Hi, I can do some basic double integrals but I am having trouble tackling this, any help would be appreciated.
Thanks in advance.
$\displaystyle I = \int _0^1\int _0^x\left(\frac{y}{x}\right)^2 \exp \left(\frac{y}{x}\right)dydx$
let u = y/x, v = y, then y = v, x = v/u
Compute the Jacobian J = dx/du dy/dv - dx/dv dy/du = -v/u^{2} and |J| = v/u^{2}
Now the limits are tricky. On the outer integral v goes from 0 to 1. On the inner integral though u goes from v to 1.
when you make your substitutions and add |J| to the integral you end up with
$\displaystyle I = \int _0^1\int _v^1v \exp (u)dudv$ and I'll let you to solve that. I got $\displaystyle \frac{1}{2} (e-2)$
take a look at http://personal.maths.surrey.ac.uk/s...double_int.pdf, page 9, section 0.15 if above is confusing.
or ring back and I'll try to answer further.
Hi, I get the integral at the end, however how do the vertices of the triangle relate to the limits?
Also the only part that confused me was "when you make your substitutions and add |J| to the integral you end up with"
What do you mean by this?
Thanks a lot for the reply.
when you make a two variable substitution like this you have to add multiply your functions by the Jacobian of your substitution transformation. Did you take a look at that link I included?
Can you understand the integration limits before the substitution? i.e. {0,1} and {0,x}
Ok, it took me a bit to see it too.
First off you have to get the limits on the pre-transformed integral.
You have a triangle (0,0), (0,1) (1,1).
You are integrating x on the outside, y on the inside.
So you go out x a bit and you integrate up to the top of the triangle. That top is given by the line y=x so your top limit is x, your bottom limit is 0.
You need to be able to get that so study it for a bit.
Now after the transform....
The deal is that now you are integrating u on the inside and v on the outside.
But v = y. So you are integrating up the y axis on the outside.
Ok so you go up y a bit but your triangle doesn't start until y up the x axis, i.e. x = y
So your lower limit on the inner integral is y. But in the new coordinates y is v (that was part of the transform).
So u is going from v to 1.
If you're tired I wouldn't try to understand this right now, and I'd definitely draw yourself a picture.