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