I have 2 questions on stokes and divergence theorem each. I think I have done both and I just want to make sure that I did them correctly.
Question 1
Let $C$ be the boundary of the surface $S={(x,y,z):z=4-x^2-y^2,x^2+y^2\le4, x,y\ge0}$ with orientation related by the right-hand rule to the upward orientation of $S$. For $E(x,y,z)=[3yz,zx,2xy]^T$, apply Stoke's Theorem to calculate the closed line integral $\int E\cdot Tds$
My Work
So by Stokes' theorem I know $\int E\cdot Tds = \iint(\nabla \times E)\cdot dS$ where $dS=[-\frac{\partial z}{\partial x}, -\frac{\partial z}{\partial y}, 1]$. So $\nabla \times E=[2x,0,-2x]$ and $dS=[-2x,-2y,1]$ which leads to the double integral $\iint -4x^2-2x \mathrm{d}x\mathrm{d}y$ but we can change this into polar coordinates. So we have $\int_0^{2\pi}\int_0^2 (-4(r\cos\theta)^2-2(r\cos\theta))rdrd\theta$ which equals $-\frac{32\pi}{3}$.
Question 2
Apply the Divergence Theorem to calculate the normal surface integral of $E$ over $S$, where $S$ is the sphere of radius 3 centered at $(0,0,0)$ oriented outward and $E(x,y,z)=[x^3+yz,y^3+zx,z^3+xy]^T$
My Work
Divergence theorem says $\iint E\cdot \vec ndr=\iiint(\nabla \cdot E)dxdydz.$ So the $\nabla \cdot E = 3(x^2+y^2+z^2)$ which screams spherical coordinates. So my triple integral is now $\int_0^3\int_0^{2\pi}\int_0^\pi (3p^2)p^2\sin(\phi)d\phi d\theta dp= \frac{2916\pi}{5}$
Are both of those correct?