Finding Volume of Solid Cut by Cylindrical Coordinates: Is My Solution Correct?

[squeeky]

Homework Statement

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.

Homework Equations

Sphere = x²+y²+z²=a³

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?

 

[tiny-tim]

Welcome to PF!

Sphere = x²+y²+z²=a³

Hi squeeky! Welcome to PF!

erm … you need oiling! …

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

it should be x²+y²+z²=a² ​

 

[HallsofIvy]

Homework Statement

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.

Homework Equations

Sphere = x²+y²+z²=a³

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?

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.

 

[tiny-tim]

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.

Hi HallsofIvy!

No, that would be r = a sintheta.

 

[HallsofIvy]

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

 

[squeeky]

Hi squeeky! Welcome to PF!

erm … you need oiling! …

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

it should be x²+y²+z²=a² ​

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?

 

[tiny-tim]

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?

yes, that looks right …

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

 

[squeeky]

yes, that looks right …

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

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.

 

[tiny-tim]

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.

Yes!

The circle has centre (a/2,0) …

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