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

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