Solutions - Homework sections

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1 olutions - Homework sections valuate xy d, where is the triangle with vertices (,, ), (,, ), and (,, ). The three points - and therefore the triangle between them - are on the plane x + y + z, and their projection onto the xy-plane forms the triangle between the points (,, ), (,, ), and (,, ). o is the portion of the plane z x y over the region {(x, y) : x, y x}. ince we have z as a function of x and y on our surface, we can use formula 4 from this section: xy d xy ( ) + ( ) + da x 6 6. [ xy xy dydx ] x dx (x x + x ) dx [ x x + 4 x4 ] 9. valuate yz d, where is the surface with parametric equations x u, y u sin v, z u cos v, for u, v π/. Our parametrization is r(u, v) u, u sin v, u cos v, so: r u u, sin v, cos v ; r v, u cos v, u sin v ; r u r v u sin v u cos v, u sin v, u cos v u, u sin v, u cos v ; r u r v u + 4u 4 sin v + 4u cos v u + 4u 4 o, using formula from this section, yz d u + 4u. π/ π/ y(u, v)z(u, v) r u r v da u cos v u sin v u 4u + dvdu sin v cos v u 4u + dvdu u 4u + u 4u + du [ ] π/ sin v du

2 ubstitute : t 4u +, u 4 (t ), dt u du : changing the bounds, we get: (t ) t 8 dt t / t / dt [ 64 5 t5/ t/ ] 5. valuate x z d, where is the part of the cone z x + y between the planes z and z. The widest point of is at the intersection of the cone and the plane z, where x + y 9; its thinnest point is where x + y. Thus, is the portion of the surface z x + y over the region {(x, y) : x + y 9}. o: x z d 64 x ( ( x + y ) x x + y ) + x (x + y x + y ) x + y + da x (x + y ) da (r cos θ) (r )r drdθ r 5 cos θ drdθ [ ] r 6 dθ 6 cos θ dθ π. [ θ + 4 sin θ ] π (Use the Reference Pages in the back of your textbook to evaluate cos θ dθ.). ( y x + y ) + da valuate y d, where is the part of the paraboloid y x + z inside the cylinder x + z 4. We already have y as a function of the other two variables, and we want to use x and z as parameters. The projection of onto the xz-plane is the disk {(x, z) : x + z 4}. y d ( ) y y + x ( ) y + da z (x + z ) 4x + 4z + da r 4r + r drdθ

3 Make a substitution: u 4r + ; r 4 (u ), and du r dr: changing the bounds, this is: (u ) u 8 dudθ u / u / dudθ [ 5 u5/ u/ ] 7 dθ ( 5 75/ 7/ 5 + ) π 6 (9 7 + ). valuate (x z + y z)d, where is the hemisphere x + y + z 4, z. Parametrize the hemisphere in spherical coordinates: is parametrized by r(φ, θ) sin φ cos θ, sin φ sin θ, cos φ, where φ π/, θ π. dθ r φ cos φ cos θ, cos φ sin θ, sin φ ; r θ sin φ sin θ, sin φ cos θ, ; r φ r θ 4 sin φ cos θ, 4 sin φ sin θ, ; r φ r θ 4 sin φ. o, (x z + y z) d (x + y )z d π/ π/ π π/ [ 64π 4 sin4 φ 6π. (4 sin φ cos θ + 4 sin φ sin θ)( cos φ)(4 sin φ) dθdφ sin φ cos φ dθdφ sin φ cos φ dθdφ ] π/ dφ 6. valuate xz d, where is the boundary of the region enclosed by the cylinder y + z 9 and the planes x and x + y 5. has three parts: is the part of that lies on the cylinder, is the part of x + y 5 within the cylinder, and is the part of x within the cylinder. ince we can t come up with one parametrization that describes the entire surface, we re going to parametrize each of these three parts separately, and then combine them at the end. : is parametrized by r(u, v) u, cos v, sin v, where v π. We want the range of x to be x 5 y, so we use u 5 cos v. This gives us r u,, and r v, sin v, cos v, so r u r v, cos v, sin v, and

4 finally r u r v. o: xz d 9. 5 cos v : On, x 5 y. Use y and z as parameters: u( sin v) dudv (5 cos v) sin v dv r(y, z) 5 y, y, z ; r y,, ; r z,, ; r y r z ; r y r z. The projection of onto the yz-plane is {(y, z) : y + z 9}, so: xz d (5 y)z d : On, x, and so xz. o xz d d.. (5 r cos θ)(r sin θ)r drdθ (5r r cos θ)(sin θ) drdθ [ 5 r 4 r4 cos θ ( [ 4 8 ( ] cos θ) sin θ dθ cos θ) sin θ dθ Put the three parts together: xz d xz d + xz d + xz d + +. ] π 9. valuate F d, for F(x, y, z) xyi + yzj + zxk, where is the part of the paraboloid z 4 x y above the square x, y, with upward orientation. z g(x, y) 4 x y, g g x, and x y y. etting to be the square [, ] [, ], we can use formula from this section: F d ( xy( x) yz( y) + zx) da 7 8. (x y + y (4 x y ) + x(4 x y )) dydx ( x + x x ) dx 4

5 . valuate F d, for F(x, y, z) yi + xj + z k, where is the helicoid r(u, v) u cos vi + u sin vj + vk, u, v π, oriented upwards.. r u cos v, sin v,, and r v u sin v, u cos v,, so r u r v sin v, cos v, u. Also, F(r(u, v)) u sin v, u cos v, v. Therefore, by formula 9 from this section: F d F (r u r v ) da π π π 6 π. (u sin v u cos v + uv ) dudv u cos v + uv ) dudv ( cos v + v ) dv valuate F d, for F(x, y, z) xzey i xze y j + zk, the part of the plane x + y + z in the first octant, oriented downwards. z g(x, y) x y, and the projection of onto the xy-plane is the triangle {(x, y) : x, y x}. We can use formula 9, but because is oriented downwards in this case, we add a minus sign at the beginning.. F d x 6. ( xze y ( ) ( xze y )( ) + z) da z da ( x y) dydx ( x x + ) dx valuate F d, for F(x, y, z) xi + yj + z4 k, where is the part of the cone z x + y below the plane z, oriented downwards. z g(x, y) x + y, so g x x g and x + y y y x + y. The projection of onto the xy-plane is the disk {(x, y) : x + y }. Again, we can use formula 9, but because is oriented downwards, we add a minus sign to the 5

6 beginning: F d 6 π ( ) ( ) x y ( x y + z 4 ) da x + y x + y ( ) (x + y ) x + y + ( x + y ) 4 ( r + r 4 )r drdθ r (r 5 r ) drdθ ( 6 ) dθ da 7.8. π. valuate curl F d, for F(x, y, z) y cos zi+ex sin zj+xe y k, where is the hemisphere x +y +z 9, z, oriented upward. The boundary curve of is the circle x + y 9, z, oriented counterclockwise looking down from above. We can parametrize as r(u, v) cos t, sin t,. This gives us F(r(t)) 6 sin t,, cos t e sin t, and r (t) sin t, cos t,. Via tokes s Theorem: curl F d F dr 8π. F(r(t)) r (t) dt 8 sin t dt (To evaluate sin t dt, see the Reference Pages in the back of the textbook.). valuate curl F d, for F(x, y, z) x z i + y z j + xyzk, where is the part of the paraboloid z x + y inside the cylinder x + y 4, oriented upward. The boundary curve of is the intersection of the paraboloid and the cylinder, that is, z x + y 4: the circle of radius at the height z 4. To match the orientation of, should be oriented counterclockwise as seen from above. o, we can parametrize as r(t) cos t, sin t, 4 ; this gives us r (t) sin t, cos t,, and F(r(t)) 64 cos t, 64 sin t, 6 sin t cos t. o, by tokes s Theorem: curl F d F dr 8. F(r(t)) r (t) dt ( 8 cos t sin t + 8 sin t cos t) dt [ cos t + ] π sin t 6

7 5. valuate curl F d, for F(x, y, z) xyzi + xyj + x yzk, where is the top and four sides, but not the bottom, of the cube with vertices (±, ±, ±), oriented outward. quation in this section says that, if two surfaces and have the same oriented boundary curve, then curl F d curl F d. o, instead of working the integral over, which we would have to do in five parts, or over the boundary curve, which is a square, and would have to be done in four parts, we re going to pick another surface with the same boundary curve, and work the integral over that surface. Let be the missing bottom face of the cube: it s a solid square in the plane z. ince the top of the cube is oriented outward, we orient upwards, such that n k, so that both surfaces orient the boundary the same way. curl F x z, xy xyz, y xz. ince n k,, on, curl F n y xz x + y, since z on. o, using equation : curl F d curl F d curl F n d 8.. y dy (x + y) dxdy valuate F dr, where F(x, y, z) e x i + e x j + e z k, and is the boundary of the part of the plane x + y + z in the first octant. Let be the portion of the plane bounded by. Then is the part of the graph of z x y located over the triangle {(x, y) : x, y x}. curl F,, e x. o, using tokes s Theorem and formula from section 7.7, F dr curl F d ( + + e x ) da.a) x e x dydx ( x)e x dx (integrate by parts, u x, dv e x ) [( x)e x + e x ] e 4. valuate F dr, where F(x, y, z) x zi + xy j + z k, and is the curve of intersection of the plane x + y + z and the cylinder x + y 9, oriented counterclockwise from above. The intersection is an ellipse in the plane x + y + z ; thus, it has unit normal vector n (i + j + k). curl F, x, y, so curl F n (x + y ). 7

8 Finally, if is the part of the plane contained inside the ellipse, the projection of onto the xy-plane is the disk {(x, y) : x + y 9}. Over, is given by z g(x, y) x y, so g x g y. o, by tokes s Theorem, F dr curl F n d. 8π. (x + y ) d (x + y ) ( ) + ( ) + da (x + y ) da r drdθ 8 4 dθ Verify tokes s Theorem for F(x, y, z) y i + xj + z k, the part of the paraboloid z x + y below the plane z, oriented upward. The projection of onto the xy-plane is {(x, y) : x + y }. curl F,, y. o, by equation from section 7.7, curl F d ( + + y) da ( y) da π. ( r sin θ) r drdθ ( sin θ) dθ Also, the boundary curve of is the circle x + y, z, oriented counterclockwise from above. o a parametrization is r(t) cos t, sin t,, t π. This gives us r (r) sin t, cos t,, and F(r(t)) sin t, cos t,. o, F dr F(r(t)) r (t) dt π. (cos t sin t) dt ( + cos t) dt [t + ] π sin t ( cos t ) sin t dt [ cos t + cos t Therefore, F dr curl F d, and tokes s Theorem holds. ] π 8

9 7. A particle moves along line segments from the origin to (,, ) to (,, ) to (,, ) and back to the origin, under the force field F(x, y, z) z i + xyj + 4y k. Find the work done. If is the path the particle follows, then the work done is F dr. All four points are in the plane z y. o, if is the flat surface with boundary, is the portion of the plane z y over the rectangle [, ] [, ], oriented upward (to match the direction in which the particle moves around ). curl F 8y, z, y. o, using equation from section 7.7, the work done is F dr curl F d ( 8y() z( ) + y) da (y z) da (y y) da y da 8.. [ 4 y dx y dydx valuate (y+sin x)dx+(z +cos y)dy+x dz, where is the curve r(t) sin t, cos t, sin t, t π. ] dx This is F dr, for F(x, y, z) y + sin x, z + cos y, x, giving us curl F z, x,. is on the surface z xy; let be the part of this surface bounded by. The projection of onto the xy-plane is {(x, y) : x + y }. Looking down from above, is oriented counterclockwise; to match this, our surface needs to be oriented downwards. ince z g(x, y) xy, we can use equation from the last section, adding a minus sign on the front because of the downward orientation: F dr curl F d ( ( 4xy)(y) ( x )(x) ) da (8xy + 6x ) da π. (8r cos θ sin θ + 6r cos θ ) r drdθ ( 8 5 cos θ sin θ cos θ ) dθ [ 8 5 sin θ (sin θ sin θ) θ ] π 9

10 9. If is a sphere and F satisfies the hypotheses of tokes s Theorem, show that curl F d. On its own, a sphere doesn t have a boundary curve to which we can apply tokes s Theorem. o, cut the sphere in half: let be the upper half of the sphere, and the lower half, such that. Let be the boundary curve of, and let be the boundary curve of. Then: curl F d curl F d + curl F d F dr + F dr But and are the same circle, oriented in opposite directions: is oriented counterclockwise from above, to match the orientation of the top half of the sphere, while is oriented clockwise, to match the bottom half of the sphere. o curl F d F dr + F dr F dr + F dr. 7.9 alculate F d: 5: F(x, y, z) e x sin yi + e x cos yj + yz k; is the surface of the box bounded by x, x, y, y, z, z. 7. div F P x + Q y + R z (ex sin y) + ( e x sin y) + (yz) yz. etting to be the box of which is the boundary, we can use the ivergence Theorem: F d. div F dv yz dzdydx [ yz ] dydx 4y dydx [ y ] dx dx F(x, y, z) xy i + xe z j + z k; the surface of the solid bounded by y + z, x, and x. div F y + + z y + z. Let be the solid bounded by, and use the ivergence Theorem:

11 F d div F dv (y + z ) dv 4 [ 4 r4 r r drdθdx ] dθdx 4 π dx 4 π 9π. dθdx 9. F(x, y, z) xy sin zi + cos(xz)j + y cos zk; is x /a + y /b + z /c. div F y sin z + y sin z. o, if is the volume inside the ellipsoid,. F d. div F dv F(x, y, z) (cos z + xy )i + xe z j + (sin y + x z)k; is the surface of the solid bounded by z x + y and z 4. div F y + + x x + y. If is the solid bounded by, then we want to think about in cylindrical coordinates: the paraboloid and the plane intersect where x + y z 4. Inside, z values range from x + y r to 4. o: F d x + y dv 4 π. dv r r dzdrdθ r [ r z ] 4 drdθ r 4r r 5 drdθ [r 4 r6 6 dθ ] dθ

12 . F(x, y, z) 4x zi + 4y zj + z 4 k; is the sphere of radius R centered at the origin. 7. div F x z + y z + z z(x + y + z ). Letting be the volume inside the sphere: F d π R z(x + y + z ) dv [θ] π dθ π R6 6. ρ cos φ ρ ρ sin φ dρdφdθ π [ sin φ cos φ sin φ dφ ] π [ ] R 6 ρ6 R ρ 5 dρ Use the ivergence Theorem to evaluate F d, where F(x, y, z) z xi+( y +tan x)j+(x z +y )k and is the top half of the sphere x + y + z. by itself is not the boundary of a simple solid region - the bottom of is open. o, we re going to add to so that it does enclose a simple solid region, and then use the ivergence Theorem on that. To fill in the bottom of, we need the missing disk. Let be the disk x + y, z. ince is part of the unit sphere, is oriented outwards; to match that, we make be oriented downwards. Then if we set, is a closed surface, and F d F d F d. ince is oriented downward, the normal vector on points straight downward: n k. o F n F k x z y y, since z on. Let be the unit disk. Then: F d F n d ( y )da 4 4 π (r sin θ) r drdθ r sin θ drdθ sin θ dθ is a closed surface, so we can use the ivergence Theorem on it. div F z + y + x.

13 Let be the volume inside the half-sphere encloses. Then: F d div FdV (x + y + z )dv π ρ ρ sin φ dρdφdθ π. π sin φ dρdθ dθ o, putting the two parts together, F d F d F d 5 π ( π) 4 π. 4. Use the ivergence Theorem to evaluate (x + y + z ) d, where is the sphere x + y + z : To get this in the right form, we need to find a vector field F such that F n d (x+y+z ) d, so we want F n x + y + z. xi + yj + zk In general, on a sphere, the normal vector is (you can show this using the φ and θ x + y + z xi + yj + zk parametrization of the sphere). For, the sphere of radius, then, n xi + yj + zk. x + y + z o, if F n x + y + z, F i + j + zk. div F + +. o, let be the volume inside. Then: (x + y + z ) d F n d F d dv 4 π, since 4 π is the volume of the ball of radius.