Homework Help: Cylindrical Coordinates

1. Nov 16, 2008

squeeky

1. The problem statement, all variables and given/known data
Use Cylindrical Coordinates.
Find the volume of the solid that the cylinder $$r=acos\theta$$ cuts out of the sphere of radius a centered at the origin.

2. Relevant equations
Sphere = x2+y2+z2=a3

3. The attempt at a solution
I think that the limits are from -pi/2 to positive pi/2 for theta, 0 to acos(theta) for r, and negative (a3-r2)1/2 to positive (a3-r2)1/2. This gives me the equation:
$$\int^{\pi/2}_{-\pi/2}\int^{acos\theta}_0\int^{\sqrt{a^3-r^2}}_{-\sqrt{a^3-r^2}} dzrdrd\theta$$
Solving this, I get a volume of $$\frac{4\pi}{3}a^{9/2}+\frac{8}{9}a^3$$
But is this right?

2. Nov 17, 2008

tiny-tim

Welcome to PF!

Hi squeeky! Welcome to PF!

erm … you need oiling!

i know it's three-dimensional, but still …

it should be x2+y2+z2=a2

3. Nov 17, 2008

HallsofIvy

No, it's not. The the graph of the equation $r= acos(\theta)$, in polar coordinates is a circle with center at (0, a/2) and radius a/2. Since it lies only in the upper half plane, $\theta$ ranges from 0 to $\pi$, not $-\pi/2$ to $\pi/2$.

4. Nov 17, 2008

tiny-tim

Hi HallsofIvy!

No, that would be r = a sintheta.

5. Nov 17, 2008

HallsofIvy

Oops! Well, I'll just go back and edit it so it looks like I never made that mistake!

6. Nov 18, 2008

squeeky

Re: Welcome to PF!

Ah! That's right, I don't know how I got that cube, I must have been seeing things when I looked up the formula.

And so now I get a an equation of $$\int^{\pi}_0\int^{acos\theta}_0\int^{\sqrt{a^2-r^2}}_{-\sqrt{a^2-r^2}}dzrdrd\theta$$
which (unless I did my math wrong) gives me a somewhat nice value of $$V=(\frac{2a^3}{3})(\frac{4}{3}-\pi)$$
Is this right now?

7. Nov 18, 2008

tiny-tim

yes, that looks right …

except doesn't theta go from -π/2 to π/2, as in your original post?

8. Nov 18, 2008

squeeky

That's what I thought at first, but HallsofIvy pointed out that it's actually from 0 to pi. Was I right at first then? Because it does make more sense to me if it is from -pi/2 to pi/2, since I see the limits as lying in the xz-plane.

9. Nov 19, 2008

tiny-tim

Yes!

The circle has centre (a/2,0) …

for example, the point (0,1) obviously doesn't lie on it.