Question:
Evaluate the surface integral $\int_SF\cdot dS$ where $F=\langle3x,-z,y \rangle$ and $S$ is the part of the sphere $x^2+y^2+z^2 = 4$ in the first octant, with orientation towards the origin.
We thus parametrize our surface as:
$r(u,v)=\langle 4\cos(u)\sin(v), 4\sin(u)\sin(v), 4\cos(v)\rangle$
I am going to skip the details of how to calculate the normal because for a sphere it always follows the same pattern, the normal is:
$\vec n = (-16\cos(u)\sin^2(v), -16\sin(u)\sin^2(v), -16\sin(v)\cos(v))$
We can see this is the origin pointing normal, just as requested and now:
$F\cdot\vec n = -192\cos^2(u)\sin^3(v)$ (verified with wolfram)
Thus our integral becomes:
$\int^{\pi/2}_0 \int^{\pi/2}_0-192\cos^2(u)\sin^3(v) du dv = -32\pi$
All I know is that this solution is wrong, and I am not sure where the mistake is. I am following the theorem:
$\int\int_SF\cdot dS = \int\int_D F(\phi(u,v))\cdot\vec n(u,v) du dv$
Best Answer
For $\iint_S\langle3x,-z,y\rangle\cdot d\vec{n}$, with orientation toward the origin, I get $$\begin{align} &\iint_S\langle3x,-z,y\rangle\cdot d\vec{n}\\ &=\int_0^{\pi/2}\int_0^{\pi/2}\langle12\cos(u)\sin(v),-4\cos(v),4\sin(u)\sin(v)\rangle\cdot-\langle\cos(u)\sin(v),\sin(u)\sin(v),\cos(v)\rangle\,16\sin(v)\,du\,dv\\ &=-192\int_0^{\pi/2}\int_0^{\pi/2}\cos^2(u)\sin^3(v)\,du\,dv\\ &=-192\int_0^{\pi/2}\cos^2(u)\,du\int_0^{\pi/2}\sin^3(v)\,dv\\ &=-192\cdot\frac{\pi}{4}\cdot\frac{2}{3}\\ &=-32\pi \end{align}$$
which matches your answer. Sometimes there is a bug in the problem code for these software platforms. Is this MyMathLab? WeBWorK? WebAssign? Something else?