Find the volume of the region in the first octant underneath the paraboloid $$z = 1 – \frac{x^2}{81} – \frac{y^2}{36}$$
I've been given the hint to use the change of variables $x = 9rcos(\theta)$ and $y = 6rsin(\theta)$
I know that $x > 0, y > 0, z > 0$ since we're in the first octant. Still not quite sure how to go about finding the limits, and how to use the substitution.
Best Answer
HINT
draw the region by plot on $x-y$, $z-x$, $z-y$ plane and find the limit of integration
set up the integral in cartesian coordinates $\iiint_V 1 dxdydz$
set up the integral in polar coordinates $\int_0^{\frac{\pi}2}\int_0^{1}\int_0^{r(z)} |J| dr\,dz\,d\theta$