- #1
flyusx
- 63
- 10
- 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!
$$\int_{-6}^{6}\int_{-\sqrt{36-x^{2}}}^{\sqrt{36-x^{2}}}\int_{-\sqrt{36-x^{2}-y^{2}}}^{\sqrt{36-x^2-y^2}}\text{d}z\text{d}y\text{d}x-\int_{-3}^{3}\int_{-\sqrt{9-x^{2}}}^{\sqrt{9-x^{2}}}\int_{-\sqrt{36-x^{2}-y^{2}}}^{\sqrt{36-x^2-y^2}}\text{d}z\text{d}y\text{d}x$$
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!