quickly before I even look at the actual problem
LaTeX Help
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).
quickly before I even look at the actual problem
LaTeX Help
Finding the volume of the cylinder: . '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):
and
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.
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.