# Cartesian/Polar Integration

1. Nov 17, 2013

### tolove

I'm confused with limits of integrating and which to integrate first,

I've been getting by so far with just knowing the following,
$\int_0^1\int_0^y f(x,y) dxdy = \int_0^1\int_0^{rsin\theta} f(rcos\theta,rsin\theta) rdrd\theta$

But what happens when we switch it around?
$\int_0^1\int_0^{x} f(x) dydx = ...?$

Will it still be the same? That is,
$= \int_0^1\int_0^{rcos\theta} f(rcos\theta,rsin\theta) rdrd\theta$

edit: I suppose it would be, since the assignment of the particular characters x and y are arbitrary. I'm still confused since I'm not showing it explicitly. I need to review Jacobians I suppose, very short on time this weekend though.

Last edited: Nov 17, 2013
2. Nov 17, 2013

### tiny-tim

hi tolove!
the limits and the jacobian are two separate matters

finding the correct limits is simply a question of drawing the region, and making sure the new limits give the same region

your first limits were for 0 ≤ x ≤ y ≤ 1

your second limits are for 0 ≤ y ≤ x ≤ 1 …

so how should you change them to get the same region?

3. Nov 17, 2013

### tolove

What if it's too complicated to draw by hand, or I'm unable to recognize the function?

4. Nov 17, 2013

### tiny-tim

not gonna happen …

it'll always be f(y) ≤ x ≤ g(y) and a ≤ y ≤ b

5. Nov 17, 2013

### tolove

I'm unable to recognize the function rather often. For example,

$\int_0^1 \int_0^{\sqrt{(1-x^2)}} e^{-x^2-y^2} dydx$

6. Nov 17, 2013

### tiny-tim

ok, thats 0 ≤ y ≤ √(1 - x2) ≤ 1

ie 0 ≤ y2 ≤ 1 - x2 ≤ 1,

which becomes … ?

7. Nov 17, 2013

### tolove

Alright, I got it! Thank you very much! There wasn't any graphing involved in this, though? It's all shown algebraically with those neat little inequalities. I suppose there are cases where traps are present, however.

r = 0..1, theta=0..Pi/2

Edit: Ahh! When you say graph, you mean the limits, not the actual function being integrated?

Last edited: Nov 17, 2013
8. Nov 17, 2013

### Staff: Mentor

Yes, the limits of integration. They define the region over which integration is to take place. It's very important to have a good understanding of what this region looks like, especially if you're changing from Cartesian to polar or vice-versa.