Math 53 Homework 7 Solutions

Transcription

1 Math 5 Homework 7 Solutions Section 5.. To find the mass of the lamina, we integrate ρ(x, y over the box: m a b a a + x + y dy y + x y + y yb y b + bx + b bx + bx + b x ab + a b + ab a b + ab + ab. We then integrate xρ(x, y, which gives a b x + x + xy dy a a xa x xy + x y + xy yb y bx + bx + b x bx + bx + b x 6 a b + a b + a b 6 xa x a b + 6a b + a b. Dividing by the mass of the lamina, we have Similarly, x a b + 6a b + a b (a b + ab + ab. y a b + 6ab + ab (a b + ab + ab. Therefore, the center of mass is ( a b + 6a b + a b (x, y (a b + ab + ab, a b + 6ab + ab (a b + ab + ab.

2 8. In the region bounded by y x + and y x, x ranges from to (the two roots of x x +. Thus, the mass is m We now integrate xρ(x, y: x+ Integrating yρ(x, y gives x+ x+ x kx y kx dy yx+ yx kx (x + x kx + kx 6k. x kx dy x kx y dy kx5 5 kx y x x yx+ yx kx (x + x kx5 5 + kx 6k 5. kx y yx+ yx kx6 6 kx (x + x x x k x (x x x + x + k ( x 8 8 x7 7 x6 + x5 5 + x x x 77k 56. After dividing by the mass, we get ( 56 (x, y 89, Since the density is proportional to the square of the distance from the origin, we have ρ(x, y k(x +y for some constant k. In terms of polar

3 coordinates, this is just ρ(r, θ kr. The mass is therefore m π π k kπ. kr r dr dθ To integrate xρ(x, y, we just multiply the integrand by r cos θ, so we get π kr cos θ r dr dθ. This is just since the integral of cos θ from to π is. Since the same holds for sin θ, the integral of yρ(x, y is also, so the center of mass is (,. Section 5.9. The Jacobian is. The Jacobian is ( (x, y u + v u (u, v det v uv (u + v uv uv u v + uv uv u v + uv. ( (x, y e (p, q det q pe q qe p e p e q+p pqe q+p ( pqe p+q. 8. Since x v and v ranges from to, x can take on any value from to. For fixed x, y u( + x, so as u varies from to, y varies from to + x. Thus, the image of the transformation is the region bounded by x, x, y, and y + x.. We may as well try to find a linear transformation sending the unit square to the given parallelogram. Say T (u, v (au + bv, cu + dv for some constants a, b, c, d. We want T to satisfy T (, (,, T (, (,,

4 since then T (, and T (, will automatically be sent to the other two points. These two constraints give a, b, c, d, so the change of variables is x u v, y u + v. 6. Inverting the change of variables gives u x y, v x + y. Substituting in the four vertices of the paralellogram, we get the points (,, (,, (, 8, (, 8 in the uv-plane. Since the transformation is linear, the image in the uv-plane will be a rectangle with these four vertices. The Jacobian is just so we have R x + 8y da det ( 8 8, (u + v + 8 (v u dv du 5u + v dv du 5uv + v u + du 5u + u 9. u u v8 v 9. The two lines y x and y x become u v and u v / in the uv-plane. The two hyperbolas xy and xy are just u and u v. Thus, the new region is the one bounded by u, u, u v, and u v /. The Jacobian is ( det v u v v. du This gives But R xy da u u u v u log v v u v u dv du log( u log( u log(, du.

5 so R xy da u log( du u log( log. u u Section The integrand does not depend on x, so z z dz dy y + z z y + dz dy. We can pull the /(y + out of the integrand, which allows us to write z ( z ( dz dy y + y + dy z z dz. The first integral is just log, and the second is (make the substitution u z. Thus, the value of the original integral is log. 8. We have x y xye z dz dy xy(e x y dy. We can split this up into two integrals and then factor each into a product: xy(e x y dy e ( xye x y dy xe x ( ye y dy xy dy To integrate xe x, we substitute u ( x, and it reduces to integrating e u from to. This integral is e, so the original integral is ( ( e ( (e e e e.. Since the values of x depend on y and the values of z depend on x and y, we integrate z, then x, then y. This gives xy e z/y dv e z/y dz dy E y y ye z/y zxy dy z ( ( x y dy. 5

6 y y(e x dy y(e x x (e y e e 7 6. x xy dy y[(e (e y y] dy + + y (y y ey y. First notice that y ranges from to, independently of x and z, so we should put dy on the outside of the integral. The surfaces z x and z x intersect when x ± and z, so x ranges from to. This gives the iterated integral E x y dv x x x y dz dy (x y ( x dy x x y + x y dy x x xy + x y 8 y dy y y 6. y x x. Since the only condition on x is that x + z (so x we should have the outermost integral be with respect to x. Then z varies from x to x, and finally, y varies from to z. This gives the iterated integral E dv x z x x dy dz 5 z dz x 5z z z x z x x. dy 6

7 This is just times the area of a semicircle of radius, so the volume of the solid is π π. 8. See the attached diagram.. Since x is independent of y and z, the integral with respect to x can be freely moved around. If the integral in y comes before the integral in z (from left to right, then the bounds are y, 9 y z 9 y. Similarly, if z comes before y, then the bounds are This gives the 6 integrals z, 9 z y 9 z. 9 z 9 z 9 y 9 y 9 z 9 z f(x, y, z dy dz, f(x, y, z dz dy, 9 z f(x, y, z dy dz, f(x, y, z dy dz, 9 z 9 y f(x, y, z dz dy, 9 y 9 y f(x, y, z dz dy. 9 y. Since the conditions on y and z are written in terms of x, we can swap dy and dz in the given integral while keeping the bounds the same. This gives x x f(x, y, z dz dy. If we instead have the order be dz dy, then y ranges from to, x ranges from to y, and z ranges from to x. This gives the integral y x f(x, y, z dz dy. If we use the order dy dz, then z ranges from to, x ranges from to z, and finally y ranges from to x. Thus, we get the integral z x f(x, y, z dy dz. Using the order dy dz, z and y both range from to. However, the range for x will depend on where in the unit square (y, z lies. If 7

8 we project the curve where z x and y x intersect onto the yz-plane, we get z ( y y + y, or equivalently, y z. If the point (y, z lies below this curve, then x is limited by the plane y x, so x y. If (y, z lies above the curve, then x will range from to z. Therefore, we have to write the integral as a sum: z z f(x, y, z dy dz+ z Similarly, if we take the order to be dz dy, we get y +y y f(x, y, z dz dy+ y +y y z f(x, y, z dy dz. f(x, y, z dz dy. 8