# Iterated integral in polar coordinates

## 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

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
Homework Helper
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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?

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

LCKurtz
Homework Helper
Gold Member

## 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