# Evaluate the surface integral from the paraboloid

Evaluate the surface integral $$\iint\limits_S xy \sqrt{x^2+y^2+1}\,\mathrm d\sigma,$$ where $S$ is the surface cut from the paraboloid $2z=x^2+y^2$ by the plane $z=1$.

Is it possible for the answer to be $0$ ? I am not so sure. Would anyone mind telling me the answer?

• Why would that not be possible? – Santiago Canez Nov 14 '14 at 13:37

Notice that your bounds are arranged as such: \begin{equation} 2z=x^2+y^2\bigg|_{z=1}\implies y=\pm\sqrt{2-x^2}, \end{equation} therefore the bounds on your integrals are \begin{equation} \int_{-\sqrt{2}}^{\sqrt{2}}\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}xy\sqrt{x^2+y^2+1}\:\:dy\:dx, \end{equation} because it appears as though you are integrating over a circular region and the bounds on the outer integral thus make this a definite integral. We may rewrite the bounds on the interior integral in terms of $y$ as well, which would thus change the order of integration. Now try integrating the interior using a u-substitution to get \begin{align*} \int_{-\sqrt{2}}^{\sqrt{2}}x\left(\int_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}\frac{u^{\frac{1}{2}}}{2}\:du\right)\:dx,\:\:\:\:u & =y^2+\left(x^2+1\right) \\ du & = 2y\:dy\implies\frac{du}{2y}=dy \end{align*} which gives us \begin{align*} & \int_{-\sqrt{2}}^{\sqrt{2}}x\cdot\left[\frac{\left(y^2+\left(x^2+1\right)\right)^{\frac{3}{2}}}{3}\right]_{-\sqrt{2-x^2}}^{\sqrt{2-x^2}}\:\:dx \\[4ex] & =\int_{-\sqrt{2}}^{\sqrt{2}}\left\{x\left[\frac{\left(\left(2-x^2\right)+\left(x^2+1\right)\right)^{\frac{3}{2}}}{3}-\frac{\left(\left(2-x^2\right)+\left(x^2+1\right)\right)^{\frac{3}{2}}}{3}\right]\right\}\:dx=\boxed{0,} \end{align*} since what's inside the integrand cancels. 