Triple integral: volume bound between sphere and paraboloid - cylindrical coordinates

Question

I need to calculate the volume bound between:

$x^2 +y^2 +z^2 = 5$

$z=3-x^2-y^2$ (the inside of the paraboloid).

Now, when I find the intersection points I get that $z=-1$ and $z=2$, which means there are two intersections and both are circles: $x^2+y^2 = 4$ and $x^2 + y^2 = 1$

Now, I know that if I take the projection on the $xOy$ plane I get a the area of a circle and a ring, $r\in[0,1]$ and $ r\in[1,2]$. I also know how to figure out the angle boundaries, but I can't figure out the boundaries for $z$ in both cases.

I'm not sure because, if I plot the functions on GeoGebra I get that going from the lower intersection(circle) to the top I can't figure out how $z$ changes for both cases. Could anyone help?

Answer 1

If you express your problem in cylindrical coordinates, then

• $r$ can go from $0$ to $\sqrt{3-z}$ when $z\in[-1,2]$
• $r$ can go from $0$ to $\sqrt{5-z^2}$ when $z\in\left[-\sqrt5,-1\right]\cup\left[2,\sqrt5\right]$.

So, that volume is equal to\begin{multline}\int_0^{2\pi}\int_2^{\sqrt5}\int_0^{\sqrt{5-z^2}}r\,\mathrm dr\,\mathrm dz\,\mathrm d\theta\int_0^{2\pi}\int_{-1}^2\int_0^{\sqrt{3-z}}r\,\mathrm dr\,\mathrm dz\,\mathrm d\theta+\\+\int_0^{2\pi}\int_{-\sqrt5}^{-1}\int_0^{\sqrt{5-z^2}}r\,\mathrm dr\,\mathrm dz\,\mathrm d\theta=\frac{2}{3} \left(5 \sqrt{5}-11\right) \pi+\frac{15\pi}2+\frac{2}{3} \left(5 \sqrt{5}-7\right) \pi.\end{multline}

Answer 2

You need to split in two integrals because the upper bound for $z$ is not the same surface if you consider the whole region. So you would have $$ \int_0^{2\pi} \int_0^1 \int_{-\sqrt{5-\rho^2}}^{\sqrt{5-\rho^2}} \rho dz d\rho d\theta + \int_0^{2\pi} \int_1^2 \int_{-\sqrt{5-\rho^2}}^{3-\rho^2}\rho dz d\rho d\theta. $$

Answer 3

Please note that for $0 \leq r \leq 1$, $z$ is bound between surface of the sphere (above xy-plane and below). So it is a cylinder of radius $1$ with two spherical caps on both ends.

For $1 \leq r \leq 2$, $z$ is bound below by the sphere and above by the paraboloid.

So the integral can be written as,

$ \displaystyle \int_0^{2\pi} \int_0^1 \int_{-\sqrt{5-r^2}}^{\sqrt{5-r^2}} r \ dz \ dr \ d\theta + \int_0^{2\pi} \int_1^2 \int_{-\sqrt{5-r^2}}^{3-r^2} r \ dz \ dr \ d\theta$