Math 425 Notes 9. We state a rough version of the fundamental theorem of calculus. Almost all calculations involving integrals lead back to this.

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1 Multiple Integrals: efintion Math 425 Notes 9 We state a rough version of the fundamental theorem of calculus. Almost all calculations involving integrals lead back to this. efinition 1. Let f : R R and let [a, b] be a closed interval. Let a = x 0 < x 1 < < x n = b be a division of the interval [a, b]. For each i let x i [x i, x i 1 ]. A sum of the form is called a Riemann sum. i=n f(x i )(x i x i 1 ) Notation: Keep the notation from the definition above. efine b a i=1 i=n f(x)dx = lim f(x i )(x i x i 1 ). The limit is taken over partitions of [a, b] as the length of the pieces all go to zero. Theorem 1. Fundamental Theorem of Calculus Let f : R R be continuous. Then where b a i=1 f(x)dx = F (b) F (a) df dx = f. Notation: Let x i, x i 1 R, then x i = x i x i 1 and similarly for y j. efinition 2. Let R be the rectangle [a, b] [c, d] R 2. Let f : R R be a function on R. Let x i = a + i (b a), y n j = c + j (d c). Then n R fdxdy = lim i,j=n n i,j=1 f(x i, y j ) x i y j provided the limit exists. The limit is taken over all partitions of [a, b] and [c, d] such that the lengths of all the subintervals go to zero. Theorem 2. Assume that f is continuous on a rectangle R except on points in the union of a finite number of graphs. Then the limit in the definition above exists. 1

2 Assume that z = f(x, y) is the graph of a continuous (except possibly on points in the finite union of graphs) non-negative function. Then fdxdy is the volume of the solid R above the z = 0 plane and below the graph of f. Here is a fancier case. How do we find the volume under the function f(x, y), above the plane z = 0 and within the triangle with vertices (0, 0), (1, 0), (1, 1). Apparently this is not covered by the above theorems. We extend the function f by making it equal to zero outside the triangle and using the rectangle [0, 1] [0, 1]. The singularities are all on the line y = x and hence are allowable. How do we compute these double integrals. These integrals are the limits of sums in a rectangular array. Here are two ways of organizing the summing of numbers in a rectangular array. The first method is to add the numbers in each column and then adding these sums. A second method is to add the numbers in each row and then adding these sums. Formally we have i,j=n i,j=1 j=n f(x i, y j ) x i y j = ( i=n j=1 i=1 ( i=n j=n = i=1 ) f(x i, y j ) x i y j ) f(x i, y j ) y j x i. If we take the limit as n, then the three sums above have the same limit provided f is bounded, and continuous outside a finite union of graphs. This is a delicate result. j=1 Theorem 3. Assume that f is nice on a rectangle R = [a, b] [c, d]. converges to fdxdy. The second converges to The third converges to These are all equal. d c b a R ( b a ( d c ) fdx dy. ) fdx dy. The first sum Example 4. Let R = [ 1, 2] [3, 4], f(x, y) = 2x 2 + xy + y 2. Then 4 2 f(x, y)dxdy = ( f(x, y)dx)dy. R 3 1 We separate out the inside integral to get 2 1 (2x 2 + xy + y 2 )dx = x 3 + (x 2 y)/2 + xy 2 x=2 x= 1 = 5y 2 + (3/2)y

3 Figure 1: Tetrahedron y z + x = 1 and the whole integral is y=4 y=3 We can also integrate with respect to y first. (5y 2 + (3/2)y + 9)dy = LT S. Example 5. We find the volume of the tetrahedron bounded by the planes x = 0, y = 0, z = 0, y z + x = 1. The first step is to sketch this. We sketch the intersection of the plane y z + x = 1 with each of the planes x = 0, y = 0, z = 0. We find that the tetrahedron lies below the z = 0 plane. Indeed, it lies below the triangle T in the z = 0 plane with vertices (0, 0), (1, 0), (0, 1). Up to this point we have performed double integrals only over rectangles. We deal with integrating over a triangle theoretically by extending the integrand by the value 0. We are only able to anti-differentiate functions given by a single formula such as e x, x 2 7x + 1. Here is how we get out of this difficulty. Assume that f(x) given by a formula from x = a to x = b and is given by zero from x = b to x = c. Then c fdx = b fdx. a a Let f(x, y) = 1 x y. The function f gives the height of the tetrahedron. We integrate with respect to y first and then x. We get x=1 ( y=1 x ) fdy dx. x=0 y=0 We could also integrate with respect to x first and then with respect to y. In this case we get y=1 ( x=1 y ) f(x)dx dy. y=0 x=0 3

4 Figure 2: Limits of Integration We state Fubini s Theroerm. Let R be the rectangle [a, b] [c, d]. Assume that f is bounded everywhere in R and continuous except at points in a finite union of graphs. The integrals below are all equal. f(x, y)dxdx R y=d ( x=b ) f(x, y)dx dy y=c x=a x=b ( y=d x=a y=c ) f(x, y)dy dx. We can calculate integrals over more general regions than rectangles. If a region can be describe as {(x, y) φ 1 (x) y φ 2 (x), a x b} We say it is y-simple. If a region can be describe as {(x, y) φ 1 (y) x φ 2 (y), a y b} We say it is x-simple. If a region is either of these we say it is elementary. We can integrate over a y-simple region using the formula x=b x=a ( y=φ2 (x) y=φ 1 (x) f(x, y)dy)dx. Example 6. How do you integrate over the annulus A = {(x, y) a 2 x 2 + y 2 b}. You break it up into a number of elementary regions, integrate over each, and then sum the results. In this case you could use the regions (there are many ways of doing this): R 1 = {(x, y) A a x b, a 2 x 2 y a 2 x 2 } R 2 = {(x, y) A a x a, y 0, a 2 x 2 y b 2 x 2 R 3 = {(x, y) A b x a, } R 4 = {(x, y) A a x a, y 0, } 4

5 Figure 3: Annulus Example 7. We set up an integral to calculate the volume of a right cone of radius R and height H. We center the cone so that its central axis is the z axis. First we find a formula for the height of the cone over the point (x, y). By symmetry this only depends on r = x 2 + y 2 so we express h(x, y) in terms of r and then express r in terms of (x, y). Make a vertical slice through the center of the cone and the point (x, y). We get two similar triangles. One has sides of length H and R and the other has sides h and R r. We obtain h R r = H R. From this we can get h = (R r)h. R To obtain the volume of the cone we integrate h(x, y)dxdy, where S is the disk of radius R centered at the origin. The volume is S x=r y= R 2 x 2 x= R y= R 2 x 2 h(x, y)dxdy. Example 8. We find the volume of the solid inside the cylinder x 2 +4y 2 = 4, above z = 0, and below the plane x + 2y + 2z = 2. The first step is to sketch the cylinder x 2 + 4y 2 = 4, and, in particular, see where it intersects the x and y axis. The second step is to sketch the plane, and, in particular, to see how it intersects the plane z = 0, and also see the three points where it intersects the three axes. Let h(x, y) be the height of the plane above a point (x, y, 0), for those points where the plane is above the x y plane. We have h(x, y) = 2 x 2y. This means that the volume is 2 h(x, y)dxdy where C is the region in the plane z = 0 of points (x, y, 0) such that C 5

6 Figure 4: Cone the plane x + 2y + 2z = 2 is above the z = 0 plane, and (x, y) is inside the ellipse x 2 + 4y 2 = 4. We can integrate with respect to x first or we can integrate with respect to y first. Either choice requires breaking the region into two pieces. We choose to integrate with respcet to x first. We just give the limits of integration of the two regions. The total volume is the sum of these two volumes. The first is The second is y=1 x=2 2y y=0 h(x, y)dxdy. x= 4 4y 2 y=1 x= 4 4y 2 y= 1 h(x, y)dxdy. x= 4 4y 2 Example 9. Let f(x, y) = ( x) y 2. We set up the limits of integration in order to find the integral of f over the region x > 0, y > x 2, y < 10 x 2 in two ways. The first step is to sketch the region. We integrate with respect to y first. We obtain x= 5 y=10 x 2 x=0 y=x 2 f(x, y)dydx. We can also integrate with respect to x first. We need to break the region into two parts in this case, namely the region above the line y = 5 and the region below y = 5. For the region above y = 5 we obtain y=10 x= 10 y y=5 For the region below y = 5 we obtain x=0 y=5 x= y y=0 x=0 The integral is the sum of the two integrals above. f(x, y)dxdy. f(x, y)dxdy. 6

7 Figure 5: Cylinder Figure 6: Two Parabolas 7

8 Figure 7: Change of Variables: One Variable Change of Variables The One Variable Case Here is how we change variables while integrating in the one variable case. We start with a function f(x) and an interval [a, b]. Let x = g(u) be a change of variable and assume that g maps the interval [A, B] to [a = g(a), b = g(b)] in a one-to-one onto fashion. We have dx du = dg du = g (u) so we can write dx = g (u)du. We get b a f(x)dx = B A f(g(u))g (u)du. We look at this with more care. An integral is the limit of sums. We look with more care at a single summand in one of the sums. Let u and x be corresponding points so g(u) = x. Write u for a small interval around u and and x for image of that interval with respect to the map g. We have that We also have that f(x) = f g(u). Hence x u g (u). f(x) x f g(u)g (u) u. Taking sums and taking the limit we conclude that b a f(x)dx = B Example with Plus and Minus: We integrate A f(g(u))g (u)du. by changing variables. Let x=3 x=1 (3x + 2)dx x = 3 u 8

9 so We get that x = 1 = u = 2, x = 3 = u = 0, and dx = du. x=3 x=1 Note the limits of integration. (3x + 2)dx = u=0 u=2 (3(3 u) + 2)( du). Geometry of Maps R 2 R 2 efinition 3. Let f : R 2 R 2 and let R 2. We say that f( ) = {f(p ) P } is the image of under f. We say that f is one to one on provided f(p ) = f(q), P, Q implies P = Q. We say that f : is onto provided S implies there is a P so that f(p ) = S. Example 10. I claim that a linear map R 2 R 2 which is one-to-one maps parallelograms to parallelograms. To see this the first step is to describe a parallelogram as {p R 2 p = b + su + tv, b, u, v R 2, s, t [0, 1]}. Let A be a one-to-one linear map R 2 R 2. Apply A to p and we get A(p) = A(b) + sa(u) + ta(v), a parallelogram with one vertices at A(b), A(b)+A(u), A(b)+A(v), A(b)+A(u+v). Where do we use one-to-one in this argument? We see that A multiplies area by the absolute value of det(a). To see this we first verify it for the unit square with vertex at the orgin. Second we verify it for an arbitray square. Third we fill up a parallelogram with many small squares and use approximation. Note that if a linear map f is not one-to-one, then f maps a parallelogram to a point or to a line segment. Example 11. We consider the map p : R 2 R 2, p : (r, θ) (x = r cos(θ), y = r sin(θ)). This map is not one-to-one. Indeed p(r, θ) = p(r, θ + 2π). The map p maps (0 r < ) (0 θ < 2π) in a one-to-one fashion onto R 2. It maps 1 r 2 in a many-to-one way onto the annulus with inner radius 1 and outer radius 2. The map p takes 0 θ π/4 onto a sector corvering 1/8-th of the plane. Let be a subset of R 2 and be subset of another copy of R 2. Let T be a one-to-one differentiable map T :. Let f : R. How are the integrals of f and f T related? 9

10 Figure 8: Change of Variables: Two Variables, Points Example 12. Let f(x) 1, and let T : R 2 R 2, ( ) x y ( ) 3x. y Let = [0, 1] [0, 1] and = [0, 3] [0, 1] so T is a one-to-one map from to. We have T f = area of = 1, while f = area of = 3. To see what is happening use the definition of the integral. To find f we break up into many small squares S, pick a point P S, and we take the sum (f T )(P )( area of S) over all the squares S. To find f we break up into the squares T (S), take the point Q = T (P ) T (S), and then we add up the summands f(q)( area of T (S)). Note that in the two summands, the terms f T (P ) and f(q) are equal, but the areas are not the same. Indeed, we have the area of dett (P ) ( area of ). We conclude that f(q)( area of T (S)) = f T (P )( area of S) det(t (P ). This gives f = f T det(t (P )). This is the Change of Variables Formula. efinition 4. Let, be subsets of R n and let T : be one-to-one and differentiable. The Jacobian of T is the determinant of T. 10

11 Figure 9: Change of Variables: Two Variables, Regions Figure 10: Polar Coordinates Example 13. Let We have p : R 2 R 2, ( ) r θ ( ) r cos(θ). r sin(θ) ( ) cos(θ) r sin(θ) p(r, θ) =, sin(θ) r cos(θ) so the Jacobian is r. We find the area inside the curves r = sin(θ), θ = 0, θ = π, r = 0 Let denote the region of the r, θ plane so that 0 r sin(θ). We are asking to find the area of = p( ). To find the area of a region we integrate the function 1 over the region. By the change of variable formula this is equal to θ=π r=sin(θ 1 = detp 1 = 1 rdrdθ = π/4 θ=0 r=0 We solve this problem in a different way. Multiply r = sin(θ) by r and use r 2 = x 2 + y 2, y = sin(θ) to get x 2 +y 2 = y. Completing the squares gives us x 2 +(y (1/2)) 2 = (1/2) 2. This is a circle of radius 1/2 which has an area of π/4. Example 14. We evaluate the integral A = 11 e x2 dx.

12 We have A 2 = e x2 dx e y2 dy = lim a x2 y2 e a where a is the disk with center at the origin and radius a. We change variables to polar coordinates. This last integral is a e r2 rdrdθ = θ=2π r=a θ=0 Taking the limit we obtain r=0 e r2 rdrdθ = A 2 = π = A = π. 2π e r2 r=a 0 2 dxdy r=0dθ = 2π( 1 2 e a2 ). (1) 2 Implicitly we have used the definition of integrals of unbounded regions. We just take it to be the integral of bounded regions which eventually fill up the unbounded region. Example 15. Let T : R 2 R 2, ( ) u v ( ) x = u uv. y = uv Let be the region bounded by x = 0, y = 0, x + y = 1, x + y = 4. We calculate the integral 1 x + y dxdy using change of variables. Write f(x, y) = 1/(x + y). First we sketch the region. Next we find the region in the u, v plane so that T ( ) =. Note that change of variables allows us to compute integrals over using integrals over the region. The region is bounded by u = 1, u = 4, v = 0, v = 1. To see this note that x + y = (u uv)+uv = u. Since u 0, we can solve x = 0 u uv = 0 = u(1 v) v = 1. Similarly we see that y = ( 0 = v = ) 0. 1 v u The derivative of T is, so the Jacobian of T is u. Note that u 0 in the v u region. Is the map T one-to-one on? This is the same as asking if the equations x = u uv, y = uv have a unique solution for (x, y). We find that u = x + y, v = y/u = y/(x + y). Since x + y 0 for (x, y), there is a unique solution. The change of variables formula gives We use that u = u in. f T det(t ) dudv = 1 u u dvdu = u=4 v=1 dvdu = u=1 v=0 fdxdy 1 x + y dxdy 1 x + y. 12

13 Figure 11: Trapaziods 13