4
$\begingroup$

Calculate $$\iiint_{B} y\;dxdydz.$$

The set is $\;B=\{(x,y,z) \in \mathbb R^3$; $\; x^2+y^2+4z^2\le12$, $-x^2+y^2+4z^2\le6$, $y\ge 0 \}$.


I know that B is defined by a real ellipsoid, an elliptical hyperboloid and by the positive half-space of y, so I tried to use the cylindrical coordinate system but I can't find the correct limits of integration. How can I change the equations?

I hope you'll help me. Thanks a lot!

$\endgroup$
  • $\begingroup$ try rewriting the bounds in cylindrical coordinates. I don't know if this will work, but definitely should try it. $\endgroup$ – Callus - Reinstate Monica Jun 28 '16 at 15:27
  • $\begingroup$ It's what I have done, but unfortunately I'm not sure it's the right way. $\endgroup$ – Simone Laera Jun 28 '16 at 16:32
2
$\begingroup$

Note $$B=\{(x,y,z)|\,\,\,({{y}^{2}}+4{{z}^{2}})-6\le {{x}^{2}}\le 12-({{y}^{2}}+4{{z}^{2}})\,\,,\,y\ge 0\}$$ set $$\left\{ \begin{align} & y=2r\,\sin \theta \\ & z=r\,\cos \theta \\ \end{align} \right.\,\,\,\Rightarrow \,\,\,\,\left| \frac{\partial (y,z)}{\partial (r,\theta )} \right|dydz=2r\,drd\theta $$ $$\iiint\limits_{B}{y\,dxdydz}=4\int_{0}^{\pi }{\int_{0}^{\frac{\sqrt{6}}{2}}{\int_{\sqrt{6-4{{r}^{2}}}}^{\sqrt{12-4{{r}^{2}}}}{{{r}^{2}}\sin \theta \,dx\,drd\theta }}}+4\int_{0}^{\pi }{\int_{0}^{\frac{\sqrt{6}}{2}}{\int_{-\sqrt{12-4{{r}^{2}}}}^{-\sqrt{6-4{{r}^{2}}}}{\,{{r}^{2}}\sin \theta \,dx\,drd\theta }}}$$

$\endgroup$
  • $\begingroup$ Hi @BehoruzMaleki, I think your method is correct, so I thank you. Probably, you did a mistake when you re-wrote B pointing out x^2: if I'm not wrong, it would be (y^2+z^2)-6<x^2<12-(y^2+4z^2). Despite this I realized how to solve my doubt. $\endgroup$ – Simone Laera Jun 28 '16 at 19:15
  • $\begingroup$ Oh, I'm sorry. You have been faster than me! $\endgroup$ – Simone Laera Jun 28 '16 at 19:16
  • 1
    $\begingroup$ Please yes it was edited. $\endgroup$ – Behrouz Maleki Jun 28 '16 at 19:17
  • $\begingroup$ Also the square roots have to be changed but don't worry. $\endgroup$ – Simone Laera Jun 28 '16 at 19:19
  • 2
    $\begingroup$ @SimoneLaera If this answer gives the solution you were hoping for, or you found it to be the most useful, consider "accepting" it by clicking the little checkmark under the arrows to the left of the answer. This is the best way to show Behrouz gratitude for assisting you. :) $\endgroup$ – The Count Jun 28 '16 at 21:16
1
$\begingroup$

I would suggest starting with a rectilinear change of coordinates first. Try $x_{new}=2z$, $y_{new}=y$, and $z_{new}=x$. Now the region involves circular figures instead of elliptical ones, and the central axis of the (now) circular hyperboloid is the $z$-axis, so the region should be pretty easy to describe in cylindrical coordinates. You will have to introduce the Jacobian determinant into the integrand appropriately and look at what the transformation does to the order of integration, but since $y_{new}=y$, the part of the integrand you already have won't change with this transformation (of course, when you go to cylindrical, you will need to rewrite it). The rest should be a pretty standard matter of converting to cylindrical and evaluating the triple integral.

$\endgroup$
  • $\begingroup$ I'll try this way. I wish it is less difficult. Thanks! $\endgroup$ – Simone Laera Jun 28 '16 at 18:38

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.