Question

Evaluate the surface integral s f · ds for the given vector field f and the oriented surface s. in other words, find the flux of f across s. for closed surfaces, use the positive (outward orientation. f(x, y, z = x i ? z j y k s is the part of the sphere x2 y2 z2 = 16 in the first octant, with orientation toward the origin

Answer 1

I'm reading the vector field as

$\mathbf F(x,y,z)=x\,\mathbf i-z\,\mathbf j+y\,\mathbf k$

The surface $S$ can be parameterized by

$\mathbf s(\theta,\varphi)=4\cos\theta\sin\varphi\,\mathbf i+4\sin\theta\sin\varphi\,\mathbf j+4\cos\varphi\,\mathbf k$

with $0\le\theta\le\dfrac\pi2$ and $0\le\varphi\le\dfrac\pi2$.

Then the surface integral is given by

$\displaystyle\iint_S(\mathbf F\cdot\mathbf n)\,\mathrm dS=\iint_S\left(\mathbf F(\mathbf s(\theta,\varphi))\cdot(\mathbf s_\theta\times\mathbf s_\varphi)\right)\,\mathrm dS$
$\displaystyle-64\int_{\theta=0}^{\theta=\pi/2}\int_{\varphi=0}^{\varphi=\pi/2}(2\cos\varphi\sin^2\varphi\sin\theta+\cos^2\theta\sin^3\varphi))\,\mathrm d\varphi\,\mathrm d\theta=-\dfrac{32(\pi+4)}3$