Here is the first one.
Use the substitution t=sin theta
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!
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}
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).