Triple Integral: Volume problem, Using cylidrical coordinates

• Dec 8th 2013, 05:27 PM
Elusive1324
Triple Integral: Volume problem, Using cylidrical coordinates
Problem statement:
A cylindrical drill with radius 1 is used to bore a hole through the center of a sphere of radius 3. Find the volume of the ring shaped solid that remains using cylindrical coordinates.

My solution: 0=Theta | = integral (Sorry I don't know know Latex. If you can please direct me to a page with syntax and I will happily reformat my solution). Orient the cylindrical hole through the sphere such that the axis of the hole is parallel and on the z axis.

||| r dzdrd0

Limits of integration (left to right): 0 to 2*pi, 1 to 3, -sqrt(9-r^2) to +sqrt(9-r^2)

I would like to know why the limits of integration of z is -sqrt(9-r^2) to +sqrt(9-r^2). The radius is at most 3 and at least 1. However if r=3, z=0 which is true when r=3 on the xy-plane but a picture can be used to show that we can find a z=/=0 if r=3.

It seems sqrt(9-r^2)=z doesn't give a correct relation between r and z at all for this problem. I expect to find the max of z (z value when the cylinder intersects the sphere).
• Dec 8th 2013, 05:33 PM
romsek
Re: Triple Integral: Volume problem, Using cylidrical coordinates
quickly before I even look at the actual problem

LaTeX Help
• Dec 8th 2013, 05:37 PM
romsek
Re: Triple Integral: Volume problem, Using cylidrical coordinates
also, can you not find the volume of the part removed and subtract it from the volume of the sphere? That's got to be an easier integral.
• Dec 26th 2013, 05:57 PM
Elusive1324
Re: Triple Integral: Volume problem, Using cylidrical coordinates
Quote:

Originally Posted by Elusive1324
Problem statement:
A cylindrical drill with radius 1 is used to bore a hole through the center of a sphere of radius 3. Find the volume of the ring shaped solid that remains using cylindrical coordinates.

My solution: 0=Theta | = integral. Orient the cylindrical hole through the sphere such that the axis of the hole is parallel and on the z axis.

||| r dzdrd0

Limits of integration (left to right): 0 to 2*pi, 1 to 3, -sqrt(9-r^2) to +sqrt(9-r^2)

My solution:
$\int^{2\pi}_{0}\int^{3}_{1}\int^{\sqrt{9-r^2}}_{-\sqrt{9-r^2}} r\,dz\,dr\,d\theta$

Is the expression correct?
Edited: I used wolfram to do the integration which yields $\frac{64\pi\sqrt{2}}{3}$. I crossed check this value with the online homework system and it says that this is correct.

O.K

Next, I'm going to try to verify this using spherical coordinates.
• Dec 27th 2013, 04:38 AM
HallsofIvy
Re: Triple Integral: Volume problem, Using cylidrical coordinates
Quote:

Originally Posted by romsek
also, can you not find the volume of the part removed and subtract it from the volume of the sphere? That's got to be an easier integral.

I second this. The material removed consists of a cylinder of radius 1 and length $4\sqrt{2}$ and the two "caps" at each end of the cylinder. Calculating their volume is the only "hard" part.
• Dec 27th 2013, 02:02 PM
Elusive1324
Re: Triple Integral: Volume problem, Using cylidrical coordinates
Finding the volume of the cylinder: $V=\pi(r^2)h=\pi(h) = 4\pi\sqrt{2}$. 'r', the radius of the cylinder is 1. 'h' = 4√2 is given in the diagram in the attachment.

Volume of 1 Cap (Cavalieri's Principle):

$dV=\pi(r^2)*dx$ and $r=f(x)=\sqrt{9-x^2}$
$Vcap = \pi\int^{3}_{2\sqrt{2}} 9-x^2\,dx = 18-\frac{38\sqrt{2}}{3}$

$2*Vcap + 4\pi\sqrt{2} \neq \frac{64\pi\sqrt{2}}{3}$

Did I miss something?

Edited: Oh wait... subtract this from the volume of the sphere.
Edited: I evaluated (4/3)(pi)(27)-4pi*sqrt(2)-2*[18-38*sqrt(2)/3] == 64pi*sqrt(2)/3 in wolfram alpha to check if the two values are equal but they aren't. So something in my calculation went wrong.
• Dec 27th 2013, 05:41 PM
Elusive1324
Re: Triple Integral: Volume problem, Using cylidrical coordinates
Edited: Forgot the pi in front of a term. This expression is true: (4/3)(pi)(27)-4pi*sqrt(2)-2pi*[18-38*sqrt(2)/3] == 64pi*sqrt(2)/3
On the left hand side it's the volume minus the cylinder and the two caps. The right side is obtained by triple integrate in cylindrical coords.