A9_soln

# A9_soln - Math 237 Assignment 9 Solutions π 0 π −y (x 0...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 237 Assignment 9 Solutions π 0 π −y (x 0 1. Use T (x, y ) = (x + y, −x + y ) to evaluate Solution: We have u = x + y and v = −x + y . The region of integration is 0 ≤ x ≤ π − y and 0 ≤ y ≤ π . Thus, the region is bounded by the lines x = 0, y = 0 and x = π − y . Under the mapping T we get: LINE 1: x = 0, 0 ≤ y ≤ π gives v = y = u with 0 ≤ u ≤ π . LINE 2: y = 0, 0 ≤ x ≤ π gives v = −x = −u with −π ≤ u ≤ 0. + y ) cos(x − y ) dx dy . LINE 3: x + y = π , 0 ≤ x ≤ π gives u = π . We have u − v = 2x, hence v = u − 2x = π − 2x so −π ≤ v ≤ π . Drawing rectangle vertically in the region we get −u ≤ v ≤ u and 0 ≤ u ≤ π . The Jacobian is 1 1 ∂ (x, y ) −1 2 = 1 1 2 = = 0. ∂ (u, v ) 2 2 2 Hence, since the transformation has continuous partial derivatives we get by the change of variables theorem π 0 0 π −y π u −u π 0 π 0 (x + y ) cos(x − y ) dx dy = 0 u cos(−v ) u 1 dv du 2 du −u π 1 = 2 1 = 2 −u sin(−v ) 2u sin u du = −u cos u + sin u =π 0 2. Find a linear transformation that maps x2 + 4xy + 5y 2 = 4 onto a unit circle. Hence show that the area enclosed by the ellipse equals 4π . Solution: Completing the square we get x2 + 4xy + 5y 2 = 4 is (x + 2y )2 + y 2 = 4. Hence, we let u = x+2y and v = y . Thus, the inverse transformation is x = 2u − 4v , y = 2v . Which 2 2 maps the unit circle onto the ellipse. The Jacobian is ∂ (x, y ) 2 −4 = = 4 = 0. 02 ∂ (u, v ) Hence since the mapping has continuous partial derivatives the change of variables theorem gives Area = D 1 dA = Duv 4 dA = 4 Duv 1 dA = 4π, since Duv 1 dA just calculates the area of circle in the uv -plane. 2 3. Evaluate and y = x. R x2 + y dV where R is the region bounded by x + y + z = 2, z = 2, x = 1 Solution: Integrating wrt to z ﬁrst gives 2 − x − y ≤ z ≤ 2 and then the region Dxy is bounded by x = 1, y = x and the intersection of x + y + z = 2 and z = 2 so x + y = 0. Drawing the region in the xy -plane we see we want to draw rectangles vertically to get −x ≤ y ≤ x and 0 ≤ x ≤ 1. Hence we get 1 x 2 x2 + yz dV = R 0 1 −x x −x x −x 2−x−y x2 + y dz dy dx (x2 + y )(2 − (2 − x − y )) dy dx x3 + xy + yx2 + y 2 dy dx 0 1 = 0 1 = = 0 2 2x4 + x3 dx 3 17 = 30 4. Dxyz ex−y+z dV , where Dxyz is bounded by the planes x − y + z = 2, x − y + z = 3, x + 2y = −1, x + 2y = 1, x − z = 0 and x − z = 2. Solution: Let u = x − y + z , v = x + 2y , w = x − z then Dxyz is mapped to in a one-to-one fashion onto the region Duvw given by 2 ≤ u ≤ 3, −1 ≤ v ≤ 1, 0 ≤ w ≤ 2. We now ﬁnd that 1 −1 1 ∂ (u, v, w ) 0 = −5 = 0. =1 2 ∂ (x, y, z ) 1 0 −1 ∂ x,y,z 1 Hence ∂ ((u,v,w)) = 5 . Since the mapping has continuous partial derivatives, by the change of variables theorem we get ex−y+z dV = Dxyz eu 1 5 Duvw 2 1 0 −1 1 dV 5 3 = eu du dv dw 2 1 2 13 = e − e2 dv dw 5 0 −1 4 = (e3 − e2 ) 5 3 5. Show that the region D in the ﬁrst quadrant bounded by ay = x3 , by = x3 , cx = y 3 and √ √√ √ 1 dx = y 3 has area 2 ( b − a)( d − c). 3 3 Solution: Let u = x and v = yx . Then the region D is mapped in a one-to-one fashion y onto the region Duv given by a ≤ u ≤ b, c ≤ v ≤ d. We have, ∂ (u, v ) = ∂ (x, y ) 3x2 y −y 3 x2 −x 3 y2 3y 2 x = 8xy = 0. √ ∂ (x,y ) 1 1 Also uv = x2 y 2 so xy = uv, and hence ∂ (u,v) = 8xy = 8√uv . Since the mapping has continuous partial derivatives, by the change of variables theorem we get A= D 1 dA = 1 2v 1/2 8 d c = 1d 1 √ du dv = 8 uv 8c Duv b √√ √ 1√ 2u1/2 = ( d − c)( b − a) 2 a 1· b u−1/2 v −1/2 du dv a 6. Evaluate the following integrals. a) D (x2 + y 2 + z 2 )−3/2 dV where D is the bounded by x2 + y 2 + z 2 = a2 and x2 + y 2 + z 2 = b2 with 0 < b < a. Solution: We observe that the region is bounded by spheres so we use spherical coordinates. In spherical coordinates we have the spheres ρ = a and ρ = b so we get 2π π 0 2π π b a (x2 + y 2 + z 2 )−3/2 dV = D 0 (ρ2 )−3/2 ρ2 sin φ dρ dφ dθ ln 0 0 = √ 2−x2 −y 2 √2 2 dz dy dx. x +y a a sin φ dφ dθ = 4π ln b b b) 1 0 √ 0 1−x2 √ Solution: The region is bounded by x2 + y 2 ≤ z ≤ 2 − x2 − y 2 , 0 ≤ y ≤ 1 − x2 , 0 ≤ x ≤ 1. Thus the region is that which is above the cone z = x2 + y 2 and under the hemisphere z = 2 − x2 − y 2 and in the ﬁrst octant (since x, y, z ≥ 0). Since we have √ spheres and cones we put these in spherical coordinates to get φ = π and ρ = 2 and since 4 we are in the ﬁrst octant 0 ≤ θ ≤ π . Hence by the change of variables theorem 2 √ √ √ 1−x2 2−x2 −y 2 π /2 π /4 1 2 π√ dz dy dx = ρ2 sin φ dρ dφ dθ = ( 2 − 1) √ 3 x 2 +y 2 0 0 0 0 0 4 c) 1 0 √ 0 1−y 2 √ Solution: The region is bounded by 3x2 + 3y 2 ≤ z ≤ 3, 0 ≤ x ≤ 1 − y 2 and 0 ≤ √ y ≤ 1. Thus, we have the region that lies under the plane z = 3, and above the cone z = 3(x2 + y 2 ) in the ﬁrst octant. Since there √ symmetry about the z axis we put these is √ into cylindrical coordiantes to get z = 3, z = 3r and 0 ≤ θ ≤ π . Thus, integrating wrt 2 √ √ √ to z ﬁrst√ get 3r ≤ z ≤ 3, then the maximum radius is at the intersection of z = 3 we and z = 3r so 0 ≤ r ≤ 1 and since we are in the ﬁrst octant we get 0 ≤ θ ≤ π . Hence we 2 get by the change of variables theorem √ √ √ 1 1−y 2 3 π /2 1 3 0 0 √3 √ 3x2 +3y 2 dz dx dy . √ dz dx dy = 3x2 +3y 2 0 π /2 0 1 0 √ √ r3 r dz dr dθ = 0 π /2 √ √ r 3 − r 2 3 dr dθ = √ π3 = 12 0 3 dθ 6 7. Find the volume of the region bounded by the surfaces. a) z = x2 + y 2, x2 + y 2 = 4, z = 0. Solution: We have symmetry around the z axis so we want to use cylindrical coordinates. In cylindrical coordinates we have z = r , r = 2 and z = 0. Thus, integrating wrt z ﬁrst gives 0 ≤ z ≤ r , then the maximum radius is at the intersection of z = r and r = 2 so 0 ≤ r ≤ 2, and no restriction on θ gives 0 ≤ θ ≤ 2π . Thus by the change of variables theorem we get 2π 2 0 0 r V= R 2π 1 dV = 0 2 |r | dz dr dθ = 0 2π 0 r 2 dr dθ 8 dθ 3 = 0 16 =π 3 5 b) x + y + z = 2, x2 + y 2 = 1, z = 0. Solution: We again think to try cylindrical coordinates because of the cylinder x2 + y 2 = 1. The surfaces in cylindrical coordinates are r cos θ + r sin θ + z = 2, r = 1 and z = 0. Thus, integrating wrt z ﬁrst gives 0 ≤ z ≤ 2 − r cos θ − r sin θ, then all possible radius are 0 ≤ r ≤ 1, and no restriction on θ gives 0 ≤ θ ≤ 2π . Thus by the change of variables theorem we get 2π 1 0 0 V= R 2π 1 dV = 0 1 0 2π 2−r cos θ −r sin θ |r | dz dr dθ = 0 2r − r 2 cos θ − r 2 sin θ dr dθ 1 1 cos θ − sin θ dθ = 2π 3 3 = 0 1− c) Inside x2 + y 2 + z 2 = 2 but outside x2 + y 2 = 1. Solution: Since we have a sphere we think spherical coordinates (although cylindrical coor√ dinates would also work nicely). In spherical coordinates we have ρ = 2 and ρ2 sin2 φ = √ 1 1 ⇒ ρ = sin φ since ρ, sin φ ≥ 0. Hence, we integrating wrt to ρ ﬁrst we get csc φ ≤ ρ ≤ 2. Then φ must go from 0 to the intersection of ρ = csc φ and ρ = 2 hence where φ = π and 4 3π π , so π ≤ φ ≤ 34 . Finally there is no restriction on θ so 0 ≤ θ ≤ 2π . Thus, we get by the 4 4 change of variables theorem 2π 3π/4 π/4 √ 2 V= R 1 dV = 0 3π/4 π/4 2π csc φ ρ2 sin φ dρ dφ dθ = = 1 3 2π 0 √ 2 2 sin φ − csc2 φ dφ dθ 1 30 4π = 3 2 dθ ...
View Full Document

## This note was uploaded on 04/18/2010 for the course MATH 237 taught by Professor Wolczuk during the Spring '08 term at Waterloo.

Ask a homework question - tutors are online