Math 54 - Vector Calculus Notes 3. - 3. Double Integrals Consider f(x, y) = 8 x y. Let s estimate the volume under this surface over the rectangle R = [, 4] [, ] in the xy-plane. Here is a particular estimate: We can split R into 8 equal-area sub-rectangles using gridlines x =,,, 3, 4 and y =,,, and by evaluating f at upper-left corners (any sample point really) we can get an estimate f over each sub-rectangle. So an estimate for the volume is: Such a sum is called a Riemann sum. V (7 + 6 + 6 + 5 + 3 + + 8 + 7) = 4. Formally: Let f(x, y) be defined over a rectangular region R in the xy-plane. If the limit, lim n n f(x i, y i ) A i, i= exists for all partitions of R into n sub-rectangles, where (x i, y i ) is a sample point from the ith sub-rectangle, and A i (such that A i ) represents the area of the ith sub-rectangle, then the limit is the double integral of f over R. Notation: R f(x, y)da. Fubini s theorem (proof not given) implies that we can iterate the integral when f is continuous over R (let R = [x, x ] [y, y ]): R f(x, y)da = For example, (8 x y)da = = 4 [,4] [,] ( 34 x ) dx = x y x 4 y f(x, y)dydx = (8 x y)dydx = y x y 4 x f(x, y)dxdy. ( 8y x y.5y y= y=) dx = (34x 3 x3 ) 4 = 8/3 93.3 so the 4 is an overestimate ( %).
Contributed by Bruce Torrence: The double integral being estimated is [,] [,] x + y da. In the following figures, Riemann sum estimates with various numbers of equal-sized subrectangles, and midpoints as sample points, are compared against the actual value of the integral. 5 sub-rectangles: Rie m a n n s u m 4.9 6 d ou b le in te g ra l 4.6 6 6 7 8 6 4
sub-rectangles: Rie m a n n s u m 4. 4 d ou b le in te g ra l 4.6 6 6 7 8 6 4
4 sub-rectangles: Rie m a n n s u m 4.5 6 d ou b le in te g ra l 4.6 6 6 7 8 6 4
6 sub-rectangles: Rie m a n n s u m 4.6 4 d ou b le in te g ra l 4.6 6 6 7 8 6 4 You can really start to make-out the paraboloid! Note: All these figures can be found (and played with) on the Bruce Torrence s web page (he does some neat stuff with Mathematica): http://demonstrations.wolfram.com/author.html?author=bruce+torrence
valuate the double integrals: () () (3) π 3 3xy cos (x )dxdy. x(x + y 4 ) dydx. x ( + xy) dxdy. The average value of a function over a region R is given by f avg = f(x, y)da. Area(R) R Find the average value of f(x, y) = x y + xy x on [, ] [, 5]. The definition of a double integral for a rectangular region can be easily extended to general regions that are closed and bounded. Basic types of regions: () R is bounded by a x b and g(x) y h(x). Then b h(x) a g(x) f(x, y)dydx. () R is bounded by c y d and l(y) x m(y). Then d m(y) c l(y) f(x, y)dxdy. A region might need to be split into these types of regions above! Note that if the integrand is, the double integral recovers the area of the region.
valuate the double integrals: () () R xy + da where R is bounded by x =, y =, and x + y = 4. x x x 8x 3 ydydx. (3) yda. x +y (4) (5) x xe y y dydx. x sin (y 4 )da where R is bounded by x =, y = x, and y =. R Compute the volume bounded by the paraboloids z = x + y and z = x y. These intersect when z = and the shadow in the xy-plane is the unit circle. Hence, the volume is = ( x y (x +y ))da = 4 x x +y ( x y )da (by factoring and symmetry!) =
3.3 - Double Integrals in Polar Coordinates The integral, x ( x y )da, can be much more easily done in polar coordinates. But first we need some theory. A polar sub-rectangle can be given by the bounds, r dr r r + dr and α θ α+dθ. To compute the area of these sub-rectangles we use the formula for the area of a sector: The area bounded by r dr r r + dr : A = θ π (πr ) = r θ. da = dr (r ) dθ (r + dr ) dθ = rdrdθ. As dr, r r. Thus in polar coordinates, f(r, θ)da = β b R α a f(r, θ)rdrdθ. Of course f(x, y) might need to be converted to f(r, θ) by replacing x, y in terms of r, θ. So, x ( x y )da = π ( r )rdrdθ = A double integral in polar coordinates can be extended to more general regions. One can also consider a polar region R given by g(θ) r h(θ) and α θ β. In that case: R f(r, θ)da = β h(θ) α g(θ) f(r, θ)rdrdθ.
Lets find the area of the region bounded by one leaf of the rose r = cos (5θ): y.5..5.5..5.5..5. x.5..5 Since cos (5θ) is an even function, we just need to figure out the first time after θ = that the function equals. That occurs at θ = π. Hence the area is... A = π π cos (5θ) rdrdθ = π = 4 π π ( cos (5θ)) dθ = π 4 cos (5θ)dθ = ( + cos (θ))dθ = ( π ) = π 5.
Let s use polar coordinates to determine the integral: x x x x x + y dydx =... Let s use polar coordinates to determine the integral: 3 4 x 3 dydx =... x + y You may find the following indefinite integral useful: csc (x)dx = ln csc (x) + cot (x) + C. Now we prove a result used in statistics... The graph of f(x) = e x determine: has the classic bell-shape to it. It is important to be able to e x dx. t t t e x dx = lim e x dx = lim e x dx e y dy = t t t t t = lim t = lim t t π t t t e x y dxdy = lim t π t e r rdrdθ = ( ) (e t ) dθ = lim π (e t ) = π. t
3.4 Triple Integrals Let f(x, y, z) be defined over a solid region in the xyz-space. If the limit, lim n n f(x i, y i, z i ) V i, i= exists for all partitions of into n sub-solids, where (x i, y i, z i ) is a sample point from the ith sub-solid, and V i (such that V i ) is the volume of the ith sub-solid, then the limit is the triple integral of f over. Notation: f(x, y, z)dv. Fubini s theorem (proof not given) implies that we can iterate the integral when f is continuous over. There are six basic types of solids (sometimes a solid has to be broken into basic pieces): () : a x b, g(x) y h(x), u(x, y) z v(x, y). Then f(x, y, z)dv = b h(x) v(x,y) a g(x) u(x,y) f(x, y, z)dzdydx. () : a x b, g(x) z h(x), u(x, z) y v(x, z). Then f(x, y, z)dv = b h(x) v(x,z) a g(x) u(x,z) f(x, y, z)dydzdx. (3) : a y b, g(y) x h(y), u(x, y) z v(x, y). Then f(x, y, z)dv = b h(y) v(x,y) a g(y) u(x,y) f(x, y, z)dzdxdy. (4) : a y b, g(y) z h(y), u(y, z) x v(y, z). Then f(x, y, z)dv = b h(y) v(y,z) a g(y) u(y,z) f(x, y, z)dxdzdy.
(5) : a z b, g(z) x h(z), u(x, z) y v(x, z). Then f(x, y, z)dv = b h(z) v(x,z) a g(z) u(x,z) f(x, y, z)dydxdz. (6) : a z b, g(z) y h(z), u(y, z) x v(y, z). Then f(x, y, z)dv = b h(z) v(y,z) a g(z) u(y,z) f(x, y, z)dxdydz. If the integrand is then the triple integral finds the volume of. xamples: () 3 xy zdzdydx. () Find the volume inside the parabolic cylinder y = x between the planes z = 3 y and z =. (3) 4 z +x +z y dydxdz. (4) On this last one... 4 4z π z cos yz dydxdz. x 3 -Switch the order of the x and y iterated integrals, which is easy as in any z-plane between and 4 the x, y-limits form a rectangle! -valuate the x-integral and simplify. -Switch the order of y and z iterated integrals in the remaining double integral, which is also over a rectangular region. The last integral is improper, but converges. The result is 8. This should also prove useful: ( ) sin ( yz) = cos yz. z y z
Average value of a function f of three variables over a solid : f(x, y, z)dv. Volume() Find the average value of f(x, y, z) = xyz in the region bounded by z = and z = 9 x y. Let s setup and evaluate an integral that determines the volume of the solid bounded by the surface y = z and the planes x = and z = x y. Hint: Iterate in the dxdydz order with x z y, z y z, and z. valuate the integral after switching the order of integration to dxdzdy: x y x + y dzdydx. First switch the order for the outer two integrals (dydx to dxdy), then switch the order of the inner two integrals (dzdx to dxdz).
3.5 Triple Integrals in Cylindrical and Spherical Coordinates This picture helps visualize cylindrical and spherical coordinates: Cylindrical coordinates: (r, θ, z) where (r, θ) are polar coordinates for the xy-plane. Note: Whenever convenient, one can permute the role of z with either x, y. From rectangular to cylindrical: r = x + y, θ = tan y x (adjusted to quadrant) or ± π, and z = z. From cylindrical to rectangular: x = r cos θ, y = r sin θ, z = z.
Spherical coordinates: (ρ, θ, φ) where ρ is the distance to the origin, θ is the polar angle made in the xy-plane, and φ is the angle the point regarded as a vector makes with the positive z axis. From rectangular to spherical: ρ = x + y + z, θ = tan y x z, and φ = cos x. +y +z ± π (adjusted to quadrant) or From spherical to rectangular: x = ρ sin φ cos θ, y = ρ sin φ sin θ, and z = ρ cos φ. From cylindrical to spherical: ρ = r + z, θ = θ, and φ = cos From spherical to cylindrical: r = ρ sin φ, θ = θ, and z = ρ cos φ. z r. +z Integration in cylindrical coordinates: Let f(r, θ, z) be defined over the solid. If lim n n f(r i, θ i, z i ) V i, i= exists for all partitions of into n sub-solids, where (r i, θ i, z i ) and V i (such that V i ) are a sample point and volume of the ith sub-solid respectively, then the limit is the triple integral of f over. When f is continuous, the integral may be iterated. For example, when g(θ) r h(θ), α θ β, and G(x, y) z H(x, y), then β h(θ) H(r cos θ,r sin θ) α g(θ) G(r cos θ,r sin θ) f(r, θ, z)dz(rdrdθ). So dv = rdzdrdθ.
So, for example, π 4 r rdzdrdθ = Describe the solid region for which we just computed the volume! Integration in spherical coordinates: Let f(ρ, θ, φ) be defined over the solid. If lim n n f(ρ i, θ i, φ i ) V i, i= exists for all partitions of into n sub-solids, where (ρ i, θ i, φ i ) and V i (such that A i ) are a sample point and volume of the ith sub-solid respectively, then the limit is the triple integral of f over. A note on dv : suppose a sub-solid is made up of small changes in ρ, θ and φ. The volume of the box is approximately dρ times ρdφ times ρ sin φdθ. So dv = ρ sin φdρdθdφ. When f is continuous, the integral may be iterated. xamples: () 3 9 x 3 x +y xydzdydx () Compute the volume bounded by the hyperboloid z = + x + y and z = 5. (3) π π 4 sec φ ρ sin φdρdφdθ. We just calculated the volume of the cone of height and radius.
(4) π π 6 3 sec φ ρ sin φdρdφdθ. We just calculated the volume of... (5) x + y dv, where is the the solid bounded by the paraboloid z = 4 x y and the xy-plane. (6) ln (x + y + z )dv, where is the the solid between the spheres of radii and subject to inequality z x + y. (7) x x y ( x + y ) 3 dzdydx. x
3.6 Integrals for Mass Calculations Consider objects of weights m and m sitting on a see-saw. To balance the two objects the distance from the balance point x to the objects, say they are located at x and x, must satisfy: m (x x) = m (x x ) m (x x) + m (x x) = x =?? Let objects of masses m, m,...,m n be arranged along a line with coordinates x, x,...,x n with respect to a balance point or center of mass, x = x. ach mass contributes a moment of m i x i x and to balance, n m i (x i x) = x =?? i= Now consider a continuous density function for a linear object: ρ(x). In that case, one can take n subintervals of the rod, each of width x, such that n m i (x i x) = i= n ρ(x i ) x(x i x) = i= lim n n ρ(x i ) x(x i x) = i= Thus the balance condition becomes, b ρ(x)(x x)dx = b xρ(x)dx x b a a a ρ(x)dx = x = ba xρ(x)dx ba ρ(x)dx. Note that the denominator is the mass! The numerator is called the moment.
Suppose a rod situated on [, ] has a linear density function ρ(x) = x, what is its mass and center of mass? Now let a plate R have a continuous density function ρ(x, y). To compute the center of mass, (x, y), consider how to compute x (other one can be done by symmetry): You could collapse all vertical lines to the x-axis. Then the f(x ) g(x ) x ρ(x, y)dy (or a bunch of these integrals) is the moment contribution of the vertical line x = x. Integrating, over x, all these moments, gives us the moment about the y-axis. M y = Similarly, the moment about the x-axis is M x = R R xρ(x, y)da. yρ(x, y)da. Thus: x = M y m and y = M x m where the mass is m. Let R be the region in the first quadrant where x + y 4 with the density function ρ(x, y) = 5 xy. Find the mass and the center of mass. Hint: By symmetry, x = y.
In three dimensions, moments are about planes and thus: x = M yz m, y = M xz m, z = M xy m, where m = ρ(x, y, z)dv is the mass and M yz = M xz = M xy = xρ(x, y, z)dv, yρ(x, y, z)dv, zρ(x, y, z)dv, Calculate the z-coordinate of the center of mass for the hemisphere z 4 x y that has constant density. Note: It doesn t matter what the density is! What is the center of mass?
3.7 Change of Variables in Multiple Integrals Consider the single variable integral: 4 e x x dx. We can make a change of variables by setting u = dx. Then the differential of u is du = x x and the integral becomes: 4 e x x dx = 4 e u du = Note that [, 4] in the x-axis transforms to [, ] in the u-axis and the transformation 4 is one-to-one. In fact, letting x = g(u), we can write what we just did symbolically: b a f(x)dx = β α f(g(u)) dx du where a = g(α), b = g(β). du A change of variables in a double integral is a transformation that relates two sets of variables, (u, v) and (x, y), where the region R in the xy-plane corresponds to a region S in the uv-plane. Suppose x = g(u, v) and y = h(u, v) defines transformation where g, h have continuous first order partials over S and is one-to-one on the interior of S. The Jacobian determinant is computed in the following way: J(u, v) = (x, y) (u, v) = det x u y u x v y v = x y u v x y v u. Calculate the Jacobian determinant for x = r cos θ and y = r sin θ... of course we have seen this before!
Theorem (the proof is beyond the scope of this course...) Under the preceding assumptions and the assumption that f is continuous on R... f(x, y)da = R S f(g(u, v), h(u, v)) (x, y) (u, v) da. Note that (x,y) (u,v) measures the magnification or reduction of a differential area in the uvplane compared to the transformed differential area in the xy-plane... xamples: ) Compute (x + y) da x + y using a nice change of variables... [DRAW RGION!] Hint: u = x + y and v = x y. ) Compute R y da where R is the region in the first quadrant bounded by xy =, xy = 4, y x = and y x = 3... [DRAW RGION!] Hint: u = xy and v = y x.
In 3 variables... Suppose x = g(u, v, w), y = h(u, v, w) and z = l(u, v, w) is a transformation of a solid B in uvw-space into a solid in xyz-space that on the interior is one-to-one and has continuous partials. The Jacobian determinant is computed in the following way: J(u, v, w) = (x, y, z) (u, v, w) = det x u y u z u x v y v z v x w y w z w. Calculate the Jacobian determinant for x = ρ sin φ cos θ, y = ρ sin φ cos θ, and z = ρ cos φ... of course we have seen this before! Theorem (the proof is beyond the scope of this course...) Under the preceding assumptions and the assumption that f is continuous on... f(x, y, z)dv = B f(g(u, v, w), h(u, v, w), l(u, v, w)) (x, y, z) (u, v, w) dv. Note that (x,y,z) (u,v,w) measures the magnification or reduction of a differential volume in the uvw-space compared to the corresponding differential volume in the xyz-space... Use a change of variables to calculate the volume bounded by the ellipsoid: Hint: Use x = au, y = bv, and z = cw. x a + y b + z c =.