Problem: Find the volume enclosed by the cone $$x^2 + y^2 = z^2$$ and the plane $$2z – y -2 = 0$$
So I know that I need to do a triple integral over this region, and the integrand will be 1.
My problem is with finding the bounds for the integral. I set the $z$s equal to each other and found the intersection is $x^2 + \frac{3}{4}(y-\frac{4}{9})^2 = \frac{13}{9}$ (although I may have made a mistake here.)
I believe that the bound for $z$ is from $\sqrt{x^2 + y^2}$ to $\frac{y+2}{2}$?
Now I am not sure what to do. How can I find the bounds for $x$ and $y$? Also, I think I should do a change of variable so that the plane lies flat across the cone instead of slanted, to make the upper bound for $z$ constant. Is this a good idea? How could I do it? I am really lost with this problem.
Best Answer
It might help to convert this into cylindrical coordinates:
$$ (x, y, z) = (r \cos \theta, r \cos \theta, z) $$
With $ r \ge 0$ and $0 \le \theta \le 2\pi$
So we have the surfaces:
$$ z = \pm \; r $$ $$ z = \frac{r\cos \theta + 2}{2} $$
Since the plane cuts above the double cone, $z \ge 0$ and the first equation simplifies to $ z = r$. We now have the bounds for $z$.
Substitute the first equation into the second to find the intersection curve (ellipse) $$ 2r - r \cos \theta - 2 = 0 $$ $$ r(2 - \cos \theta) = 2 $$ $$ r = \frac{2}{2 - \cos \theta} $$
Since we're integrating from the origin, $r = 0$ is the lower bound and the the ellipse is the upper bound
Finally, the bounds for $\theta$ is just $0$ to $2\pi$ since we're taking the entire ellipse
Put it all together:
$$ V = \int_{0}^{2\pi} \int_{0}^{2/(2 - \cos \theta)} \int_{r}^{(r\cos \theta + 2)/2} r\,dz \, dr \, d\theta$$