Double Integral - Evaluate ∫∫D xy dA

[shards5]

Homework Statement

Evaluate the double integral ∫∫_D xy dA where D is the triangular region with vertices (0,0) (6,0) (0,1).

Homework Equations

The Attempt at a Solution

0 <= x <= -\frac{1}{6}x+1
0 <= x <= 6
the first integral would be the integral from 0 to -1/6x+1 of xy with respect to y
which gives me \frac{x*y²}{2} which I plug in -1/6x + 1 into to get
\frac{1/36x^3-12x^2+1/36x}{2}
Integrating that I got x^4/288 - x^3/72 + x^2/4, which is wrong. I am not sure what I did wrong.

 

[Mark44]

Fixed up your LaTeX. Tip: Don't use [ sup] and [ /sup] inside [ tex] tags.

Homework Statement

Evaluate the double integral ∫∫_D xy dA where D is the triangular region with vertices (0,0) (6,0) (0,1).

Homework Equations

The Attempt at a Solution

0 <= x <= -\frac{1}{6}x+1
0 <= x <= 6
the first integral would be the integral from 0 to -1/6x+1 of xy with respect to y
which gives me \frac{x*y^2}{2} which I plug in -1/6x + 1 into to get
\frac{1/36x^3-12x^2+1/36x}{2}

You should get (1/2)[(1/36)x³ - (1/3)x² + x]

Integrating that I got x^4/288 - x^3/72 + x^2/4, which is wrong. I am not sure what I did wrong.

 

[shards5]

Whoops, I forgot to add in a step. However, when I integrate that I get what I have shown in my question and when I evaluate it on the interval from 0 to 6 I am getting the wrong answer which makes me think that maybe I am doing something else wrong.

 

[Mark44]

There are lots of opportunities for arithmetic mistakes. This is what I got.
(1/2)\int_0^6 (x^3/36 - x^2/3 + x)dx = (1/2)\left.(x^4/144 -x^3/9 + x^2/2)\right|_0^6

After evaluating at 6 and 0, I ended up with 3/2.

 

[shards5]

Oh, I see what I did wrong, I thought that 1/6+1/6 was 1/12. Thanks a lot for checking my work.