I have a problem that I can't seem to figure out. The problem is as follows:
Let $D$ be the part of the cone $z = \sqrt{x^2+y^2}$ that satisfies $x^2 +y^2 \leq 2ax$. Calculate $\iint_{D}^{}(z^2-2ax)dS$.
What I've tried is the following:
$dS = \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2}dxdy = \sqrt{2}dxdy$
Then I tried to parametrize the cone using:
$x = r\cos\theta$, $y = r\sin\theta$, $z = r$ with $ 0 \leq \theta \leq 2\pi$ and $0 \leq r \leq \sqrt{2ax}$
Now filling this in the previous integral would give me:
$\int_{0}^{2\pi}\int_{0}^{\sqrt{2ar\cos\theta}}(\sqrt{2}r^2-2ar\cos\theta)rdrd\theta$
or
$\sqrt{2}\int_{0}^{2\pi}\int_{0}^{\sqrt{2ar\cos\theta}}(r^3-2ar^2\cos\theta)drd\theta$
But if I integrate this with respect to $r$ I get a pretty large expression to integrate over $\theta$, which I'm pretty sure is not intended for this problem.
The only things I think are right in this entire problem are $dS = \sqrt{2}dxdy$ and the parametrization, but everything else feels like I'm doing it wrong. Can someone give me a hint on what I'm doing wrong or tell me how I should tackle this problem?
Best Answer
The region
$$x^2+y^2 \leq 2ax \iff (x-a)^2+y^2\le a^2$$
is a circle with radius $a$ centered at $(a,0)$.
For the surface we can use the parametrization
$$ r(t,\theta) =(t\cos \theta, t\sin \theta, t) \implies dS =|r_\theta\times r_t| \;dt\;d\theta= \sqrt2t\;dt\;d\theta $$
therefore we have
$$\iint_{D}^{}(z^2-2ax)dS=\sqrt 2\int_{-\frac \pi 2}^{\frac \pi 2} \int_0^{2a\cos \theta}t^3-2at^2\cos \theta \;dt \;d\theta =-\frac{\sqrt 2}{2}\pi a^4$$