- #1

ramb

- 13

- 0

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

[tex]\int\int_{B}2x+3y dA[/tex]

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

## 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 separate double integrals.

This is what I got.

[tex]\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[/tex]

With the first integral I got

[tex]\dfrac{-651}{8}[/tex]

for the second integral i got

[tex]\dfrac{-4739}{12}[/tex]

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, [tex]\frac{11431}{24}[/tex] 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