Iterated integral in polar coordinates

[e^(i Pi)+1=0]

Homework Statement

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

Homework Equations

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

z=√(16-r²)

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.

 

[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?

 

[e^(i Pi)+1=0]

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

 

[LCKurtz]

Homework Statement

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

Homework Equations

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

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

 

[haruspex]

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

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?