$$\vec{\phi}(x,y,z)=\biggl(\frac{\ln(x^5y^3z^2+1)}{z^7+y^4+5}-yx^2e^{2z}, \ \ \cos^2(\pi z)xy^2, \ \ e^{x^2y^2}+\cos z\sin z^3 \biggr)$$
Then let $S$ be the upper-hemisphere $x^2+y^2+z^2=4$ with outward $\vec{n}$. Using Stokes' Theorem, find $$\iint_S(\nabla \ \times \ \vec{\phi})\cdot d\vec{S}$$
I asked this question in another thread, but I want to know another way of solving this.
Apparently we can transform this integral to:
$$\int_0^{2\pi}\int_0^2r^3drd\theta$$
But, I don't the know the process to get there.
There is a corollary to Stokes's theorem that allows us to change the surface to a more convenient one, if it shares the same boundary as the given one.
Let $D$ be the disk of radius $2$ in the $xy$-plane, thought of as a surface in space. Then $\partial S$ and $\partial D$ are the same curve $C$, the circle of radius $2$ in the $xy$-plane. By Stokes' Theorem twice, $$ \iint_S (\nabla \times \vec \phi)\cdot d\vec S =\oint_C \vec\phi\cdot d\vec s = \iint_D (\nabla \times \vec \phi)\cdot d\vec S $$ On $D$, $z=0$, so the vector field simplifies to $\vec\phi = \left(-yx^2,xy^2,e^{x^2y^2}\right)$. And the normal vector on $D$ is just $\hat k$. When dotting with $\hat k$, all components of $\nabla\times\vec\phi$ are killed except the last one. So $$ \iint_D (\nabla \times \vec \phi)\cdot d\vec S =\iint_D \left(\frac{\partial}{\partial x}(xy^2) - \frac{\partial}{\partial y} (-yx^2)\right)\,dA =\iint_D(y^2 + x^2)\,dA $$ Now switch to polar coordinates on $D$: $$ \iint_D(y^2 + x^2)\,dA = \int_0^{2\pi} \int_0^2 r^2\,r\,dr\,d\theta $$ This is how we get to your “apparent” expression. And as for that: $$ \int_0^{2\pi} \int_0^2 r^3\,dr\,d\theta = \int_0^{2\pi} 1\, d\theta \int_0^2 r^3\,dr = (2\pi) \left(\frac{2^4}{4}\right) = 8\pi $$