# Surface integral over parabolic cylinder that lies inside another cylinder

To be precise, I'm given the following:

Find $\iint_K {xdS}$ over the part of parabolic cylinder $z = \frac{{{x^2}}}{2}$ that lies inside the first octant part of the cylinder $x^2+y^2=1$.

In my attempt, I tried to parametrize as following: $x=r\cos(\theta),y=r\sin(\theta),z=r^2\cos^2(\theta)/2$. The surface element turned out to be $dS=r\sqrt{r^2\cos^2(\theta)+1}drd\theta$, which means that considering the substitution, the integral (now, a double integral) is:

$$\iint\limits_D {{r^2}\cos (\theta )\sqrt {{r^2}{{\cos }^2}(\theta ) + 1} drd\theta },\qquad D = \{ 0 \leqslant r \leqslant 1,0 \leqslant \theta \leqslant \pi /2\}$$

Numerically, this turns out to be $\pi/8$, but I don't think it is clear how this can be solved analytically. Is there any way to solve these kinds of double integrals where the variables are bounded in such a way, or should a different parametrization be used instead?

## 1 Answer

It is easier using the $xy$ coordinates:

$$\int^1_0 x \sqrt{1+x^2} \int^{\sqrt{1-x^2}}_0 dy dx\\ =\int^1_0 x \sqrt{1-x^4}dx$$

Use substitution $u=x^2$, you can get

$$\frac{1}{2}\int^1_0 \sqrt{1-u^2} du$$

Now making use the fact that this is half of the area of a quarter of a unit disk, it gives you $\pi/8$.