1. ompute the surface integral M255 alculus III Tutorial Worksheet 11 x + y + z) d, where is a surface given by ru, v) u + v, u v, 1 + 2u + v and u 2, v 1. olution: First, we know x + y + z) d [ ] u + v) + u v) + 1 + 2u + v) r u r v da, where is the domain of the parameters u, v given by u 2, v 1. We have r u 1, 1, 2 and r v 1, 1, 1. Then, r u r v 1, 1, 2 1, 1, 1 3, 1, 2. o, r u r v 3, 1, 2 3 2 + 1 2 + 2) 2 14. Thus, x + y + z) d 1 2 11 14. 4u + v + 1) 14 du dv 2. Let be the portion of the graph z 4 2x 2 3y 2 that lies over the region in the xy-plane bounded by x, y, and x + y 1. Write the integral that computes x 2 + y 2 + z ) d. olution: First, we need a parametrization of the surface. ince is a surface given by the equation z 4 2x 2 3y 2, we can choose x and y to be the parameters. o, rx, y) x, y, 4 2x 2 3y 2, and the domain of the parameters x, y is given by the region in the xy-plane bounded by x, y, and x + y 1 see picture below)
Now, r x 1,, 4x and r y, 1, 6y. o, r x r y 4x, 6y, 1 and r x r y 4x, 6y, 1 16x 2 + 36y 2 + 1. Thus, x 2 + y 2 + z ) d x 2 + y 2 + 4 2x 2 3y 2 ) r x r y da 1 x+1 4 x 2 2y 2 ) 16x 2 + 36y 2 + 1 dy dx. 3. ompute F d, where F yi xj + zk and is a surface given by x 2u, y 2v, z 5 u 2 v 2, where u 2 + v 2 1. has downward orientation. olution: We have ru, v) 2u, 2v, 5 u 2 v 2, so r u 2,, 2u and r v, 2, 2v and so r u r v 2,, 2u, 2, 2v 4u, 4v, 4. Note that r u r v 4u, 4v, 4 gives unit normal vectors pointing upward z-component is positive). But, has downward orientation so F d F r u r v ) da. u 2 +v 2 1 Now, F ru, v) ) 2v, 2u, 5 u 2 v 2. o F r u r v ) 2v, 2u, 5 u 2 v 2 4u, 4v, 4 2 4u 2 4v 2.
Thus, F d u 2 +v 2 1 1 polar 18π. 2 4u 2 4v 2 ) da. 2 4r 2 )r dr dθ 4. ompute the flux of the vector field F xi + yj + zk over the part of the cylinder x 2 + y 2 4 that lies between the planes z and z 2 with normal pointing away from the origin. olution: We want to compute F d, where is the part of the cylinder x 2 + y 2 4 that lies between the planes z and z 2 with normal pointing away from the origin. First, we parametrize : let x 2 cos u, y 2 sin u, z v. Then ru, v) 2 cos u, 2 sin u, v, domain is u 2π, v 2. Then, r u 2 sin u, 2 cos u, and r v,, 1. o, r u r v 2 sin u, 2 cos u,,, 1 2 cos u, 2 sin u,. Now, let s check our orientation. Let s take the point where u π/2 and v 1, ie x, y, z), 2, 1). At the point, 2, 1), the unit normal vector points in the direction of the vector r u r v )π/2, 1), 2,. This means the unit normal vector is pointing away from the origin. o, our parametrization of gives the correct orientation for. Moving on! Now, F ru, v) ) r u r v ) 2 cos u, 2 sin u, v 2 cos u, 2 sin u, 4. Thus, F d 2 16π. F r u r v ) da 4 da. 4 dv du
Alternative solution using divergence theorem We cannot apply the divergence theorem directly to the cylinder because it is not a closed surface. But we can apply the divergence theorem to the cylinder with the top and bottom added to eliminate all computations. The divergence theorem says Top Bottom ide We want to find F d. ides The R is 3dV The L is x 2 +y 2 4 Top 2k kda+ F d inside of cylinder F nd + x 2 +y 2 4 Bottom ence F nd 24π 8π 16π. ide inside of cylinder F )dv 3dV 3 volume of cylinder) 24π F nd + k)da+ ide ide F nd F nd 8π+ ide F nd 5. Find the flux of the vector field Fx, y, z), z, 1 across the hemi-sphere x 2 + y 2 + z 2 4, z with orientation away from the origin. olution: If we do this problem from scratch, we need to start by parametrizing the hemi-sphere: xφ, θ) 2 sin φ cos θ, yφ, θ) 2 sin φ sin θ, zφ, θ) 2 cos φ, where φ π/2 and θ 2π. Then rφ, θ) 2 sin φ cos θ, 2 sin φ sin θ, 2 cos φ, where φ π/2 and θ 2π. And we get r φ r θ 4 sin 2 φ cos θ, 4 sin 2 φ sin θ, 4 sin φ cos φ We now want to check the orientation of the surface. Let φ π/4 and θ π/2, then at the point, 2, 2), we get the vector r φ r θ π/4, π/2), 2, 2 points away from the origin. Thus, our parametrization gives the correct orientation of the surface.
Then, we have the flux of F across the given hemi-sphere can be compute using the formula F d F rφ, θ) ) r φ r θ ) da. F rφ, θ) ), 2 cos φ, 1 and φ π/2 θ 2π F rφ, θ) ) r φ r θ ) 8 sin 2 φ cos φ sin θ + 4 sin φ cos φ Thus, F d π/2 8 8 sin 2 φ cos φ sin θ + 4 sin φ cos φ ) dφ dθ 3 sin3 φ π/2 sin θ + 2 sin 2 φ π/2 ) 8 3 sin θ + 2 8 3 cos θ 2π + 2 2π 4π dθ ) dθ Another olution: If you already know that for a sphere of radius 2 with orientation 1 away from the origin, its unit normal vector is given by n 2 x, 1 2 y, 1 2 z and r φ r θ 4 sin φ, then we could use the definition of the flux integral to compute F d as follows: F d F n d 1, z, 1 2 x, 1 2 y, 1 2 z d 1 2 yz + 1 ) 2 z d 2 sin φ cos φ sin θ + cos φ) r φ r θ da φ π/2 θ 2π π/2 π/2 4π 2 sin φ cos φ sin θ + cos φ) 4 sin φ dφ dθ 8 sin 2 φ cos φ sin θ + 4 sin φ cos φ ) dφ dθ
Yet another solution using divergence theorem Again, we cannot apply the divergence theorem directly to the hemisphere because it is not a closed surface. But we can apply the divergence theorem to the hemisphere with the inside of the bottom disk added. The divergence theorem says emisphere Bottom F d We want to find F d. emisphere The R is. The L is F nd + emisphere emisphere F nd + ence F nd 4π emisphere x 2 +y 2 4 inside of hemisphere Bottom F nd k k)da F )dv emisphere F nd 4π Each of the problem below can be solved using one of these theorems: Green s Theorem, tokes Theorem, or ivergence Theorem 6. Let be the surface defined as z 4 4x 2 y 2 with z and oriented upward. Let F x y, x + y, ze xy. ompute curl F d. int: use one of the theorems you learned in class.) olution: This question uses tokes theorem: is a surface with boundary, and we are taking the flux integral of the curl of F. The boundary of is given by z, 4x 2 + y 2 4, and since is oriented with upward orientation, the boundary of has counterclockwise orientation when viewed from above. Thus, a parametrization of the boundary is given by rt) cos t, 2 sin t,, t 2π.
Thus, by tokes Theorem, we have curl F d F dr 4π. Frt)) r t)dt cos t 2 sin t, cos t + 2 sin t, sin t, 2 cos t, dt sin t cos t + 2 sin 2 t + 2 cos 2 t + 4 sin t cos t ) dt 2 + 3 sin t cos t) dt 2t + 3 ) 2π 2 sin2 t 7. Evaluate x 4 y 5 2y)dx + 3x + x 5 y 4 )dy where is the curve below and is oriented in clockwise direction. olution: This problem uses Green s theorem. One main clue is the shape of the curve it has 8 pieces!). Let be the region enclosed by the curve. And since
the orientation of is clockwise, instead of counterclockwise, we have [ x 4 y 5 2y)dx + 3x + x 5 y 4 )dy 3 + 5x 4 y 4 ) 5x 4 y 4 2) ] da 5 da 5 1 da 5 Area) 5 9 45. 8. Let be the boundary surface of the region bounded by z 36 x 2 y 2 and z, with outward orientation. Find F d, where F xi + y 2 j 2yzk. olution: This is a closed surface, so the divergence theorem works nicely here. div F x x) + y y2 ) + 2yz) 1 + 2y 2y 1 z all the solid since it s half of a ball). o, the divergence theorem gives F d 1 dv volume) The solid is half of the ball of radius 6, and so its volume is volume) 1 ) 4 2 3 π6)3 2 216π) 144π. 3 9. Let be the boundary curve of the part of the plane x + y + 2z 2 in the first octant. has counterclockwise orientation when viewing from above. ompute F dr, where F e sin x2, z, 3y.
olution: Note: To compute F dr directly, we need to do 3 integrals since consists of 3 pieces. But, because we know is the boundary curve of the surface x + y + 2z 2 in the first octant, we can try to use tokes Theorem. By tokes Theorem, F dr curl F d, where is the part of the plane x + y + 2z 2 in the first octant. ince the equation x + y + 2z 2 or z 1 1x 1 y determines, a parametrization of is given by 2 2 rx, y) x, y, 1 1 2 x 1 2 y, where x, y). The domain is given by the projection of onto the xy-plane:
1. A hallenging Problem) Evaluate y 3 + cos x)dx + sin y + z 2 )dy + x dz where is the closed curve parametrized by rt) cos t, sin t, sin 2t with counterclockwise direction when viewed from above. int: the curve lies on the surface z 2xy.) olution: If you rewrite this integral as F dr and note that the curve lies in R 3 and not in the plane otherwise we d use Green s theorem), we see that tokes theorem applies to it. The hint provides the surface to fill in the curve with. First, we need to parametrize the surface z 2xy: px, y) x, y, 2xy, x, y) { x, y) x 2 + y 2 1 } as the parametrization. has counterclockwise orientation when viewed from above, so this means that the surface, call it, we fill it in with must have upward orientation. p x 1,, 2y p y, 1, 2x p x p y 2y, 2x, 1 Notice that p x p y points upward, since the ˆk-component is positive, so this is the correct choice for the orientation. Now, we need the curl of F î ĵ ˆk curl F x y y 3 + cos x sin y + z 2 x z 2z, 1, 3y 2
Finally, we apply tokes Theorem y 3 + cos x)dx + sin y + z 2 )dy + x dz 1 curl F) d curl F) p x p y )da 4xy, 1, 3y 2 2y, 2x, 1 da 8xy 2 + 2x 3y 2) da 8r 3 cos θ sin 2 θ + 2r cos θ 3r 2 sin 2 θ ) r drdθ 8 5 r5 cos θ sin 2 θ + 2 3 r3 cos θ 3 ) 1 4 r4 sin 2 θ 8 5 cos θ sin2 θ + 2 3 cos θ 3 ) 4 sin2 θ dθ 8 5 cos θ sin2 θ + 2 3 cos θ 3 1 cos 2θ 4 2 8 15 sin3 θ + 2 3θ sin θ 3 8 + 3 ) 2π sin 2θ 16 3 4 π dθ )) dθ