I was being given a problem:
"Find the vector field flux $\Phi$ of the vector field $\vec{F}=2x \vec{i} - y\vec{j} + z \vec{k}$ through the surface of $S: x^2+y^2+z^2 \leq 4, \ 3z \leq x^2 +y^2$".
That's what I did:
- We know that $\Phi = \int_{S} (\vec{F}, \vec{n})dS = \underset{S}{\int\int\int} div\vec{F} dxdydz$. Also it's easy to see, that in this particular case $div\vec{F}=2$.
- We're left with an integral $2\underset{S}{\int\int\int} dxdydz$. So at this point, for finding an answer, it suffices to find the volume of $S$.
- Moving to spherical coordinate system: \begin{cases} x=r\cos\phi \cos \psi \\ y=r\sin\phi \cos \psi \\ z=r\sin\psi \end{cases} we get (by putting this substitution to the $(x,y,z)$ form of the surface, we obtain that $r\leq 2$ and $r\geq \dfrac{3\sin \psi}{\cos^2 \psi}$, also since we have an idea of how the surface looks like, which is an intersection of a sphere with paraboloid, so we have $\phi $ from $0$ to $2\pi$ and $\psi$ from $0$ to $\pi/4$): \begin{equation} \int_{0}^{2\pi} d\phi \int_{0}^{\pi/4} d\psi \int_{\frac{3sin\psi}{\cos^2\psi}}^{2} r^2 \sin \psi dr \end{equation}
- We evaluate the integral: $\int_{0}^{2\pi} d\phi \int_{0}^{\pi/4} d\psi \int_{\frac{3sin\psi}{\cos^2\psi}}^{2} r^2 \sin \psi dr =[Wolfram]=$ non-correct answer.
4'. We also know that the figure we have is a result of spinning some curve around $z$ axis. So I tried to find the volume like so as well (we let $x=0$): $\pi \int_{0}^{1} (\sqrt{4-y^2})^2 - (\dfrac{y^2}{3})^2 dy = [Wolfram] =$ non-correct answer.
The correct answer, according to the book, is $15 \pi$.
Can anyone point to me where I'm making a mistake, please?