# Thread: coordinate transformation double integral

1. ## coordinate transformation double integral

Hey everyone, I am stuck on this question, I just can't get anymore done. Any help would be great.

Evaluate $\int_1^2 \int_{1/y}^y \ \sqrt{\frac{y}{x}} \cdot e^{\sqrt{xy}} \ dx \ dy$ by making the coordinate transformation $x=\frac{u}{v}$ and $y=uv$.

My working so far:
$xy=u^2$

$\frac{y}{x}=v^2$

Jacobian $\ = \frac{2u}{v}$

So:
$\int_1^2 \int_{1/uv}^{uv} \ \sqrt{v^2} \cdot e^{\sqrt{u^2}} \cdot \frac{2u}{v} \ du \ dv$

But I don't get how to go from here, with the limits of one of the integrals having terminals with u and v in it.

If anyone can help that would be fantastic!

Thanks

2. The limits of integration, in terms of x and y are: y= 1, y= 2, and x= 1/y, x= y.

Since x= u/v and y= uv, those become uv= 1, uv= 2, and u/v= 1/uv, u/v= uv. The first two give v= 1/u and and v= 2/u while the last two reduce to $u^2= 1$ and $v^2= 1$.

Drawing those lines on a uv- graph, you can see that u= 1 and v= 1 intersect at (1, 1) on the graph of v= 1/u and intersect the graph of y= 2/v at u= 1, v= 2 and u= 2, v= 1. Essentially, then, the region you want to integrate over, in the uv-plane, is bounded by u= 1, v= 1, and v= 2/u. If u ranges from 1 to 2 then, for each u, v ranges from 1 to 2/u.

3. Originally Posted by HallsofIvy
The limits of integration, in terms of x and y are: y= 1, y= 2, and x= 1/y, x= y.

Since x= u/v and y= uv, those become uv= 1, uv= 2, and u/v= 1/uv, u/v= uv. The first two give v= 1/u and and v= 2/u while the last two reduce to $u^2= 1$ and $v^2= 1$.

Drawing those lines on a uv- graph, you can see that u= 1 and v= 1 intersect at (1, 1) on the graph of v= 1/u and intersect the graph of y= 2/v at u= 1, v= 2 and u= 2, v= 1. Essentially, then, the region you want to integrate over, in the uv-plane, is bounded by u= 1, v= 1, and v= 2/u. If u ranges from 1 to 2 then, for each u, v ranges from 1 to 2/u.
Oh yes, thanks HallsofIvy, so the double integral is: $\int_1^2 \int_1^{2/u} 2u \cdot e^u \ dvdu$

4. Originally Posted by Nguyen
Oh yes, thanks HallsofIvy, so the double integral is: $\int_1^2 \int_1^{2/u} 2u \cdot e^u \ dvdu$

Yes...yet the original integral is pretty easy to evaluate directly.

Tonio

5. Originally Posted by tonio
Yes...yet the original integral is pretty easy to evaluate directly.

Tonio
Thanks Tonio!