y ds y(θ, z) = 3 sin(θ) 0 z 4 Likewise the bottom of the cylinder will be a disc of radius 3 located at a fixed z = 0

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1 1. Let denote the closed cylinder with bottom given by z and top given by z 4 and the lateral surface given by x 2 + y 2 9. Orient with outward normals. Determine the urface Integral y d (a) Is this a vector or a scalar surface integral? (b) Over what surface are we integrating? Can you write it with parametric equations? (c) Find a outward facing normal vector (d) Compute the integral olution: (a) The function F(x, y, z) y is a scalar valued function, so we are computing a scalar surface integral. (b) We are integration over the cylinder. We can write it with parametric equations, we need three different equations in order to write it, one for the top of the cylinder, one for the bottom, and one for the lateral surface. We will use cylindrical coordinates for all of them. Here is the top of the cylinder, it is just a disc of radius 3, located at a fixed z 4 x(θ, r) r cos(θ) θ 2π y(θ, r) r sin(θ) r 3 z(θ, r) 4 Liewise the bottom of the cylinder will be a disc of radius 3 located at a fixed z x(θ, r) r cos(θ) θ 2π y(θ, r) r sin(θ) r 3 z(θ, r) And the lateral surface will be given by x(θ, z) 3 cos(θ) θ 2π y(θ, z) 3 sin(θ) z 4 z(θ, z) z

2 (c) Now we find a normal vector. We proceed in the usual way, by taing a cross product of the partial derivatives. ince there are three parametric equations, we need three normal vectors Computing the normal for the top section which we will call N 1 : det cos(θ) sin(θ) (,, r) r sin(θ) r cos(θ) Computing the normal for the bottom section which we will call N 2 : det r sin(θ) r cos(θ) (,, r) cos(θ) sin(θ) Computing the normal vector for the lateral sides which we will call N 3 : det 3 sin(θ) 3 cos(θ) (3 cos(θ), 3 sin(θ), ) 1 (d) Now it is easy to compute the surface integral. We first compute the magnitude of each of the vectors N 1 r N 2 r N 3 3 o the surface line integral becomes 4 2π 3 9 sin θdθdz + 2 2π r 2 sin θdθdr + 2. Using the same surface, repeat a-d for the following surface integral (xi + yj)d

3 olution: (a) This time we are computing a vector surface integral, since the integrand (xi+yj) is a vector valued function (b) Nothing has changed from the setup for the previous problem, so we compute the with the same parametric equations. We will use cylindrical coordinates for all of them. Here is the top of the cylinder, it is just a disc of radius 3, located at a fixed z 4 x(θ, r) r cos(θ) θ 2π y(θ, r) r sin(θ) r 3 z(θ, r) 4 Liewise the bottom of the cylinder will be a disc of radius 3 located at a fixed z x(θ, r) r cos(θ) θ 2π y(θ, r) r sin(θ) r 3 z(θ, r) And the lateral surface will be given by x(θ, z) 3 cos(θ) θ 2π y(θ, z) 3 sin(θ) z 4 z(θ, z) z (c) The computations for the normal vector will match our previous wor as well. Computing the normal for the top section which we will call N 1 : det cos(θ) sin(θ) (,, r) r sin(θ) r cos(θ) Computing the normal for the bottom section which we will call N 2 : det r sin(θ) r cos(θ) (,, r) cos(θ) sin(θ)

4 Computing the normal vector for the lateral sides which we will call N 3 : det 3 sin(θ) 3 cos(θ) (3 cos(θ), 3 sin(θ), ) 1 (d) This time around we don t need to compute the magnitude because this is a vector line integral. We compute the integral as (xi + yj) π 3 2π 3 2π 4 2π 72π (3 cos(θ), 3 sin(θ), ) (3 cos(θ), 3 sin(θ), )dθdz+ (r cos(θ), t sin(θ), ) (,, r)dθdt+ (r cos(θ), t sin(θ), ) (,, r)dθdr 9dθdr 3. Let C be the boundary of the surface z x 2 + y 2 with x 2 and y 1, oriented with upward facing normal. Define F (x, y, z) (sin(x 3 ) + xz, x yz, cos(z 4 )) and evaluate F d C olution: This problem is given as a line integral but we will compute the line integral using toes s theorem. To being we will compute the curl of F. curlf det x y z (y, x, 1) sin(x 3 ) + xz x yz cos(z 4 ) Which is a much nicer function to wor with.

5 Now we will need a parametric equation for the surface. ince the surface is the graph of a function we can easily find the parametric equation X(u, v) ( u, v, u 2 + v 2) Which has the upward facing normal vector of ( 2u, 2v, 1) Now we can setup the integral ( v, u, 1) ( 2u, 2v, 1)dudv 2uv 2uv + 1dudv 1 2 4uv + 1dudv 2 4. The helicoid surface is parametrized by X(s, t) (s cos(t), s sin(t), t) for s 1 and t π/2. Compute the line integral F df For the function F(x, y, z) zi + xj + y olution: This problem is stated as a line integral, and ass you to use toes theorem. o the way we will use the theorem is setup the integral as a flux integral. First we need to compute F, you will recall that this is the curl and is computed with x y z (1, 1, 1) z x y We will also need to compute the normal vector r sin(t) cos(t) (sin(t), cos(t), s) cos(t) sin(t) 1

6 Then the integral is: (1, 1, 1) (sin(t), cos(t), s) 1 π/2 sin(t) cos(t) + sdtds π 4 5. Let be the hemisphere x 2 + y 2 + z 2 4 with z oriented upwards. Let F(x, y, z) (x 2 e yz, y 2 e xz, z 2 e xy ) be a vector field. Evaluate: curl F d olution: First we can chec that curl F so we expect the integral to be zero. But we want to do this using toes Theorem. o we replace curl F d With the other side of toes Theorem, Namely F ds The boundary of is a circle of radius 2 in the z plane. We create a parametric equation for this circle as (2 cos(θ), 2 sin(θ), ) Then computing the line integral we have: C 2π 2π F (x, y, z) 2π ( x 2 e yz, y 2 e xz, z 2 e xy) ( sin(θ), cos(θ), ) ( cos 2 (t), sin 2 (t), ) ( sin(θ), cos(θ), ) cos 2 (t) sin(t) + sin 2 (t) cos(t)

7 6. Let F(x, y, z) (xy, e z2 + y, x + y) and let be the graph of the function y x 2 /9 + z 2 /9 1 with z oriented so that the normal vector has positive y component. Compute the integral F d olution: This problem is setup in order to use toes s Theorem. We will convert the given flux integral by means of a line integral. All we must do is figure out what the boundary is. We can recognize this surface as a paraboloid opening along the y axis. Thus its boundary will lay on the y plane. When y we now x 2 /9 + y 2 /9 1 or x 2 + y 2 9, which is the equation for a circle of radius 3. The parametric equation for this circle is x(θ) (3 cos(θ),, 3 sin(θ)) for θ 2π. When we setup a line integral we also need to compute the derivative x (θ) which is easily found to be x (θ) ( 3 sin(θ),, 3 cos(θ)) Now we can use toes s theorem to setup F d C Fds The right hand side of this can be written more exactly as Fds (xy, e z2 + y, x + y)ds C C (3 cos(θ), e (3 sin(θ)2 +, 3 cos(θ) + ) ( 3 sin(θ),, 3 cos(θ)) C 2π 2π 9 cos 2 (θ)dθ 9 cos 2 (θ)dθ 9π 7. Use toes Theorem to evaluate F ds where F(x, y, z) (y, z, x) and C is the triangle with vertices (,, ), (2,, ) and (, 2, 2) oriented counterclocwise when viewed from above. olution: In this problem we will use toes s theorem to convert a line integral into a surface integral. The first step will the be to find a parametric equation for a surface whose boundary is the triangle with vertices (,, ), (2,, ) and (, 2, 2).

8 The parametric equations we come up with are u Φ(u, v) u v With the bounds on u and v being u 2 u 2 v Having the parametric equations in hand, we can proceed to computing the normal vector We also must compute the curl of F. curlf ( 1, 1, 1) Finally, we can setup and solve the integral. F ds curlfd ( 1, 1, 1) (1, 1, )d 2 2dvdu 4 u 2 8. The height of a mountain at a point (x, y) is given by z 2 x 2 y 2. You wal counterclocwise around the mountain on the boundary of the region R defined by 1 x 1 and 1 y 1. A magnetic field is exerting force F(x, y, z) (x 2 y, 2x(y 2 +z), z 3 ) on your compass. Use toes theorem to calculate the wor done by the magnetic field as you wal around the mountain. olution: Once again we are going to convert a line integral into a surface integral to mae integration easier. The surface here is already given to use, we simply need to find parametric equations for it. ince this is the graph of a function, parametric equations are easy to find. u Φ(u, v) v 2 u 2 v 2

9 With the bounds 1 u 1 1 v 1 We can also compute the normal vector 1 2u 1 2v 2u 2v 1 (here we have taen care to pic the upward facing normal vector). We also must compute the curl of F. curlf ( 2x,, 2y 2 + z x 2 ) o we can now setup and compute the integral F ds curlfd ( 2u,, 2u 2 + 2(v 2 ) + 2 u 2 v 2 ) (2u, 2v, 1)d u 2 dudv Let F(x, y, z) (xy 2, yx 2, z 2 ), calculate the total flux out of a canister W, which can be expressed by the equations x 2 + y 2 9 and 1 z 3. olution: We will solve this problem using the divergence theorem. As the problem is stated we want compute a flux (surface) integral over the boundary of W. That is we want to compute W Using the divergence theorem this becomes It is not difficult to compute the div of F W Fd Div W da

10 Div F ( ) xy 2 + ( ) yx 2 + ( ) z 2 x y z y 2 + x 2 2z o we setup the integral according to the divergence theorem, and writing the region in cylindrical coordinates W y 2 + z 2 2z 3 2π 3 1 (r 2 2z)rdrdθdz 9π 1. Use the Divergence Theorem to compute the value of the flux integral Fd olution: Where F(x, y, z) (y 3 +3x, xz+y, z+x 4 cos(x 2 y)) and is the boundary of the region bounded by x 2 + y 2 1, x, y and z 1 Using the divergence theorem we will compute this rather difficult integral into a (hopefully) simpler triple integral over the divergence of F. To begin let us compute Div F ( Div (F) x, y, ) (y 3 + 3x, xz + y, z + x 4 cos(x 2 y) ) 5 z We then can setup the triple integral as W 5dV Looing at the region we want to compute in cylindrical coordinates as 1 π 2 1 5rdrdθdz 5π 4 Which is the value of the surface integral, by the Divergence theorem.

11 11. The department of transportation is building a elliptical road around mount dull, which they plan to name the Tedium Trail (as in has no exits, only on ramps). The road is to be bounded by the curves x y y2 9 4 while the height of the land at the point (x, y) is given by x 2 z h(x, y) 1 x 2 2y 2 (a) Find a parametric equation for the road surface (b) ince Mount Dull is located in North America, its orientation is given using a upward facing normal vector. Find this upward facing normal vector. (c) Find parametric equations for the edge of the road, which are consistent with the orientation. What does this say about what side of the road people drive on. (d) The people of Australia decide to build a copy of Tedium Trail, but down under, where they prefer to use a downward facing normal vector. How does the different normal vector change your answers to parts (b) and (c). olution: (a) To parametrize the surface, we will use a modification of polar coordinates which has scaled the x coordinates scaled by a factor of 2 and the y coordinates scaled by a factor of 3. We will the write the z coordinate in terms of the x and y parametrization. x(r, θ) 2r cos(θ) Φ(r, θ) y(r, θ) 3r sin(θ) z(r, θ) 1 4r 2 cos(θ) 2 9r 2 sin(θ) 2 Notice that when r 1 this satisfies the equation x2 + y2 1 and when r this satisfies the equation x2 + y2 4, so we have the bounds on r of 1 r (b) To compute the normal vector we begin by computing the two tangent vectors, then compute their cross product. We don t now, apriori what direction the normal vector will face, but we can fix that later. 2 cos(θ) T r 3 sin(θ) 8r cos(θ) 2 18r sin(θ) 2 2r sin(θ) T θ 3r sin(θ) 1r 2 cos(t) sin(t) 24r 2 cos(θ) N Φ 36r 2 sin(θ) 6r

12 We notice the 6r so the normal vector is upward facing. (c) The boundary of the road will be two ellipses and nowing a parametric equation for the road, its not difficult to find a parametrization for the edges of the road. By looing at r 1 and r 2 we easily find C inside (θ) Φ(1, θ) 2 cos(θ) 3 sin(θ) 1 4 cos(θ) 2 9 sin(θ) 2 4 cos(θ) C outside (θ) Φ(1, θ) 6 sin(θ) 1 16 cos(θ) 2 36 sin(θ) 2 However, in order to obey the right hand rule, the inside curve must have a clocwise orientation, so we will modify the sign of the y coordinate. C inside (θ) Φ(1, θ) 2 cos(θ) 3 sin(θ) 1 4 cos(θ) 2 9 sin(θ) 2 This means that when you are traveling clocwise, you should be driving on the right side of the road. (d) This reverses the orientation of both boundary curves, so we have C inside (θ) Φ(1, θ) 2 cos(θ) 3 sin(θ) 1 4 cos(θ) 2 9 sin(θ) 2 4 cos(θ) C outside (θ) Φ(1, θ) 6 sin(θ) 1 16 cos(θ) 2 36 sin(θ) 2 Notice that now, when driving in a clocwise direction, you are on the left side of the road.