Cartesian and polar coordinates - integrals

1. Mar 10, 2008

Niles

1. The problem statement, all variables and given/known data
When dealing with an integral integrated with respect to dxdy, I can convert this to polar coordinates, and then integrate with respect to dr d\theta. But I have to multiply with a "r" before integrating.

If I am dealing with an integral with respect to dydz, I can substitute this with d\theta dz, but not have to multiply with "r". Why is that? (And is what I wrote even correct?)

2. Mar 10, 2008

Niles

- or do I have to multiply with "r" (being the radius, btw) every time, but when it's d\theta dz, I just insert the value for r, but when it's drd\theta, I have to integrate it?

3. Mar 10, 2008

tiny-tim

Hi Niles!

I'm not sure what your y and z are …

But anyway, if you think about the dimensions, x y z and r (and dx dy and dz) are all lengths, but theta (and dtheta) is only a number.

So, to keep everything the right dimensions, you'll always need an extra r for every theta or dtheta, to make it a length.

(That's not a proof, of course … just a useful check when you're not sure you have the right formula!)

4. Mar 10, 2008

Dick

The area of the rectangle between x,y and x+dx,y+dy is dx*dy. The area of a the region from r,theta to r+dr,theta+dtheta is r*dr*dtheta. So if you are integrating a function over an area, that's where the extra r comes from.

5. Mar 10, 2008

HallsofIvy

Staff Emeritus
I seriously doubt that yo can "substitute d$\theta$dz" without multiplying by r but I don't understand what coordinates you are using. In converting from Cartesian, x, y, z, coordinates to cylindrical r, $\theta$, z the "differential of volume is r dr d$\theta$ dz. The only kind of integral I can think of that would involve "d$\theta$ would be on the surface of a cylinder. And even then, you would multiply by the radius of the cylinder. That is because lines of constant $\theta$ grow farther apart as r increases.