Finding the bounds on a triple integral

Question

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.

Answer 1

The intersection is indeed an ellipse, but you seem to have made some algebra mistakes: \begin{align*} x^2 + y^2 &= (\tfrac{y + 2}{2})^2 \\ x^2 + y^2 &= \tfrac{1}{4}y^2 + y + 1 \\ x^2 + \tfrac{3}{4}(y^2 - \tfrac{4}{3}y) &= 1 \\ x^2 + \tfrac{3}{4}(y^2 - \tfrac{4}{3}y + \tfrac{4}{9} - \tfrac{4}{9}) &= 1 \\ x^2 + \tfrac{3}{4}(y^2 - \tfrac{4}{3}y + \tfrac{4}{9}) - \tfrac{1}{3} &= 1 \\ x^2 + \tfrac{3}{4}(y - \tfrac{2}{3})^2 &= \tfrac{4}{3} \\ \end{align*} Using the shadow method, notice that the above ellipse is the desired shadow that we want to integrate our $x$ and $y$ over. Solving for $x$ (for example) in the above ellipse yields: $$ x = \pm \sqrt{\tfrac{4}{3} - \tfrac{3}{4}(y - \tfrac{2}{3})^2} $$ A quick plot tells us that the min/max values of $y$ occur at the $y$-intercepts, where $y = \tfrac{2}{3} \pm \tfrac{4}{3}$. so our triple integral is: $$ V = \int_{-2/3}^2 \int_{-\sqrt{\frac{4}{3} - \frac{3}{4}(y - \frac{2}{3})^2}}^{\sqrt{\frac{4}{3} - \frac{3}{4}(y - \frac{2}{3})^2}} \int_\sqrt{x^2 + y^2}^{\frac{y + 2}{2}} \, dz \, dx \, dy $$

To simplify this calculation consider using polar coordinates.

Answer 2

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:

$$ V = \int_{0}^{2\pi} \int_{0}^{2/(2 - \cos \theta)} \int_{r}^{(r\cos \theta + 2)/2} r\,dz \, dr \, d\theta$$