Triple integrating a function inside a sphere but below a paraboloid

Question

I want to integrate a function $f(x,y,z)$ inside the sphere of radius 4 centered at the origin but below the paraboloid $z = x^2 + y^2$ opening upward.

I think switching to spherical coordinates is best since cylindrical might take more than two integrals and rectangular might be too complicated to integrate.

Now $\theta$ should go from $0$ to $2\pi$. And for the upper half of the sphere $\rho$ should start at the paraboloid and end at the sphere, so from $\csc \phi \cot \phi$ to $2$.

But I am stuck on the starting bound for $\phi$. For the upper half of the sphere $\phi$ ends at $\pi/2$. But how to solve for $\phi$ when starting at a point on the paraboloid? I only get results that I can't solve for $\phi$.

Answer 1

Use cylindrical coordinates:

$0 \leq \theta \leq 2 \pi$

$0 \leq r \leq 2$

$-r^2 \leq z \leq {\rm Min}[\sqrt{4 - r^2}, r^2]$

$$\int\limits_{\theta = 0}^{2 \pi} \int\limits_{r=0}^2 \int\limits_{z=-\sqrt{4 - r^2}}^{{\rm Min}[\sqrt{4 - r^2}, r^2]} f(r,\theta,z)\ r\ d\theta\ dr\ dz$$

where you must express $f$ in cylindrical coordinates.

Answer 2

So if you want the region inside a sphere of radius 4 centred at the origin then you have: $$x^2+y^2+z^2\le 4^2$$ and below the paraboloid would be $$z\le x^2+y^2$$ now if we want it upward im assuming you mean that $z\ge 0$.

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Spherical coordinates would be difficult for this, I think cylindrical might work well as you would have: $$z^2\ge \sqrt{4^2-r^2}$$ $$z\le r^2$$ or something similar

Answer 3

If you are integrating using cylindrical coordinates, there are two choices - integrate wrt $dz$ first or $dr$ first.

At the intersection of the paraboloid and the sphere,

$z = x^2 + y^2 = 16 - z^2 \implies z = \dfrac{\sqrt{65}-1}{2} = a$ (say)

$r = \sqrt{z} = \sqrt{\dfrac{\sqrt{65}-1}{2}} = b$ (say)

So you can set up the integral in the order $dr \ dz$ as,

$\displaystyle \int_0^{2\pi} \int_0^a \int_{\sqrt{z}}^{\sqrt{16-z^2}} f(x, y, z) \ r \ dr \ dz \ d\theta \ \ + $

$ \displaystyle \int_0^{2\pi} \int_{-4}^0 \int_0^{\sqrt{16-z^2}} f(x, y, z) \ r \ dr \ dz \ d\theta$

And in the order $dz \ dr$ as,

$\displaystyle \int_0^{2\pi} \int_0^b \int_{-\sqrt{16-r^2}}^{r^2} f(x, y, z) \ r \ dz \ dr \ d\theta \ \ + $

$ \displaystyle \int_0^{2\pi} \int_b^4 \int_{-\sqrt{16-r^2}}^{\sqrt{16-r^2}} f(x, y, z) \ r \ dz \ dr \ d\theta$