# use triple integral to find volume of solid bounded by sphere and hyperboloid

• May 10th 2014, 06:58 PM
SNAKE
use triple integral to find volume of solid bounded by sphere and hyperboloid
bounded below by - x^2 - y^2 + z^2=1 and above by x^2+y^2+z^2=19, for z > 0

would this be better in cylindrical? z^2 -r^2 =1 and r^2 + z^2 =19

limits:
theta: 0 to 2pi
r: 0 to sqrt 19
z: sqrt(1+r^2) to sqrt(19-r^2)
• May 10th 2014, 07:29 PM
romsek
Re: use triple integral to find volume of solid bounded by sphere and hyperboloid
I know this is going to sound like crazy talk, especially as it's Saturday night, but this is the sort of problem where the best thing you can do is to work it both ways and see what happens. This is how you develop intuition for math.
• May 10th 2014, 08:10 PM
SNAKE
Re: use triple integral to find volume of solid bounded by sphere and hyperboloid
That does sound crazy, but I will give it a shot :) Are my limits correct, however?
• May 11th 2014, 02:15 PM
SNAKE
Re: use triple integral to find volume of solid bounded by sphere and hyperboloid
can someone please check my limits to see if they are correct?
• May 11th 2014, 06:36 PM
hollywood
Re: use triple integral to find volume of solid bounded by sphere and hyperboloid
The sphere and the hyperboloid intersect, but the sphere is above and below the enclosed volume. Do you mean to have a cylindrical wall on the side?

- Hollywood
• May 11th 2014, 08:47 PM
SNAKE
Re: use triple integral to find volume of solid bounded by sphere and hyperboloid
I think since it says for z > 0, I am looking for the volume inside the cone made by the hyperboloid which is capped by a hemisphere. I don't see where there is a cylindrical wall.
• May 12th 2014, 06:15 AM
hollywood
Re: use triple integral to find volume of solid bounded by sphere and hyperboloid
Ok, I see it now. I'm not sure what I was thinking before.

So, yes, cylindrical coordinates would work well. The intersection of the two surfaces is the circle x^2+y^2=9 in the $\displaystyle z=\sqrt{10}$ plane. So your limits are correct except r should go from 0 to $\displaystyle \sqrt{10}$.

- Hollywood
• Jun 3rd 2014, 12:34 PM
rubcksy
Re: use triple integral to find volume of solid bounded by sphere and hyperboloid
the region bounded by the paraboloid z=sqrt(x^2+y^2) and above by the sphere x^2+y^2+z^2=6
• Jun 3rd 2014, 05:39 PM
hollywood
Re: use triple integral to find volume of solid bounded by sphere and hyperboloid
You should start a new thread for a new problem.

The intersection is the circle $\displaystyle x^2+y^2=3$ in the $\displaystyle z=\sqrt{3}$ plane.

So $\displaystyle z(x,y)$ goes from $\displaystyle r$ to $\displaystyle \sqrt{6-r^2}$. Now we need to define the "floor plan", which is the circle $\displaystyle x^2+y^2=3$.

If cylindrical coordinates work for you, your integral is

$\displaystyle \int_0^{2\pi} \int_0^{\sqrt{3}} \int_r^{\sqrt{6-r^2}} r\,dz\,dr\,d\theta$

The extra $\displaystyle r$ is required for cylindrical coordinates.

- Hollywood
• Jun 4th 2014, 05:59 AM
rubcksy
Re: use triple integral to find volume of solid bounded by sphere and hyperboloid
The region in the first octant bounded above by the cylinder z=1-y^2 and lying between the vertical planes x+y=1 and x+y=3
• Jun 4th 2014, 06:18 AM
hollywood
Re: use triple integral to find volume of solid bounded by sphere and hyperboloid
If you go to Calculus - Math Help Forum, there's a button near the bottom that says "+Post New Thread".

- Hollywood
• Jun 4th 2014, 08:48 AM
rubcksy
Re: use triple integral to find volume of solid bounded by sphere and hyperboloid
thanks
• Jun 4th 2014, 09:46 AM
rubcksy
Re: use triple integral to find volume of solid bounded by sphere and hyperboloid
the region bounded by the paraboloid z=x^2+y^2 and above by the sphere x^2+y^2+z^2=6