# Homework Help: How to evaluate int 2x-3y dA using change of variables

1. Dec 2, 2004

### AeroFunk

Let R be the region bounded by the graphs of x+y=1, x+y=2, 2x-3y=2, and 2x-3y+5. Use the change of variables:
$$x=1/5(3u+v)$$
$$y=1/5(2u-v)$$
to evaluate the integral:
$$\iint(2x-3y)\,dA$$

I found the jachobian to be -1/5
and the limits of integration to be
1<=u<=2
2<=v<=5

so i set up the integral like this:
$$\frac{-1}{5}\int_{2}^{5}\int_{2}^{1} vdv$$

and I get -21/5 which doesn't seem right(a negitive number??),what am I doing wrong?

2. Dec 2, 2004

### Justin Lazear

I'm getting -21/10, but that's just because you forgot a 1/2 in the integration. I am wondering, though, why your bounds of integration on u are going from 2 to 1 instead of 1 to 2, especially since you don't evalute the integral as such.

What's wrong with a negative number? The integral in u-v obviously isn't going to be negative, and the Jacobian is in fact negative, so obviously the answer should be negative. From the x-y integral, we're integrating 2x-3y. If y is sufficiently large compared to x, the integral will be negative, as in this case.

--J

3. Dec 2, 2004

### AeroFunk

Oh sorry the 2 to 1 is just a typo , and lol yea it would be negitive (have been doing to many volume problems lately)

thanks for the help