F Tds. You can do this either by evaluating the integral directly of by using the circulation form of Green s Theorem.

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1 MATH 223 (alculus III) - Take Home Quiz 6 olutions April 22, 25. F. Ellermeyer Name Instructions. This is a take home quiz. It is due to be handed in to me on Wednesday, April 29 at class time. You may work on this quiz alone or in a group of one or two other people. (If you work in a group, then you will hand in only one paper and all in the group will receive the same grade on the quiz.) You may use any books or other resources that you need to do the quiz with the exception of consulting other people. Your solutions must include sufficient detail so that I am able to understand your reasoning process in solving the problems. The paper that you hand in should be written neatly and use correct mathematical notation and writing. Writing, notation and neatness will taken into account in my grading of the quiz. All papers must be stapled together (not paper clipped or folded). Points will be deducted for no staple.. (worth 2 points) Let be the curve y x 2 and let 2 be the curve y x (pictured below) and let 2 with the counterclockwise orientation as shown by the arrows in the picture. Let F be the vector field Fx,y x 3 x 2 yi xy 2 j. Find the value of the flow (circulation) integral F Tds. You can do this either by evaluating the integral directly of by using the circulation form of Green s Theorem.

2 olution: We will use Green s Theorem which says that F Tds N x M y da. We see that N x M y da x 2x y 2 x 2 dydx (worth 2 points) Let be the curve y x 2 and let 2 be the curve y x (pictured below) and let 2 with the counterclockwise orientation as shown by the arrows in the picture. Let F be the vector field Fx,y x 3 x 2 yi xy 2 j. Find the value of the flux integral F nds. You can do this either by evaluating the integral directly of by using the flux form of Green s Theorem.

3 olution: We will use Green s Theorem which says that F nds M x N y da. We see that M x N y da x 2x 3x 2 dydx (worth 2 points) Below is a picture of the surface : z 2xy x,y unit disk. (The surface looks something like a horse s saddle or perhaps like an exaggerated Pringle s potato chip.)

4 Find the surface area of this surface. olution: We parameterize the surface as x rcost y rsint z 2r 2 sintcost r 2 sin2t t 2 r. The surface area is Note that 2 d rr r t dr. rr,t rcosti rsintj r 2 sin2t k r r costi sintj 2rsin2tk Thus and r t rsinti rcostj 2r 2 cos2tk. r r r t 2r 2 sintcos2t 2r 2 costsin2ti 2r 2 cos2tcost 2r 2 sin2tsintj rk 2r 2 sint 2ti 2r 2 cos2t tj rk 2r 2 sinti 2r 2 costj rk

5 r r r t 4r 4 sin 2 t 4r 4 cos 2 t r 2 4r 4 r 2 r 4r 2. The surface area is thus r 4r 2 dr 5.33 square units Below is a picture of the surface : z 2xy x,y unit disk which has boundary curve : x cost y sint z 2sintcost sin2t t 2. Let n nx,y,z be the upward pointing normal vector field on the surface and let F be the vector field Fx,y,z xi yzj 3xk. You should do both parts a and b below (for practice) but you only need to hand in either part a or part b (not both). tokes Theorem says that you should get the same answer in part a and part b. a. (worth 2 points) Evaluate curlf nd.

6 b. (worth 2 points) Evaluate F Tds. olution to part a: First note that curlf yi 3j. Also, the equation for the surface can be written as z 2xy and thus is a level surface of the function fx,y,z z 2xy. Note that f 2yi 2xj k and this does point in the upward direction with respect to the oriented surface. Also f z and thus nd f da 2yi 2xj kda. f z We now see that curlf nd yi 3j 2yi 2xj kda 2y 2 6xdA 2y 2 3xdA. Thus curlf nd 2y 2 3xdA where is the unit disk. ue to the nature of this problem, we are probably better off converting to polar coordinates. This gives curlf nd 2 2 r 2 sin 2 3rcosrdrd. The inner integral is r 3 sin 2 3r 2 cosdr 4 sin2 cos and thus curlf nd 2 2 olution to part a: Using the parameterization for : x cost y sint z sin2t t 2. we have 4 sin2 cos d 2.

7 Thus F Tds 2 dx sint dy cost dz 2cos2t. x dx yz dy 3x dz Before proceeding, let us simplify the integrand: x dx yz dy 3x dz cost sint sintsin2t cost 6costcos2t sintcost sintsin2tcost sintcost 6costcos2t Thus we have 2 sin2t 2 sin2 2t sin2t 6costcos2t 2 sin2t 2 sin2 2t 6costcos2t sin2t 4 cos4t 6cost 2sin2 t sin2t 4 4 cos2t 6cost 2sin2 tcost F Tds 2 sin2t 4 4 cos2t 6cost 2sin2 tcost 2. As expected from tokes Theorem, curlf nd F Tds..