volume inside sphere but outside hyperboloid

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.

• Hint: Although it's rarely advisable, try integrating in the order $d\theta dr dz$. – Ted Shifrin Mar 17 '14 at 0:09 Just a picture, not an answer.

• pretty picture, doesn't solve the problem though. – Guy Mar 17 '14 at 7:26
• I get $64 \pi /3$ by another method, rotating the region produced by the intersection of hyperbola and circle around the axis . – Alan Mar 17 '14 at 23:45
• I checked Alan's computations:$$\int_{-2}^2 2\pi(4-z^2)dz=64\pi/3$$ -check! – LeoTheKub Mar 18 '14 at 1:27
• Could you comment what I am doing wrong?. You seem to make sense. In my method, Could you let me know how the volume of the sphere above the hyperboloid is found? – Satish Ramanathan Mar 18 '14 at 1:44
• Somewhere in my calculation, I hit your answer. But I do not know where I got lost – Satish Ramanathan Mar 18 '14 at 1:45

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$= \frac{80}{3}\pi - \frac{16\pi}{3} =\frac{64\pi}{3}$

• Are you sure? This is a hyperboloid of one sheet. – Ted Shifrin Mar 17 '14 at 0:07
• its supposed to be the donut shape that wraps around the hyperboloid, inside the sphere. – user2253455 Mar 17 '14 at 0:20
• @user2253455, Atlast, I found the correct setup. See if it is helpful from the picture that is drwan by the other responder. Goodluck – Satish Ramanathan Mar 17 '14 at 22:00