Let $S$ be the surface formed by intersection of the cone $ z = \sqrt{x^2 + y^2}$ and the cylinder $x^2 + y^2 = 2x$. Calculate the following integral over the surface $S$.
$$\iint_S x^2y^2+ y^2z^2+ x^2z^2\,\mathrm dS$$
My attempt: The easiest way to parametrize it for me was cylinderical coordinate.The the conditions boils down to $ r \le 2\cos(\theta)$, $z = r$.
Now we have the parametrization $R(t) = (r\cos(\theta), r\sin(\theta), r)$ so the surface element $\mathrm dS$ will be equal to
$$\left|\frac{\partial R}{\partial r} \times \frac{\partial R}{\partial \theta}\right| = \sqrt{2} r,$$
since only for $\theta \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ the value of $\cos(\theta)$ is non-negative we have to compute
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \int_0^{2\cos(\theta)} \sqrt{2} r^5( 1 + \cos^2(\theta)\sin^2(\theta))\,\mathrm dr\right) \mathrm d\theta,$$
which isn't much complicated but to ugly to compute. Does anyone know a better methond resulting in a simpler integral?
Thanks in advance!
Best Answer
Here is a trick.
The begining is straight forward:
$\sqrt 2\int_{-\frac {\pi}{2}}^{\frac {\pi}{2}}\int_0^{2\cos\theta} r^5(1+\sin^2\theta\cos^2\theta)\ dr\ d\theta\\ \sqrt 2\int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} \frac {1}{6} (2\cos\theta)^6(1+\sin^2\theta\cos^2\theta)\ d\theta\\ \frac {32\sqrt 2}{3}\int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} \cos^6\theta + \sin^2\theta\cos^8\theta\ d\theta\\ \frac {32\sqrt 2}{3}\int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} \cos^6\theta + \cos^8\theta-\cos^{10}\theta\ d\theta\\ $
Now we have two roads we can go by. We can say $\cos^2\theta = \frac 12(1 + \cos 2\theta)$ and keep reducing the order and simplifying. This is what they taught you in Calc II.
The tricky way:
$\cos\theta = \frac {e^{i\theta} + e^{-i\theta}}{2}\\ \cos^n\theta = \frac {(e^{i\theta} + e^{-i\theta})^n}{2^n} = \frac {e^{in\theta} + {n\choose 1}e^{i(n-2)\theta} + {n\choose 2}e^{i(n-4)\theta} + \cdots +{n\choose n} e^{-in\theta}}{2^n}$
We can then pair $e^{in\theta}$ terms with $e^{-in\theta}$ terms to turn back into $\cos n\theta$ terms. However, it is not really necessary as $\int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} \cos 2n\theta \ d\theta = 0$
This means we just need the constant terms.
$\frac {32\sqrt 2}{3}\int_{-\frac {\pi}{2}}^{\frac {\pi}{2}} \frac {1}{2^6} {6\choose 3} + \frac {1}{2^8}{8\choose 4} - \frac {1}{2^{10}}{10\choose 5} d\theta$