Finally we choose r 0 small enough that s r can be

This preview shows 4 out of 5 pages.

Info icon Subscribe to view the full document.

You've reached the end of this preview.

Unformatted text preview: Finally, we choose r > 0 small enough that S r can be completely enclosed by the ellipsoid S . Since the region between S r and S does not contain the origin, we can apply the Divergence Theorem, as above, and get integraldisplay S F · n dS = integraldisplay S r F · d S = 0. 8. Since the radiation field F satisfies the inverse square law, we can apply Gauss’ Law. (a) Since the ellipsoid contains the origin, we have integraldisplay ellipsoid F · n dS = 4 π by Gauss’ Law. LParen1 a RParen1 x y z LParen1 b RParen1 x y z (b) The piece of the sphere x 2 + y 2 + z 2 = 25 with z ≥ 3 and the disk x 2 + y 2 ≤ 16, z = 3, together form a closed surface S in R 3 . Since the origin is not con- tained within S , Gauss’ Law gives that the outward flux across S is 0; that is, integraldisplay S F · n dS = 0. Now integraldisplay S F · n dS = integraldisplay top of sphere F · n s dS + integraldisplay disk F · n d dS = 0 = ⇒ integraldisplay top of sphere F · n s dS =- integraldisplay disk F · n d dS . Since S is oriented by the outward nor- mal, both n s and n d are pointing out from the region enclosed by S . Hence integraldisplay top of sphere F · n s dS = integraldisplay disk F · (- n d ) dS = ⇒ flux across the top of the sphere = flux across the disk. (What entered across the disk exited across the top of the sphere – there is no net gain or loss.) MATB42H Solutions # 9 page 5 9. (a) ω = yx 2 dx + xz dy + y 2 dz . dω = d ( yx 2 ) ∧ dx + d ( xz ) ∧ dy + d ( y 2 ) ∧ dz = a24 a24 a24 a24 a24 a24 a24 a58 =0 2 yxdx ∧ dx + x 2 dy ∧ dx + z dx ∧ dy + xdz ∧ dy + 2 y dy ∧ dz = ( z- x 2 ) dxdy + (2 y- x ) dy dz . η = sin z dxdy + x 2 yz dy dz + xz 3 dz dx . dη = d (sin z ) ∧ dxdy + d ( x 2 yz ) ∧ dy dz + d ( xz 3 ) ∧ dz dx = cos z dz ∧ dxdy + 2 xyz dx ∧ dy dz + a24 a24 a24 a24 a24 a24 a24 a24 a58 = 0 x 2 z dy ∧ dy dz + a24 a24 a24 a24 a24 a24 a24 a24a58 =0 x 2 y dz ∧ dy dz + a24 a24 a24 a24 a24 a24 a24a24 a58 =0 z 3 dx ∧ dz dx + a24 a24 a24 a24 a24 a24 a24 a24a24 a58 = 0 3 xz 2 dz ∧ dz dx = (2 xyz + cos z ) dxdy dz . (b) (i) ω ∧ η = ( yx 2 dx + xz dy + y 2 dz ) ∧ (sin z dxdy + x 2 yz dy dz + xz 3 dz dx ) = a24 a24 a24 a24 a24 a24 a24 a24 a24 a24 a24 a58 =0 yx 2 sin z dx ∧ dxdy + x 4 y 2 z dx ∧ dy dz + a24 a24 a24 a24 a24 a24 a24 a24 a24 a58 = 0 x 3 yz 3 dx ∧ dz dx + a24 a24 a24 a24 a24 a24 a24 a24 a24 a24 a58 = 0 xy sin z dy ∧ dxdy + a24 a24 a24 a24 a24 a24 a24 a24 a24 a58 =0 x 3 y 2 z dy ∧ dy dz + x 2 yz 3 dy ∧ dz dxy 2 sin z dz ∧ dxdy + a24 a24 a24 a24 a24 a24 a24 a24 a24 a58 =0 x 2 y 3 z dz ∧ dy dz + a24 a24 a24 a24 a24 a24 a24 a24 a24 a58 = 0 xy 2 z 3 dz ∧ dz dx = x 4 y 2 z dxdy dz + x 2 yz 3 dy dz dx + y 2 sin z dz dxdy = ( x 4 y 2 z +(- 1) 2 x 2 yz 3 +(- 1) 2 y 2 sin z ) dxdy dz = ( x 4 y 2 z + x 2 yz 3 + y 2 sin z ) dxdy dz . (ii) ω ∧ dω = ( yx 2 dx + xz dy + y 2 dz ) ∧ (( z- x 2 ) dxdy + (2 y- x ) dy dz ) = a24 a24 a24 a24 a24 a24 a24 a24 a24 a24 a24 a24a24 a58 = 0 yx 2 ( z- x 2 ) dx ∧ dxdy + yx 2 (2 y- x ) dx ∧ dy dz + a24 a24 a24 a24 a24 a24 a24 a24 a24 a24 a24 a24 a58 =0 xy ( z- x 2 ) dy ∧ dxdy + a24 a24 a24 a24 a24 a24 a24 a24 a24 a24 a24 a24a58 = 0 xy (2 y- x ) dy ∧ dy dz + y 2 ( z- x 2 ) dz ∧ dxdy + a24 a24 a24 a24 a24 a24 a24 a24 a24 a24 a24 a24 a58 =0 y 2 (2 y- x ) dz ∧ dy dz = yx 2 (2 y- x ) dxdy dz + y 2 ( z- x 2 ) dz dxdy = ( yx 2 (2 y- x ) + (- 1) 2 y 2 ( z- x 2 ) ) dxdy dz = ( yx 2 (2 y- x ) + y 2 ( z- x 2 ) ) dxdy dz ....
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern