I am stuck on the following question about finding the flux:
Find the flux of $\vec F= (x,y,z)$ upward through the part of the surface $z=a-x^2-y^2$ lying above the plane $z=b<a$
My attempt:
Assumptions
- $z=a-x^2-y^2$ is an infinite paraboloid?
- $z = b < a$ is a plane less than $a$? does this make sense?
Since it is difficult to visualize this problem, I tried to go the "hard way", i.e. by computing the surface element without any consideration of the geometrical properties. i.e.:
$$\left\{ \begin{gathered} x = r\cos \theta \hfill \\ y = r\sin \theta \hfill \\ z = a - r^2 \hfill \\ \end{gathered} \right.$$ Thus,
$$d\vec S = \widehat NdS = \pm \frac{{\partial (x,y,z)}}{{\partial (r,\theta )}} = (0,0,r)$$ Now, by taking the field $\vec F$ times the surface element, we get: $$(x,y,z)*(0,0,r) = z*r = (a - 1)r$$ Thus, $$\iint\limits_D {(ar - r)drd\theta }$$
The difficulty I am having is in my assumption list. I am unable to find the boundaries. I think we should look at the intersection between the surfaace and the plane, although I am not sure.
Edit: The answer is $\pi(3a^2-4ab+b^2)/2$ (according to the key).
Let's use spherical coordinates to evaluate the surface integral. Note that $\vec F=\hat xx+\hat yy+\hat zz=\hat rr$ in spherical coordinates and that the outer unit normal of the sphere is $\hat n = \hat r$. Thus, the flux is
$$\text{flux}=\int_S \vec F \cdot \hat n dS=\int_0^{2\pi} \int_0^{\theta_0}(\hat ra)\cdot \hat r a^2\sin \theta d\theta d\phi$$
where $\theta_0 = \arccos (b/a)$. Thus,
$$\text{flux}=2\pi a^3(1-b/a)$$