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 = 25 in the first octant, with orientation toward the origin

Answer 1

Parameterize $S$ by

$\vec s(u,v)=5\cos u\sin v\,\vec\imath+5\sin u\sin v\,\vec\jmath+5\cos v\,\vec k$

with $0\le u\le\dfrac\pi2$ and $0\le v\le\dfrac\pi2$.

Take the normal vector to $S$ to be

$\vec s_u\times\vec s_v=-25\cos u\sin^2v\,\vec\imath-25\sin u\sin^2v\,\vec\jmath-25\cos v\sin v\,\vec k$

Then the integral of $\vec F$ across $S$ is

$\displaystyle\iint_S\vec F\cdot\mathrm d\vec S=\int_0^{\pi/2}\int_0^{\pi/2}\vec F(x(u,v),y(u,v),z(u,v))\cdot(\vec s_u\times\vec s_v)\,\mathrm du\,\mathrm dv$

$\displaystyle=-125\int_0^{\pi/2}\int_0^{\pi/2}(\cos u\sin v\,\vec\imath-\cos v\,\vec\jmath+\sin u\sin v\,\vec k)\cdot(\cos u\sin^2v\,\vec\imath+\sin u\sin^2v\,\vec\jmath+\cos v\sin v\,\vec k)\,\mathrm du\,\mathrm dv$

$=\displaystyle-125\int_0^{\pi/2}\int_0^{\pi/2}\cos^2u\sin^3v\,\mathrm du\,\mathrm dv$

$=\displaystyle-125\left(\int_0^{\pi/2}\cos^2u\,\mathrm du\right)\left(\int_0^{\pi/2}\sin^3v\,\mathrm dv\right)=\boxed{-\frac{125\pi}6}$