Integral over the surface of a paraboloid

Question

The integral I'm trying to solve is the following: $$ \iint_S xyz \,d\sigma $$ Where $z=x^2+y^2$ and $0<z<1$. So I transform the integral to a double integral: $$\iint_Sxy(x^2+y^2)\sqrt{1+4(x^2+y^2)}\,dx\,dy$$ After that I transform it to polar coordinates with jacobian $J=\rho$ and get the following: $$\int_{0}^{1}\int_{0}^{2\pi}\rho^5\cos(\phi)\sin(\phi)\sqrt{1+4\rho^2}\,d\phi\,d\rho$$ However the $\phi$ part of the integral evaluates to zero and I can't figure where's my mistake. The answer should be: $\frac{125\sqrt5-1}{420}$

Answer 1

You are correct, the given integral is zero. Note that the $\rho$-part yields $$\int_{0}^{1}\rho^5\sqrt{1+4\rho^2}\,d\rho=\left[(1+4\rho^2)^{3/2} \frac{30\rho^4 - 6\rho^2 + 1}{840}\right]_0^1=\frac{125\sqrt{5}-1}{840}$$ which is half of the given result. So I guess that the possible typo could be in the statement of the problem.

Indeed, using your approach, we have that $$\begin{align} \iint_S |xyz| \,d\sigma&=\int_{0}^{1}\int_{0}^{2\pi}\rho^5|\cos(\phi)||\sin(\phi)|\sqrt{1+4\rho^2}\,d\phi\,d\rho\\ &=\frac{125\sqrt{5}-1}{840}\int_{0}^{2\pi}|\cos(\phi)||\sin(\phi)|\,d\phi\\ &=\frac{125\sqrt{5}-1}{840}\cdot 4\int_{0}^{\pi/2}\cos(\phi)\sin(\phi)\,d\phi \\ &=\frac{125\sqrt{5}-1}{840}\cdot 4\left[\frac{\sin(\phi)^2}{2}\right]_0^{\pi/2}=\boxed{\frac{125\sqrt{5}-1}{420}}. \end{align}$$ Of course, also $\iint_S |xy|z \,d\sigma$ gives the same result.