# Thread: double integral, triangular region

1. ## double integral, triangular region

I need to solve the double integral for y^3, where the region (D) is a triangle with vertices (0,2) (1,1) (3,2).

So first I drew out these points, made the triangle, and then solved the equation of each of the three triangle lines.
So for (0,2) to (1,1), i got y=-x
for (3,2) to (0,2), i got y=2
for (3,2) to (1,1), i got y= x/2 +1/2

so then i decided to integrate with respect to x first. i solved the two line equations for x, and found that -y < x < 2y-1
and 1 < y < 2. so those would be my bounds for x and y.

then i integrated y^3 dxdy. which gives me first: xy^3. plug in the bounds, and i got 2y^4-y^3+y^4.

then integrated that for y. plugged in the bounds and solved and got 297/20. i'm told the answer is 147/20. i have tried doing it with y first, and using two different sets of bounds for y but i'm not getting it. i don't know what other way there is to try...

what am i doing wrong here?

2. So for (0,2) to (1,1), i got y=-x
the equation of this line is y=-x + 2

if you carry that change forwards will you get the right answer?