2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Why would that not be possible? $\endgroup$ – Santiago Canez Nov 14 '14 at 13:37
2
$\begingroup$

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.

Note the symmetry of the graph:

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.