ESE 318-02, Spring 2016 Homework Set #13 Due Tuesday, Apr. 26 Don’t use Wolfram-Alpha or computer to do your integrals. You may use a Table of integrals, but you shouldn’t need any other than the inside front cover of your text, and the hints below. 1. Zill 9.12.6. 2. Zill 9.12.13 3. Zill 9.12.15 and 9.12.5. (Do 9.12.15 first. Answer in back of book has typo.) 4. Zill 9.12.30. 5. Find the flux of F over the given surface, S. 881034,,:0,5,2  vuvuvurSyxF 6. Find the flux of F over the given surface, S.  vuuvuvurSyzxyzxF 030,sin,cos:,, 7. Find the flux of F over the given surface, S.  vuuvuvurSzyxF 40,sin,cos:,,2 8. Zill 9.13.38. (Hint: note that the surfaces are flat. This is more of a thinking problem than a calculating one. Don’t just start evaluating integrals.) 9. In class I developed these two expressions. Differential Surface Area: dudvrrdSvu Flux Integral (F a vector field):    RvuRSdudvrrvuFNdudvFndSF , I also showed that for the case where the surface can be parameterized as z = f(x,y), dS and the flux integral can be expressed as follows: Differential Surface Area: dxdyffdSyx221  Flux Integral: RSdxdyRyzQxzPndSF (See also Zill problem 9.13.43.) (a) Derive similar formulas for a surface parameterized as y = g(x,z). (Orient N so that it is positive in the y direction.) (b) Derive similar formulas for a surface parameterized as x = h(y,z). (Orient N so that it is positive in the x direction.)ESE 318-02, Spring 2016 In the remaining problems, the surface can be parameterized as z = f(x,y). You may use whatever formula you prefer to calculate the flux integrals. 10. Zill 9.13.29. 11. Zill 9.13.31. The region of integration is the circle in the xy plane. Hint: polar coordinates. 12. Zill 9.13.34. Hint: Ceuduueuu)1(. 13. Zill 9.13.36. The total outward flux is the sum of the outward flux. For the bottom piece outward is downward. General tip on surface integrals: Take your time understanding the surface and region of integration. Draw the region of integration, which is always in the uv or xy plane for these problems. (Some 3D pictures are difficult to draw/picture; what’s important is the region in the uv or xy plane.) 1. Zill 9.12.6.    42222yxyregionboundsCdyyxdxyxCQP            525654515512255142224322422246413222322168416164242422 xxxxdxxxxdxyxydxdyyxyxPQxyxyxESE 318-02, Spring 2016 2. Zill 9.12.13 It will be easier to integrate over x first.  2223311,,0.,1: yxyxyquadstCdyxyxydxyC       8181412/104212212/1032/10322/1012/101222220212/11212222 yydyyydyyyydyxyydxdyydAyyyyyyyyPQyxyxyyRCyx 3. Zill 9.12.15 and 9.12.5. (Do 9.12.15 first. Answer in back of book has typo.) 9.12.15:      RofareaAAabdAabdAabbxdyaydxRRC 9.12.5 Now use result of 9.12.15, seeing that R is a circle of radius 5.   7553325522AAxdyydxC 4. Zill 9.12.30. See Figure 9.12.14 in Zill. Radius goes from 1 to 2. yxQxyPyjxixyF2222        215116cos4sincossincos4422212142/02212132/02/021 rdrrdrdrdrrdrFxyxyxyPQCyxESE 318-02, Spring 2016 5. Find the flux of F over the given surface, S. 881034,,:0,5,2  vuvuvurSyxF      6488441541541581581,3,40,5,21,3,43104013,1,04,0,10,5,288221588881028810vvdvvdvuvududvvuFluxvuvuNFkjirrNrrvuFuvuvu 6. Find the flux of F over the given surface, S.  vuuvuvurSyzxyzxF 030,sin,cos:,,                      18290coscos3cossinsinsinsincoscossinsincoscossinsincoscossin1sincossincoscos1,sin,cossin1,cossin,1cos,sin,cossin1,cossin,1cos,sin,cos0cossin1sincos0,cos,sin1,sin,cossin1,cossin,1cos3310221303310302030222222vvvudvvvvvduududvvvvvuFluxvvvvuvvvvvvuvvvvvvuuvuvuvuvvuvuNFuvuvuvuvuvvkjirrNvuvurvvrvuvvuvuFvuvuESE 318-02, Spring 2016 7. Find the flux of F over the given surface, S.  vuuvuvurSzyxF 40,sin,cos:,,2    12824sin2cos2,sin2,cos2,sin,cos,sin2,cos20cossin2sincos0,cos,sin2,sin,cos,sin,cos340441403403332323222222vudvduududvuFluxuuvuvuuvuvuuvuvuNFuvuvuvuvuuvvkjirrNvuvuruvvruvuvuFvuvu 8. Zill 9.13.38. (Hint: note that the surfaces are flat. This is more of a thinking problem than a calculating one. Don’t just start evaluating integrals.) First, the total flux is the sum of the 6 fluxes out of each face. Secondly, use the “trick” for when the flux is constant over each surface. Use the original definition of the flux integral for each side. Looking at Figure 9.13.14 of Zill, see that n1 = k, n2 = -i, n3 = j, etc. For n1, z = 1. For n2, x = 0. For n3, y = 1. Etc.     .00122221111etcdSxdSdSizkyjxindSFAdSzdSkdSzkyjxindSFzkyjxiFSSSSxySSSS See that the flux out of each face is the area of a side (always one) times the value of the x, y or z coordinate of that side (either zero or one). So only three of the six sides contribute, and the answer is 3. (We will soon have an easier way to do