# 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?