volume inside sphere but outside hyperboloid

Question

I am trying to find the volume inside the sphere $x^2 + y^2 + z^2 = 9$, but outside the hyperboloid $x^2 + y^2 - z^2 = 1$. by using a triple integral. for some reason i just cant seem to come up the bounds of integration for this problem. To be more precise, its the region lying to the side of the hyperboloid, that wraps around it, creating a sort of donut shape.

Answer 1

Hint:

The volume inside sphere but outside hyperboloid can be seen as the volume of hyperboloid subtracted from volume of the sphere. In the cartesian coordinates, the volume of the hyperboloid could be written as the below iterated integrals.

Equating both conics such as $$z^2 = 9 - x^2 -y^2 = x^2 + Y^2 -1 => 2(x^2 + y^2) = 10$$ $$ => x^2 + y^2 = 5, implies z^2 = 9-5 = 4 => z = +2, -2$$. Let us slice the solid in the xy plane and we get the limits of z is (-2,2). Each slice is the disk enclosed by a circle $x^2 + y^2 = z^2 + 1$, which is the circle of radius $\sqrt{z^2 + 1}$. Now finding the limits of x and y, we slice the hyperboloid in the vertical direction and this amounts to slicing $$[-\sqrt{z^2 + 1},\sqrt{z^2 + 1}]$$ on the x-axis. Along each slice, y goes from bottom of the circle $y = -\sqrt{z^2 + 1 - x^2}$ to the top of the circle $y = \sqrt{z^2 + 1 - x^2}$.

Putting this altogether, the volume of the hyperboloid =

$$ V_{hyperboloid} = \int_{-2}^{2} \int_{-\sqrt{z^2 + 1}}^{\sqrt{z^2 + 1}} \int_{-\sqrt{z^2 + 1 - x^2}}^{\sqrt{z^2 + 1 - x^2}} dydxdz$$

Now the volume of the sphere with a radius of 3 $$V_{sphere} = \frac{4}{3}\pi 3^3$$

Thus the volume inside sphere but outside hyperboloid $$= V_{sphere} - V_{hyperboloid} - V=36\pi - \frac{28}{3}\pi - V = \frac{80}{3}\pi - V $$ where V is the volume of shphere around z = 2 to 3. The volume of which is

$$ \int_{0}^{2\pi} \int_{2.236}^{3} (\sqrt{9-r^2})rdrd\theta = \frac{16\pi}{3}=16.756$$

Thus the required volume$ = \frac{80}{3}\pi - \frac{16\pi}{3} =\frac{64\pi}{3}$

Answer 2

Just a picture, not an answer.