Iterated integral in polar coordinates

1. Nov 10, 2014

e^(i Pi)+1=0

1. The problem statement, all variables and given/known data
Use polar coordinates to find the volume of the solid inside the hemisphere z=sqrt(16-x^2-y^2) and inside the cylinder x^2+y^2-4x=0

2. Relevant equations
z=sqrt(16-x2-y2)
x2+y2-4x=0
x=rcos(Θ)
y=rsin(Θ)

z=√(16-r2)
3. The attempt at a solution

∫∫ r√(16-r2) dr dΘ

The problem is the bounds; because the circle isn't centered it's throwing me off. Would dr be from 2 to 4? That's the start and end of the radius as it's a circle centered at (2,0) with a radius of 2. Of course I'm assuming that dΘ is from 0 to 2pi. I tried integrating with dr from 0 to 2 and from 2 to 4, but both times the answer was different than Wolfram's.

2. Nov 11, 2014

haruspex

Do you understand what the object looks like? Can you post a decent picture, or describe it well in words?
The first trick is to get the order of integration right, and you have done that (z first here). But you need r to be a radius of the circle. How can you arrange that?

3. Nov 11, 2014

e^(i Pi)+1=0

It's an offset vertical cylinder under a hemisphere. I don't follow what else you're saying.

4. Nov 11, 2014

LCKurtz

So what do you get for the polar coordinate equation of that cylinder? You need that.

5. Nov 11, 2014

haruspex

Yes, it's offset, but can you be more precise about how the cylinder fits within the hemisphere? Draw a diagram for z=0.
In principle, you have a triple integral, ∫∫∫dxdydz. In what you posted, you have already done the dz integral, reducing it to ∫∫zdxdy. That was good. So now you can think of z as just a function of x and y, and forget the reality of the hemisphere.
In your conversion to polar, you have set r = √(x2+y2). That represents a radius of the hemisphere, which is no longer interesting. What will work better?