Question

find the volume of the solid by double integration in polar coordinates. the solid is below the paraboloid z = 2-x^2-y^2, and above the cone z=sqrt(x^2+y^2)

Answer 1

Do you mean cylindrical or spherical?

Answer 2

I don't know, what I wrote is all of the information that was written in the problem

Answer 3

I only ask because polar wouldn't work, and that you probably meant one of cylindrical or spherical. In any case, cylindrical would probably be best. I'm still trying to figure out the limits, so hold on for a bit.

Answer 4

Here's what the bounded region looks like:

Answer 5

and from a different angle:

Answer 6

I also don't think this is a double integral problem, but rather a triple integral.

After some work, you should find that the intersection is the circle \(x^2+y^2=1\) when \(z=1\), though I'm not entirely sure how to get that answer (refer to WolframAlpha). I somehow managed to get \(y=\pm\sqrt{1-x^2}\), but something tells me my method doesn't hold up...

Anyway, if we call the region \(R\), the volume would be given by the following integral:
\[V=\int\int\int_RdV=\int_{-1}^{1}\int_{-\sqrt{1-x^2}}^\sqrt{1-x^2}\int_{\sqrt{x^2+y^2}}^{2-x^2-y^2}dz~dy~dx\]

Answer 7

When converting to cylindrical you use
\[\begin{cases}x=r\cos\theta\\y=r\sin\theta\\
z=z\\
x^2+y^2=r^2\\
dV=r~dr~d
\theta~dz \end{cases}\]
So right away, you can convert your limits:
\[\int_{0}^{2\pi}\int_{0}^1\int_{r}^{2-r^2}r~dz~dr~d\theta\]

Answer 8

I'm not really sure... we've only done double integration problems and this is our first homework assignment on it so I think it should only require 2 integrals in polar coordinates

Answer 9

Well, if you insist on that, you can try setting the problem up as if it were a solid-of-revolution type of problem. Do you know how you could do that?