Question

use the divergence theorem to compute integral integral f * ds when T is the unit sphere x^2+y^2+z^2=1 and f(x,y,z)=

Answer 1

\[divergence of F, \div F=\delta(z)/\delta x +\delta (y)/\delta y +\delta(x)/\delta z =1\]

Answer 2

thats you divergence of F, div F=1

Answer 3

\[\int\limits_{}^{} \int\limits_{}^{} F *ds=\int\limits_{}^{} \int\limits_{}^{} \int\limits_{}^{} di v F=\int\limits_{}^{} \int\limits_{}^{} \int\limits_{}^{} 1 dV\]

Answer 4

\[\int\limits_{}^{}\int\limits_{}^{}\int\limits_{}^{}1dV=\int\limits_{}^{}\int\limits_{E}^{}\int\limits_{}^{}f(x,y,z)dV\]

Answer 5

and \[\int\limits_{}^{}\int\limits_{E}^{}\int\limits_{}^{}f(x,y,z)dV=\int\limits_{}^{}\int\limits_{D}^{}[\int\limits_{u1(x,y)}^{u2(x,y)}f(x,y,z)dz]dA\]

Answer 6

\[ or that will be \int\limits\limits_{a}^{b}\int\limits\limits_{g1(x)}^{g2(x)}\int\limits\limits_{u1(x,y)}^{u2(x,y)}dzdydx\] \]

Answer 7

the answer is =\[4\pi/3\]

Answer 8

that is also finding the flux of the vector field F(x,y,z)=Zi+Yj+Xk across the unit sphere x^2+y^2+z^2=1

Answer 9

use the parametric \[r(\phi,\theta)=\sin \phi \cos \Theta i +\sin \phi \sin \Theta j +\cos \phi k\]

Answer 10

\[F(r(\phi,\Theta))=\cos \phi i +\sin \phi \sin \Theta j +\sin \phi \cos \Theta k\]

Answer 11

\[r _{\phi} x r _{\Theta}=\sin^2 \phi \cos \Theta i +\sin^2 \phi \sin \Theta j+\sin \phi \cos \phi k\]

Answer 12

therefore
\[F(r(\phi,\Theta))*(r _{\phi }xr _{\Theta} )=\cos \phi \sin^2 \phi \cos \phi +\sin^2 \phi \sin^2 \Theta + \sin^2 \phi \cos \phi \cos \Theta\]

Answer 13

using the parametric surface formula
\[\int\limits_{}^{}\int\limits_{\S}^{}F*ds=\int\limits_{}^{}\int\limits_{D}^{}F*(r _{\phi}x r _{\Theta})dA\]

Answer 14

\[=\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi} (2 \sin^2 \phi \cos \phi \cos \Theta +\sin^2 \phi \sin^2 \Theta)d \phi d \Theta\]

Answer 15

\[\[=2\int\limits\limits_{0}^{\pi} \sin^2 \phi \cos \phi d \phi \int\limits\limits_{0}^{2\pi} \cos \Theta d \Theta +\int\limits\limits_{0}^{\pi}\sin^3 \phi d \phi \int\limits\limits_{0}^{2\pi}\sin^2 \Theta d \Theta\]\]

Answer 16

since\[\int\limits_{0}^{2\pi}\cos \Theta d \Theta =0\] therefore

Answer 17

\[=0+\int\limits_{0}^{\pi}\sin ^3\phi d \phi \int\limits_{0}^{2\pi} \sin^2 d \Theta=4\pi/3 \] ans..........