Computing a double integral with given vertices

[ramb]

1. The problem statement, all variables and given/known data[/b]

Use the transformation that takes the unit square to a triangle to compute the integral

\int\int_{B}2x+3y dA

Where B is a triangular region with vertices (0,0), (5,2), and (3,4).


3. The attempt at a solution

What I did was I drew the region on an xy plane, I split the triangle up into two triangles and found my limits of integration by drawing the lines made by connecting the vertices. Because I split the triangle up into two, I needed to add two seperate double integrals.

This is what I got.

\int_{x=0}^{x=3}\int_{y=5x/2}^{y=4x/2}(2x+3y) dydx + \int_{x=3}^{x=5}\int_{y=5x/2}^{y=-x+7}(2x+3y) dydx

With the first integral I got

\dfrac{-651}{8}

for the second integral i got

\dfrac{-4739}{12}

I figured that if I add both of them together I would get the volume underneath that region (the whole thing), but the number I got, \frac{11431}{24} seems to large.

I also don't think I'm doing it the method wanted.
Can anyone please direct me where to go from here, i'm somewhat lost.

Thank you

[LCKurtz]

I didn't check your work because, as you suspect, that isn't the method you have been asked to use. Your question refers to "the" transformation that takes the unit square to a triangle. I'm guessing your text has given you that example. Look it up and share the equations of that transformation with us and then we can talk.:smile: