# Thread: Help with double/iterated integrals please

1. ## Help with double/iterated integrals please

Hey guys. I'm having trouble with 3 questions:
1. I(0->pi/2) I(0->cos(theta)) e^(sin(theta)) dr d(theta)...evaluate

I get as far as I(0->pi/2) cos(theta) e^(sin(theta)) d(theta)...and then, I guess you have to use integration by parts, but no matter how I try to set it up I end up with something complicated.

2. I(0->1) I(3y->y) e^(x^2) dxdy...evaluate the integral by reversing the order of integration.

Same problem as #1.

Any degree of help is appreciated. Thanks!

2. Here is the first one.
Use the substitution t=sin theta

3. Originally Posted by ParadigmShift

2. I(0->1) I(3y->y) e^(x^2) dxdy...evaluate the integral by reversing the order of integration.
You are integrating the the region:

R = {(x,y) | 0<=y<=1 and y<=x<=3y} ---> See below

In order to change order we need to find the top and bottom curves.

The top curve is y=1 and bottom is y=(1/3)x

Hence we can think of the region alternatively as,

R = {(x,y) | 0<=x<=3 and (1/3)x<=y<=1}

4. Originally Posted by ThePerfectHacker
Here is the first one.
Use the substitution t=sin theta
I had already gotten as far as your partial solution shows. But I think from that point I start to do it incorrectly. You mean I have to let t=sin(th) and THEN use integration by parts, because that's what I've been trying to do to no avail. I still need a nudge

Thanks for your help on the second one. It's been a long time since I did calc II so I'm rusty (plus I'm generally bad at math).

5. Originally Posted by ParadigmShift
I had already gotten as far as your partial solution shows. But I think from that point I start to do it incorrectly. You mean I have to let t=sin(th) and THEN use integration by parts, because that's what I've been trying to do to no avail. I still need a nudge

Thanks for your help on the second one. It's been a long time since I did calc II so I'm rusty (plus I'm generally bad at math).
Here.

### double integral e sin theta

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