Find the volume using triple integrals of the region bounded by
$z=x^2+9y^2$ and $z=18-x^2-9y^2$
In cartesian co-ordinates, the get the following limit -
$$\int_{y=-1}^1\int_{x=-\sqrt{9-9y^2}}^{\sqrt{9-9y^2}}\int_{z=x^2+9y^2}^{18-x^2-9y^2}dxdydz$$
Is this correct ? Can this type of integral be converted into cylindrical co-ordinates for easy evaluation ?
Yes, but elliptical cylindrical coordinates. That is, $\rho^2=x^2+9 y^2$. The Jacobian of the transformation is thus $\rho/3$. The integral is then
$$V = \frac13 \int_0^{2 \pi} d\phi \, \int_0^3 d\rho \, \rho \, \int_{\rho^2}^{18-\rho^2} dz $$
It is straightforward to show that $V = 27 \pi$.