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)
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)
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.
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.
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
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
If you go to Calculus - Math Help Forum, there's a button near the bottom that says "+Post New Thread".
- Hollywood