Math 234 Review Problems for the Final Exam

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1 Math 234 eview Problems for the Final Eam Marc Conrad ecember 13, 2007 irections: Answer each of the following questions. Pages 1 and 2 contain the problems. The solutions are on pages 3 through 7. Problem 1. olve each of the following integration problems. While solving each one, try to determine the best (i.e., fastest) way to calculate the given integral. ome may require that you calculate directly, while others will be faster to compute using integration theorems. (a) Let F zi + ( + z)j y2 z 2 k. Find the flu of F inward across the surface that consists of the portion of the cone z 2 + y 2 for 1 z 2 together with its caps 2 + y 2 1, z 1 and 2 + y 2 4, z 2. (b) Integrate the function f(, y, z) z 2 over the unit sphere 2 +y 2 +z 2 1. (c) Let C be the curve that consists of the intersection of the plane 2+y+z 5 with the square tube consisting of the planes 0, 1, y 0 and y 1, oriented counterclockwise as viewed from above. Find the circulation of the vector field F yi z2 j + 2k along C. (d) Find the flu of the vector field F (y 2 +ln(y 2 +z 2 ))i+ 2 + z 4 + 1j+ (e cos y2 + 2z)k outward across the sphere 2 + y 2 + z 2 a 2. (e) Find the flu of the vector field F yi j + zk in the direction towards from the origin across the portion of the paraboloid z 1 2 y 2 in the first octant. 1

2 Problem 2. Which of the following make sense? Of the ones that make sense, which are always equal to zero for nice functions f or vector fields F? ( f) ( f) ( F) ( F) Problem 3. tate each of the following. Be sure to include the necessary hypotheses. (a) tokes Theorem (b) The ivergence Theorem (c) The Cauchy-iemann equations Problem 4. Let f( + iy) 2 2 2y 2 + 4yi. Find the derivative of f. Problem 5. For > 0, define f( + iy) ln ( ( 2 + y 2 ) 1/2) ( y + arctan i. ) (a) Is f differentiable? (b) ewrite f as a function of the polar coordinates r and θ. (c) Now let α be a fied comple number and define the function g α (z) e αf(z). Use the polar form of f that you found in part (b) to show that if n is a natural number, then g n (z) z n. ecall that the natural numbers are 1, 2, 3,... 2

3 olutions 1a. We are asked for the flu across a closed surface, so we can apply the ivergence Theorem. Furthermore, since the surface consists of three parts, the ivergence Theorem will allow us to do fewer calculations. o we have F n in dσ F d dy dz (on t forget the minus sign to account for the fact that n in points in instead of out!) We can easily calculate that F ( 2 + y 2 )z. The integration will be simplest in cylindrical coordinates. Then F r 2 z and we re left with 2 2π z F n in dσ r 3 z dr dθ dz z1 θ0 r0 (We have r 3 instead of r 2 because of the streching factor r that comes from integration in cylindrical coordinates). If we integrate this, we will get an answer of 21π 4. 1b. Even though we have a closed surface, the ivergence Theorem doesn t apply because it only works for integrating vector fields over surfaces. It cannot help us integrate a scalar valued function over a surface. We can use spherical coordinates (keeping in mind that ρ 1 on the surface of the unit sphere) to parametrize the sphere and obtain r(θ, φ) sin φ cos θi + sin φ sin θj + cos φk 0 φ π 0 θ 2π Now the integral of f over the sphere is f(, y, z) dσ 2π π θ0 φ0 f((θ, φ), y(θ, φ), z(θ, φ)) r φ r θ dφ dθ irect computation yields r φ cos φ cos θi + cos φ sin θj sin φk and r θ sin φ sin θi+sin φ cos θj. The cross product is then sin 2 φ cos θi+sin 2 φ sin θj+ 3

4 sin φ cos φk. If we calculate the length of this, we get r φ r θ sin φ (You ll need to use the identity sin 2 + cos 2 1 a couple of times). Now f((θ, φ), y(θ, φ), z(θ, φ)) (z(θ, φ)) 2 cos 2 φ. o were left with f(, y, z) dσ 2π π θ0 φ0 Computing this integral will yield a result of 4π 3. cos 2 φ sin φ dφ dθ 1c. C is a closed curve, so we can apply tokes Theorem. Furthermore, C consists of four line segments, so using the theorem will cut down on the number of calculations we have to make. The surface bounded by C is the portion of the plane 2 + y + z 5 that lies above the square 0 1, 0 y 1. o is the graph of the function f(, y) 5 2 y. Then we have that F dr F n up dσ C (We have used the right hand rule to determine that we want n up instead of n down ). Now F zi + k and so F n up dσ ( zi + k) ( f, f y, 1) d dy is the shadow of in the y plane, so in this case is the square 0 1, 0 y 1. We also have that f 2 and f y 1 and so we re left with y0 2z + 1 dy d y y0 2(5 2 y) + 1 dy d (4 + 2y 9) dy d Computing the integral will yield an answer of 6. 1d. We are asked for the flu across a closed surface, so we can apply the ivergence Theorem. In this case, although the surface isn t particularily 4

5 bad, the vector field is, so the theorem will help us a lot. In fact, F 2 and so we have that F n out dσ F d dy dz 2 d dy dz o the answer will simply be 2 times the volume of the sphere, or 8 3 πa3. 1e. We are asked for the flu across a surface that is not closed, and so the ivergence Theorem won t be of much use. o we are left with directly calculating the flu. is the graph of the function f(, y) 1 2 y 2 and we have that the shadow of in the y plane is the quarter of the circle 2 + y 2 1 that lies in the first quadrant. We also note that the normal that points towards the origin in this case is n down and so we have F n down dσ (y,, z) (f, f y, 1) d dy (y,, z) ( 2, 2y, 1) d dy z d dy ( 2 + y 2 1) d dy where is the shadow of in the y plane and we have replaced z with f(, y) in the last step. This integral is best done in polar coordinates, where we will have F n down dσ 1 π/2 r0 θ0 1 π/2 r0 Calculating this integral will give us π 8. θ0 (r 2 1)r dθ dr r 3 r dθ dr 2. The curl of the gradient of a scalar valued function f is ( f) (f zy f yz )i + (f z f z )j + (f y f y ) 0i + 0j + 0k, so it is always zero. The divergence of the gradient of a scalar valued function f is ( f) f +f yy +f zz. This makes sense, and in fact it is the Laplacian of f, denoted 5

6 f or 2 f. Its value will depend on the specific function given, so it is not necessarily zero. The divergence of a vector field F is a scalar valued function, and the curl only operates on vector fields, not on functions, so the curl of the divergence ( F) doesn t make sense. The divergence of the curl of a vector field F Mi+Nj+P k is ( F) P y N z + M zy P y + N z M yz 0, so it is always zero. 3a. If the curve C is the boundary of the surface then F dr F n dσ C where the orientations of C and are related by the right hand rule. 3b. If the surface is the boundary of the region then F n out dσ F d dy dz where n out is the outward unit normal to. 3c. If the function f( + iy) u(, y) + v(, y)i is comple differentiable then the partial derivatives of u and v satisfy the relations u v y u y v 4. We can proceed in one of two ways. The first method is to use the rule that f ( + iy) u + v i. Here u(, y) 2 2 2y 2 and v(, y) 4y, so u 4 and v 4y. o f ( + iy) 4 + 4yi. The second method is to note that z 2 (+yi) 2 2 y 2 +2yi and therefore f(z) 2z 2. Then by the power rule f (z) 4z. (Note that 4z 4 + 4yi, so indeed both methods yield the same answer). 5a. Yes. We have u(, y) ln ( ( 2 + y 2 ) 1/2) ( y and v(, y) arctan. ) Then we have that 6

7 u u y v v y 2 + y 2 y 2 + y ( y ( y ) 2 ( y 2 ) y ) 2 ( 1 ) 1 + y2 2 + y 2 ( ) 2 + y 2 o u v y and u y v. o the Cauchy-iemann equations are satisfied and therefore f is comple differentiable. 5b. We use the relations r 2 2 +y 2 and y r sin θ tan θ (we can also work r cos θ out that y tan θ by drawing a picture of the comple number z + yi in the plane and using geometry). This gives us that ln ( ( ( 2 + y 2 ) 1/2) ln r y and arctan arctan(tan θ) θ. Therefore we have that ) 5c. We have that f(r cos θ + r sin θi) ln r + θi g n (z) e nf(z) n ln r+nθi e e ln rn +nθi e ln rn e nθi r n e nθi r n (cos(nθ) + sin(nθ)i) But we have as a consequence of Proposition 2 from handout C1 that z n r n (cos(nθ) + sin(nθ)i). Therefore g n (z) z n. 7