(st 2 )(s) e 2(s2 +st) sin(st 2 )(2st) You can generalize the chain rule to an arbitrary number of variables as well. Proposition 46: Suppose u f(x 1,..., x n ) and and for each j, x j g(t 1,..., t m ), then for each k, u u x 1 + u x u x n t k x 1 t k x 2 t k x n t k n u x j. x j t k j1 Example 47: If u xyz + xz 2, where x rse t, y rse t, and z rs, find u s. u s (yz + z2 )(re t ) + (xz)(re t ) + (xy + 2xz)(r) (r 2 s 2 e t + r 2 s 2 )(re t ) + (r 2 s 2 e t )(re t ) + (r 2 s 2 + 2r 2 s 2 e t )(r) We can generalize implicit differentiation as well. Suppose we have the equation F (x, y). Then we can consider the function as implicitly defining y as a function of x. To find dy, we take the derivative of the above equation with respect to x, and by the chain dx rule, we see that F dx x dx + F dy y dx, and hence solving for dy dx and using the fact that dx dx 1, we get the following theorem Theorem 48 (Implicit Function Theorem): Under certain hypotheses (beyond the scope of this course) where F (x, y) defines y implicitly as a differentiable function of x, then F dy dx x. F y 45

46 4.1 irectional erivatives 4 ERIVATIVES Note that one of the hypotheses of this theorem must be tht F is nonvanishing. y Also note that this theorem can generalize to an arbitrary number of variables. For example if F (x, y, z), a similar calculation shows that F z x x, Example 49: ompute dy if dx x2 + xy y 2. Let F (x, y) x 2 + xy y 2. Then F z F z y y. F z 4.1 irectional erivatives F dy dx x F y 2x + y x 2y. Before defining directional derivatives, the motivation for using them is because they re one of the natural generalization of partial derivatives. To see this, recall the definition of partial derivaitves: If z f(x, y), then z x lim h z y lim h f(x + h, y) f(x, y) h f(x, y + h) f(x,, y). h These represented the rates of change in the directions e 1 i 1, and e 2 j, 1, respectively. Suppose we wish to calculate the rate of change in an arbitrary direction u a, b. This leads to the following defintion: efinition 5: The directional derivative of the function f(x, y) in the direction u a, b at the point (x, y ) is given by f(x + ha, y + hb) f(x, y) u f(x, y ) lim. h h How do we compute directional derivatives in practice? 46

47 4.1 irectional erivatives 4 ERIVATIVES Theorem 51: u f(x, y ) a f x (x, y ) + b f y (x, y ). To see this, let g(h) f(x + ha, y + hb). Then on one hand g () lim h g(h) g() h f(x + ha, y + hb) f(x, y ) lim h h u f(x, y ). On the other hand, if we let g(h) f(x, y) where x x + ha, y y + hb, then by the chain rule and so thus showing the theorem. dg dh f dx x dh + f dy y dh a f x + b f y, g () a f x (x, y ) + b f y (x, y ), Note that with this theorem, it s easy to see that partial derivatives have an interpretation as direction derivatives. Indeed, f x e 1 f, f y e 2 f. One caveat with directional derivatives that differs with the book s definition. The book only defines directional derivatives for u being a unit vector. As we saw above, neither the definition nor the theorem requires this, so it s a superfluous hypothesis. However, since the homework uses this convention, we ll adopt it as well from this point forward. Furthermore, since we re now only dealing with the case of u being of unit-length, if θ is the angle the vector u makes with the positive x-axis, then we can write u cos θ, sin θ. 47

48 4.2 The Gradient 4 ERIVATIVES Example 52: Find the directional derivative of f(x, y) x 3 3xy + 4y 2 in the direction of the unit vector with angle θ π. 6 Since u cos(π/6), sin(π/ , 1, we see the directional derivative 3 u f(x, y) 2 (3x2 3y) + 1 (8y 3x) 2 Example 53: Find the directional derivative of f(x, y) ln(x 2 + y 2 ) at the point (2, 1) in the direction 1, 2. Normalizing the vector, we see that u 1 5 1, 2. Then [ u f(2, 1) 1 2x 5 x 2 + y + 2 ] 2y 2 5 x 2 + y (2,1) The Gradient Recalling the dot product of two vectors u u 1, u 2, v v 1, v 2 is given by u v u 1 v 1 + u 2 v 2. Then we can rewrite the directional derivative of f(x, y) in the direction u a, b as follows u f(x, y) a f x + b f y xf, y f a, b. This motivates the following definition: efinition 54: The gradient of a function f(x, y) is the vector-valued function grad (f) (x, y) f(x, y) x f(x, y), y f(x, y). 48

49 4.2 The Gradient 4 ERIVATIVES Using this notation, we see that the directional derivative can we written as u f f u. Note that this definition generealizes to any number of variables in the usual way: If f(x, y, z) is a function, then f f x, f y, f z. Example 55: ompute the gradient for f(x, y, z) x sin(yz). f sin(yz), xz cos(yz), xy cos(yz). What information does this gradient vector encapsulate? Well one natural question one might first ask about a function f(x, y) is in what direction does f change the fastest? What is the maximum rate of change? We actually already know this answer. Let u be any unit-direction. Then u f f u f u cos θ f cos θ, there θ is the angle between the vectors f and u. Thus the directional derivative is maximized when cos θ is maximized which happens when θ. That is, when f and u are in the same direction. Theorem 56: If f is a differentiable function of any number of variables, then the maximum value of the directional derivative u f is f and it occurs when u is the same direction as f. Example 57: Find the maximum rate of change of f(x, y) 4y x at the point (1, 1) and the direction at which is occurs. f(1, 1) 2y x, 4 x 2, 4. (1,1) 49

50 4.2 The Gradient 4 ERIVATIVES Thus the maximum value is and the direction is f 2 5, 1 2 2, 4. 5 There is also a close connection between gradients and tangent planes to surfaces. Suppose F (x, y, z) k is a function of three variables and k is constant. We can think of this as an implicitly defined function for z of the two variables x, y or as a level surface S of the function w F (x, y, z). Now imagine that is a curve on S passing through the point P (x, y, z ). So can be paramterized as the vector-valued function r(t) x(t), y(t), z(t). Now let s differentiate F (x, y, z) k with respect to t to see that via the chain rule Since the above equation turns into F dx x dt + F dy y dt + F dz z dt. r (t) x (t), y (t), z (t), F r (t). This tells us that the gradient of F is always orthogonal to any curve that lies on the surface S, that is, for any point P on S, F is orthogonal to every tangent vector at P. Hence F is orthogonal to every tangent plane on S. This means that F can be used to describe the tangent planes, since a normal vector and a point completely describe a plane. Proposition 58: Suppose S is the level surface defined by the equation F (x, y, z) k. Then for any P (x, y, z ) on S, the tangent plane at P is given by the equation F x (x, y, z )(x x ) + F y (x, y, z )(y y ) + F z (x, y, z )(z z ). Example 59: Find the tangent plane to ellipsoid 4x 2 + y 2 + 2z 2 4 at the point (1,, ). Letting F (x, y, z) 4x 2 + y 2 + 2z 2, we see that F (1,, ) 8x, 2y, 4z (1,,) 8,,, and so we get the plane 8(x 1), x 1. 5

51 5 VETOR FIELS 5 Vector fields A vector field is a a function that assigns to each point in R 2 or R 3 a vector. That is, as a function F and can be written in several different ways (in R 2 ) F(x, y) P (x, y)i + Q(x, y)j P (x, y)e 1 + Q(x, y)e 2 P (x, y), Q(x, y) P, Q. We ve already seen an important example of vector fields, namely the gradient vector field, that is, if f(x, y) is a function then is a vector field. f f x, f y Example 6: Sketch the vector field given by F(x, y) x, x + y. Example 61: Sketch the vector field given by F(x, y) yi + xj. 51

52 5 VETOR FIELS Example 62: Sketch the vector field given by F(x, y) ye 2. Example 63: Find and sketch the gradient vector field given by f(x, y) x 2 + y 2. The gradient is given by f 2x, 2y. Note that at each point, the gradient vector points in the direction of maximum increase of the function, and its length represents how rapidly the function is increasing. efinition 64: A vector field F is called conservative if there is a function f such that F f. If such an f exists, it is called a potential function for F. Note: Potential functions are not unique! Indeed, if f is any potential function, i.e., f F, then for any constant c, f + c is another potential function, since c, we have that (f + c) f + c f F. 52

53 5 VETOR FIELS Example 65: Is the vector field F y, x conservative? Yes. Taking antiderivatives, we see that works. f(x, y) xy Example 66: Is the vector field F y, x conservative? Taking the antiderivative of both components, we see that f must equal both xy and xy, which can t happen. Thus it s not conservative. Example 67: Is the vector field F x, y conservative? Taking the antiderivatives, we see that works. f(x, y) 1 2 (x2 + y 2 ) onservative vector fields have surprising applications in integration, that we ll come back to in detail later. 53

54 6 INTEGRATION OVER URVES 6 Integration over urves So far, we ve seen double integrals over regions in R 2 and triple integrals over solids in R 3. Let s now return to single integrals, however, instead of integrating along an interval [a, b], we wish to generalize this notation to integrating over curves which lie in either R 2 or R 3. efinition 68: Given a smooth curve in R 2, the line integral of f(x, y) over is given by f(x, y)ds, where s is the length of the curve. That is, this integral is being taken with respect to arclength. This integral can be interpreted as the area of the wall under the function f(x, y) and above the curve. Let s see how to compute line integrals in practice. Suppose is any curve in R 2 and suppose we have a parametrization of given by r(t) x(t), y(t), a t b. Recall that the length of the curve from r(a) to r(t) is given by t t (dx ) 2 ( ) 2 dy s(t) r (λ) dλ + dλ. dλ dλ a Thus, by the Fundamental Theorem of alculus (dx ) 2 ( ) 2 ds dy dt +, dt dt or rather, in differential form (dx ) 2 ds + dt a ( ) 2 dy dt. dt ds is called the arclength line differential, or just llne differential. Proposition 69: If is a smooth curve with parametrization r(t) x(t), y(t), a t b, then b (dx ) 2 ( ) 2 dy f(x, y)ds f(x(t), y(t)) + dt. dt dt a 54

55 6 INTEGRATION OVER URVES Note that this is independent of parametrization provided you you only traverse the curve once. A few notes about parametrizations: i. If is part of a circle of radius r, then one parametrization of would be r(t) r cos t, r sin t with appropriate bounds on t. ii. If is the graph of some function y f(x), then one parametrization of would be r(t) t, f(t) with appropriate bounds on t. iii. If is a line (in R 2 or R 3 ), we can use the equation of a line as in part (ii.), or we can write r(t) P + vt, where P is any point on the line and v is a direction vector. Example 7: Evaluate (4 + 2xy)ds where is the upper half of the cirve x2 + y 2 1. First, we parametrize r(t) cos t, sin t, t π. Then (4 + 2xy)ds π π 4π. (4 + 2 cos t sin t) cos 2 t + sin 2 tdt (4 + sin 2t)dt We can also do integration over piecewise differentiable curves. If 1 + 2, where 1 and 2 are differentiable curves, then f(x, y)ds f(x, y)ds + 1 f(x, y)ds. 2 Example 71: ompute xds where is the quarter of the circle x2 + y 2 4 from (2, ) to (, 2), followed by the straight line connecting (, 2) to ( 2, ). Let 1 be the circle arc parametrized by r 1 (t) 2 cos t, 2 sin t with t π 2 and let 2 be the straight line with rapametrization r 2 (t) t, 2 + t with 2 t. Then xds 1 xds + π cos t4dt + 2 xds 2 t 2dt 55

56 6 INTEGRATION OVER URVES efinition 72: If is a piecewise smooth curve with density ρ(x, y), then the mass of is m f(x, y)ds, and the center of mass (x, y), where x 1 m y 1 m xρ(x, y)ds yρ(x, y)ds. There are specific examples of line integrals which which will be useful later. efinition 73: Given a curve with r(t) x(t), y(t), a t b, we can compute the line integral of f(x, y) over with respct to x by b f(x, y)dx f(x(t), y(t))x (t)dt, and the line integral of f(x, y) over with respect to y by b f(x, y)dy f(x(t), y(t))y (t)dt. In the future, we will see these written in the form P (x, y)dx + Q(x, y)dy, or more notationally convenient a a P (x, y)dx + Q(x, y)dy. Example 74: If is the curve given by x(t) t 3, y(t) t, t 2 then compute y2 dx + xdy. y 2 dx + xdy t 2 (3t 2 )dt + 2 t 3 dt 56

57 6 INTEGRATION OVER URVES When we parametrize, we naturally choose an orientation of, meaning a way at which we travel along. If we travelled the opposite way, we would get a new parametrization. If we call this curve, then f(x, y)ds f(x, y)ds, but and f(x, y)dx f(x, y)dx f(x, y)dy f(x, y)dy. So far, all of these examples and definitions dealth with curves in R 2, but they work just the same for curves in R 3. In summary, is a curve in R 3 with parametrization r(t) x(t), y(t), z(t), a t b. Then the line integral of f(x, y, z) over with respect to arclength is given by f(x, y, z)ds b a (dx ) 2 f(x(t), y(t), z(t)) + dt We also have the inegrals with respect to x, y, and z b f(x, y, z)dx f(r(t))x (t)dt, a b f(x, y, z)dy f(r(t))y (t)dt, a b f(x, y, z)dz f(r(t))z (t)dt. a ( ) 2 dy + dt ( ) 2 dz dt. dt Example 75: ompute y sin zds where is the curve parametrized by r(t) cos t, sin t, t with t 2π. y sin zds 2π 2π. sin 2 t 2dt Now that we know how to integrate functions over curves, we can actually take this a bit further, an integrate vector fields over curves! 57

58 6.1 Integrating Vector Fields over urves 6 INTEGRATION OVER URVES 6.1 Integrating Vector Fields over urves Let s now see the importance of vector fields and their relation to integration. First, perhaps a brief motivation: In previous classes, you ve probably learned various formulas for work. That is, W F d, where W is the work done by some constant force F accross some constant distance d. If F is variable, then from Math 125, we ve learned that W b a F (x)dx. Moreover, in vectorial based physics, if F is a constant force vector and d is a displacement vector, then the scalar quantity work is given by W F d. A vector field F P, Q, R can be thought of as a force field, where the vector at each point can be considered as the direction and magnitude of force acting at that point. If we have a curve in some force field, we may be interested in computing the work required to move a particle along this curve. To find an integral formula for this, let be parametrized by r(t), and we split up to very tiny pieces, each of acrlength s. But on these little curves s, the particle travels in the direction of T(t) r (t) r (t), the unit-tangent vector to at r(t). Therefore, the work to move the particle on that small piece s is given by W F T(t) s. Summing over all the little pieces, and taking the limit over partitions as in Riemann sums, we get the following formula for work: W F(x, y, z) T(x, y, z)ds F Tds. Recalling the formula for ds with paremetrization r(t), a t b we also get the 58

59 6.1 Integrating Vector Fields over urves 6 INTEGRATION OVER URVES following integral W b a b a : F(r(t)) F(r(t)) r (t)dt F dr. r(t) r (t) r (t) dt Summarizing the aforementioned in a definition: efinition 76: If F is vector field defined along some curve with parametrization r(t), a t b, the line integral of F over is defined to be F dr b a F(r(t)) r (t)dt F Tds. Example 77: Find the work done by the force field F x 2, xy in moving a particle along the quarter circle r(t) cos t, sin t, t π/2. By the above formula W b a π/2 2 F(r(t)) r (t)dt cos 2 t, cos t sin t sin t, cos t dt π/2 2 3 cos3 t tπ/2 t 2 3. cos 2 t sin tdt Example 78: Find the work done by the force field F x 2, xy in moving a particle along the three-quarter circle r(t) cos t, sin t, t 3π/2. 59

60 6.1 Integrating Vector Fields over urves 6 INTEGRATION OVER URVES As before, W b a π/2 2 F(r(t)) r (t)dt cos 2 t, cos t sin t sin t, cos t dt π/2 2 3 cos3 t tπ/2 t 2 3. cos 2 t sin tdt Note in the previous two examples, we computed the work along two different paths both connecting the points (1, ) to (1, ) and arrived the same answer. This is not always true. But it is true for certain vector fields as we ll see shortly. Example 79: Show that a constant force field does zero work on a particle that traverses the unit circle once. Let F c 1, c 2 denote this constant force field and let r(t) cos t, sin t, t 2π be the usual parametrization. Then W 2π ( c 1 sin t + c 2 cos t)dt. We can also relate these notions in terms of line integrals with respect to x, y, and z, namely, if F P, Q, R and r(t) x(t), y(t), z(t), a t b is some curve, then F dr b a b a F(r(t)) r (t)dt (P (r(t))x (t)dt + Q(r(t))y (t)dt + R(r(t))z (t)dt) (P dx + Qdy + Rdz). Besides just computing various line integrals for force fields, there is another significant reason for introducing vector fields and their various integrals, which relate the previous examples. 6

61 6.1 Integrating Vector Fields over urves 6 INTEGRATION OVER URVES Theorem 8 (Fundamental Theorem for Line Integrals): Let be a smooth curve parametrized by r(t), a t b. Let f be a differentiable function of any number of variables. Then f dr f(r(b)) f(r(a)). This is an analogue to the usual Fundamental Theorem of alculus that we re b used to ( F (x)dx F (a) F (b)), and in fact is a orollary of it. Indeed, a Proof: f dr b a b ( f dx x dt + f dy y dt + f z ) dz dt dt d a dt (f(r(t))dt by chain rule, f(r(b)) f(r(a)) by FT. What does this theorem mean for us? Since for gradient fields f, the integral only depends on the endpoints of the curve and not the actual path taken! That is, the work done along any path connecting two points of a gradient force field is the same. This idea leads to a new notion: efinition 81: We say that the integral F dr is independent of path if F dr F dr 1 2 for any two paths 1 and 2 in the domain of F with the same start and end points. This definition leads to a corollary of our FTLI: orollary 82: If F is a conservative vector field, then F dr is independent of path. Remark: We saw in line integrals of scalar functions with respect to arclength, that the result was independent of paramtrization (as long as we didn t traverse the curve multiple times). This is no longer true when dealing with vector fields. Intuitively, this makes sense in the concept of work: Suppose we want to move a ball 61

62 6.1 Integrating Vector Fields over urves 6 INTEGRATION OVER URVES from a height of h to the ground in the presence of a gravitation force field. Then the work should be positive, since the motion of the ball in the same direction as the field. However, if we were to traverse the path backwards, that is, taking the ball from the ground straight up to a height of h, we should have a negative work done by the gravitiational field. Formally, if is any curve and f a scalar function, then fds fds, however if F is a vector field, then F dr F dr. efinition 83: A curve is called closed if its endpoint coincides with the starting point. We can use closed curves to determine if certain integrals are path independent. Indeed, given two paths 1 and 2 with the same starting and ending points, then the path 1 + ( 2 ) is a closed curve. onversely, given any closed path, we can break it up (anywhere) to obtain two paths 1 and 2 with the same starting and ending points. If for every closed curve, F dr, then F dr F dr + F dr 1 2 F dr F dr, 1 2 that is, F dr F dr. 1 2 onversely, if F dr is independent of path, then let be any closed curve, and consider for any composition of curves with the opposite starting and ending points. Then F dr F dr + F dr 1 2 F dr F dr 1 2. Thus, we ve shown the following theorem 62

63 6.1 Integrating Vector Fields over urves 6 INTEGRATION OVER URVES Theorem 84: F dr is path independent if and only if F dr for every closed curve. We can go even further with path independence, which is a partial converse to orollary 82. Theorem 85: Suppose F is a continuous vector field on a nice (open and connected, we ll revisit this shortly) domain. If F dr is path independent in, then F is conservative. See the text for the reasoning of the theorem. It s not particularly enlightening in telling us if a vector field F is conservative. That is, by a combination of Theorem 84 and Theorem 85, we have to show that F dr for all closed curves, which is computationly an unreasonable criteria, since this included an infinite number of integrations. So the question remains, how do we quickly know if a vector field F is convervative? Suppose F P, Q is conservative. That is, we have a potential function f such that f F, and so P f x, Q f y. And then, since mixed partials are equal, we have that P y 2 f y x 2 f x y Q x. In turns out that this equality is not coincidental, and that if the domain of a vector field is sufficiently nice, then the converse to the above holds. Theorem 86: Let F P, Q be a vector field on an open, simply-connected domain such that P and Q have continuous first partial derivatives and P y Q x throughout. Then F is convervative. Note that this is only applicable in R 2. We ll see criteria for conservative vector fields a bit later in the course. The theorem is a pretty straightforward criteria for determining if a vector field is conservative, however, this notion of simply-connected domain cannot be ignored, and so we need to explain it in a bit more detail. Moreover, as the technical aspects of the implications of the soon-to-be defined term simply-connected are suprising quite deep and well beyond the scope of this course, we shall not attempt to explain why this theorem is true at this time. ome back to Green. 63

64 6.1 Integrating Vector Fields over urves 6 INTEGRATION OVER URVES efinition 87: Let be a region. We say is open if every point P of, there is a disk (of appropraite dimension) centered at P completely contained in. That is doesn t contain any boundary points (e.g., (a, b) in R, x 2 + y 2 < 1 in R 2, and r < 1 in R 3 ). We say is connected if any two points in can be connected by a path that completely lies in. A simple curve is a curve that contains no self-intersection with the possible exception of its endpoints, that is, for any a < t 1 < t 2 < b, we have that r(t 1 ) r(t 2 ). We say is simply-connected if it s connected (not necessarily open) and that every simple closed curve in encloses only points that are in. Intuitively, this means contains no holes. Example 88: Is F x y, x 2 conservative? This vector field is defined everywhere, however, and hence can t be conservative. Example 89: Is F 3 + 2xy, x 2 3y 2 conservative? 1 1, 64