Triple Integral To Find Volume Between Cylinder And Sphere

[flyusx]

Homework Statement

Express in rectangular, spherical and cylindrical coordinates a triple integral returning the volume of a sphere $$x^2+y^2+z^2=36$$ that is outside a cylinder $$x^2+y^2=9$$.

Relevant Equations

I use the physics convention here with spherical $$(r,\theta,\phi)$$ and cylindrical $$(\rho,\phi,z)$$. In cylindrical coordinates, $$\rho^2=x^2+y^2$$.

I got the two relations for spherical and rectangular coordinates. In rectangular:
$$\int_{-6}^{6}\int_{-\sqrt{36-x^{2}}}^{\sqrt{36-x^{2}}}\int_{-\sqrt{36-x^{2}-y^{2}}}^{\sqrt{36-x^2-y^2}}\text{d}z\text{d}y\text{d}x-\int_{-3}^{3}\int_{-\sqrt{9-x^{2}}}^{\sqrt{9-x^{2}}}\int_{-\sqrt{36-x^{2}-y^{2}}}^{\sqrt{36-x^2-y^2}}\text{d}z\text{d}y\text{d}x$$
In spherical:
$$\int_{0}^{2\pi}\int_{\frac{\pi}{6}}^{\frac{5\pi}{6}}\int_{3\csc(\theta)}^{6}r^{2}\sin(\theta)\text{d}r\text{d}\theta\text{d}\phi$$
I'm having trouble with cylindrical. The φ bounds are $$0\leq\phi\leq2\pi$$. The z bounds are already given in the rectangular system and can be converted to ρ by the relation given in relevant equations. The inner ρ bound is 3 because the cylinder is a distance 3 away from the z-axis. We have:
$$\int_{0}^{2\pi}\int_{3}^{?}\int_{-\sqrt{36-\rho^2}}^{\sqrt{36-\rho^2}}\rho\text{d}z\text{d}\rho\text{d}\phi$$
The proper solution is for the outer ρ bound to be six (the integrals resolve to the same number). I see this as coming from the maximum possible distance between the z-axis and the bound of the sphere (specifically when z=0)$. What I don't get is that since the outer boundary is the sphere, shouldn't the outer bound be written as $$\rho^{2}+z^{2}=36$$
And then solve for ρ? This must be wrong because this wouldn't allow the integral to be resolved to a number, but I can' see where I went wrong conceptually.

Thanks in advance!

 

[Mark44]

I don't think you're using cylindrical coordinates correctly. They are the 3-dimensional counterpart of polar coordinates, with the coordinates being $z, r,$ and $\theta$. The latter is the angle formed by rays in the x-y plane, while $\phi$ is the angle relative to the z-axis.

Also, it might be simpler to exploit the symmetry of your problem. Everything is symmetric about the z-axis, so you could integrate from 0 to $\frac \pi 2$ and multiply that result by 4.

 

[Orodruin]

Homework Statement: Express in rectangular, spherical and cylindrical coordinates a triple integral returning the volume of a sphere $$x^2+y^2+z^2=36$$ that is outside a cylinder $$x^2+y^2=9$$.
Relevant Equations: I use the physics convention here with spherical $$(r,\theta,\phi)$$ and cylindrical $$(\rho,\phi,z)$$. In cylindrical coordinates, $$\rho^2=x^2+y^2$$.

What I don't get is that since the outer boundary is the sphere, shouldn't the outer bound be written as ρ2+z2=36
And then solve for ρ? This must be wrong because this wouldn't allow the integral to be resolved to a number, but I can' see where I went wrong conceptually.

You are already using that relationship for z. Now you need to find the maximal value of $\rho$ regardless of the value of z. I must say I am surprised that it causes you issues as it is completely analogous to what you did in the case of Cartesian coordinates when you found the bounds for x.