Understanding Limits of Integration: Cartesian or Polar First?

[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

Thank you for your time!

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.

 

[tiny-tim]

hi tolove!

\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 = ...?

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? ​

 

[tolove]

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? ​

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

 

[tiny-tim]

not going to happen …

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

 

[tolove]

not going to happen …

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

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

 

[tiny-tim]

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

ok, that's 0 ≤ y ≤ √(1 - x²) ≤ 1

ie 0 ≤ y² ≤ 1 - x² ≤ 1,

which becomes … ? ​

 

[tolove]

ok, that's 0 ≤ y ≤ √(1 - x²) ≤ 1

ie 0 ≤ y² ≤ 1 - x² ≤ 1,

which becomes … ? ​

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?

 

[Mark44]

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?

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.