Integration over surface

Question

Let $S$ be a part of the cone $z = \sqrt{x^2+y^2} $ above circle $\ x^2+y^2 \leq 2x$. Let $f(x,y,z) = x^2+y^2 + z^2 $. Evaluate: $\iint_S f(x,y,z)dS$

I have evaluated it by converting it to spherical coordinates as $3\sqrt2\pi $. I want to confirm if my answer is correct or not.

Answer 1

I got the same result by using polar coordinates: $$\iint_S (x^2+y^2+z^2)dS=\sqrt{2}\iint_{(x-1)^2+y^2\leq 1}(x^2+y^2+(x^2+y^2))\,dxdy\\ =2\sqrt{2}\int_{-\pi/2}^{\pi/2}\left(\int_0^{2\cos(\theta)}\rho^3 d\rho\right)d\theta=3\sqrt{2}\pi.$$

Answer 2

First let's calculate the surface element of cone in terms of $dxdy$: $$ dS=\sqrt{\left( \frac{\partial z}{\partial x}\right)^2+\left( \frac{\partial z}{\partial y}\right)^2+1}\; dxdy =\sqrt{\frac{x^2}{x^2+y^2}+\frac{y^2}{x^2+y^2}+1}=\sqrt 2 \; dxdy$$ Next I'm going to use $\Omega$ for the disc $x^2+y^2 \leq2x \iff (x-1)^2+y^2 \leq1$.

Now I'll evaluate the integral: $$ I=\iint_\Omega x^2+y^2+z^2 \; dS= \iint_\Omega (x^2+y^2+x^2+y^2) \sqrt2 \; dxdy=2\sqrt2\iint_\Omega x^2+y^2 \; dxdy$$ We can use transformation to polar coordinates:

$$ x=r \cos\varphi+1 \\ y=r \sin\varphi \\ J=\det\left(\begin{matrix} \frac{\partial x}{\partial r}&\frac{\partial x}{\partial \varphi} \\ \frac{\partial y}{\partial r}&\frac{\partial y}{\partial \varphi} \end{matrix}\right)=r $$ So the integral transforms to:

$$\begin{align}I&=2\sqrt2 \int_0^1dr\int_0^{2\pi}d\varphi\;r[(r \cos\varphi+1)^2+(r \sin\varphi)^2]\\ &=2\sqrt2 \int_0^1dr\int_0^{2\pi}d\varphi\;(r^3\cos^2\varphi+2r^2\cos\varphi+r+r^3\sin^2\varphi)\\ &= 2\sqrt2 \int_0^1dr\int_0^{2\pi}d\varphi\;(r^3+2r^2\cos\varphi+r)\\ &=2\sqrt2 \left(\int_0^1(r^3+r)dr\int_0^{2\pi}d\varphi + 2\int_0^1r\,dr\int_0^{2\pi}\cos\varphi\;d\varphi\right)\\ &=2\sqrt2\left(\left[\frac{r^4}{4}+\frac{r^2}{2}\right]_0^1\left[\varphi\right]_0^{2\pi}+0\right)\\ &=2\sqrt2\left(\frac{1}{4}+\frac{1}{2}\right)2\pi\\ &=3\pi\sqrt2 \end{align}$$

So you have probably solved the problem correctly.