# Help finding the boundaries of integration of the following function

Given the question:

Obviously this is very hard in cartesian coordinates, so we switch to spherical.

This gives us that: $z=\rho\cdot \cos(\phi)$

and thus: $z(x^2+y^2+z^2)^{-3/2}=\rho\cdot \cos(\phi)(\rho^2)^{-3/2}$

The added volume of integration is:

$dV = \rho^2 \sin(\phi)d\rho d\theta d\phi$

So putting everything together gives us:

$\int \int \int_W \rho\cdot \cos(\phi)(\rho^2)^{-3/2}\rho^2 \sin(\phi) d\rho d\theta d\phi$

$=\int \int \int_W \cos(\phi)\sin(\phi) d\rho d\theta d\phi$

Now, $\rho$ should go from the plane $z=4 \iff \rho\cdot \cos(\phi)=4 \iff \rho = 4/\cos(\phi)$ to 8

and $\phi$ should go from 0 to $\pi/4$, since that's the angle at which the sphere intersects the plane $z=4$

Then $\theta$ can just happily spin around and so my bounds of integration are:

$\int^{2\pi}_0 \int^{pi/4}_0 \int^8_{\frac{4}{\cos(\phi)}} \cos(\phi)\sin(\phi) d\rho d\phi d\theta$

Apparently this answer is wrong, and I have been breaking my head trying to understand where I went wrong.

I appreciate any help, thank you in advance.

• Double check the angle at which the $z=4$ plane intersects the $r=8$ sphere. Nov 18 '17 at 6:55
• at $z=4$ we get that in the $z$ direction we can construct a right triangle from (0,0,4) the origin and the point of intersection of the plane and the sphere, so we get that this triangle is also isoceles, that only happens at pi/4 or 3pi/4. I would draw a diagram but I don't know what tool to use :P I am confident the angle is correct, are you saying you know it's wrong, or are you simply suggesting that could be a source of error? Nov 18 '17 at 7:19
• The angle is definitely not $\pi/4$. The two points of intersection and the origin will always form an isoceles triangle no matter what the angle is. You're confusing it with an isoceles right triangle Nov 18 '17 at 7:22
• I am not talking about the 2 points of intersection I am talking about the $z$ axis coordinate (0,0,4) the origin and the point of intersection (so in the unit circle this would look like sin = 1/2) and the only 2 angles such that sin = 1/2 are $\pi/4$ and $3\pi/4$ Nov 18 '17 at 7:25
• $\sin (\pi/4) = \frac{\sqrt{2}}{2}$ Nov 18 '17 at 7:26

The sphere has $\rho = 8$. Since the plane is defined by $z = \rho\cos(\phi) = 4$, the angle of intersection has to satisfies
$$\cos\phi = \frac{z}{\rho} = \frac{1}{2}$$
which gives $\displaystyle \phi = \pm \frac{\pi}{3}$