# Double integral question

• kieranl
In summary, the conversation discusses a problem involving evaluating a double integral over the region R={(x,y): y<=x<=1, 0<=y<=1}. The conversation also touches on the order of integration and provides guidance on how to approach the problem.

## Homework Statement

Sketch the region of integration and then evaluate the double integral:

## Homework Equations

$$\int\int$$x2exydA over the region R= {(x,y), y<=x<=1, 0<=y<=1}

## The Attempt at a Solution

I have managed to do half of the problem and integrate it respect to x but then have no idea how to finish the problem.

I mangaged to get upto $$\int$$ey(2/y2-2/y3-1/y)+ey2(y-2/y+2/y3) dy

can someone please tell me how to finish it or if i have done something wrong so far??

cheers

Last edited:
Welcome to PF!

Hi kieranl! Welcome to PF!

(have a ≤ )
kieranl said:
$$\int\int$$x2exydA over the region R= {(x,y), y<=x<=1, 0<=y<=1}

I mangaged to get upto $$\int$$ey(2/y2-2/y3-1/y)+ey2(y-2/y+2/y3) dy

uhh?

start again, and integrate over dy first!

but if i integrated by dy first wouldn't i end up with an answer containing y's and not numbers??

kieranl said:
but if i integrated by dy first wouldn't i end up with an answer containing y's and not numbers??

If you integrate with respect to dy, then you eliminate y, and only have x's.

but i have to integrate over y<=x<=1 for dx so the final answer would then contain y's?

Draw a picture of the region R. Then use the picture to write R in the form

R={(x,y):a<x<b, f(x)<y<g(x)}.

Hi kieranl!

(have an integral: ∫ and a ≤ )
kieranl said:
but i have to integrate over y<=x<=1 for dx so the final answer would then contain y's?

No, you have to integrate over 0 ≤ y ≤ x ≤ 1 …

whether you do it over x first or over y first is up to you.