Triple Integral of Tetrahedron

[shards5]

Homework Statement

Evaluate the triple integral \int\int\int^{}_{E} xy dV where E is the tetrahedron (0,0,0),(3,0,0),(0,5,0),(0,0,6).

Is there a simple way to simplify the integration?

Homework Equations

The Attempt at a Solution

\frac{z}{6} + \frac{y}{5} + \frac{x}{3} = 1
z = 6 - 2x - \frac{6}{5}y
Set z = 0 and solve for y
6 = 2x + \frac{6}{5}y \rightarrow y = 5-\frac{5}{3}x
I get the following integral
\int^{3}_{0}\int^{5-5/3x}_{0}\int^{6-2x-6/5y}_{0} xy dydx
After the first integration I get
\int^{3}_{0}\int^{5-5/3x}_{0} xy(6-2x-6/5y) dydx
After multiplying it out I get
\int^{3}_{0}\int^{5-5/3x}_{0} 6xy-2x^y-6/5xy^2 dydx
And this is where it gets complicated. Integration with respect to y . . .
\int^{3}_{0} 3xy^2-x^2y^2-2/5xy^3 dx
Plugging everything I get the following
\int^{3}_{0} 3x(5-5/3x)^2 -x^2(5-5/3x)^2 -2/5x(5-5/3x)^3 dx
Which when expanded gives the following
(5-5/3x)^2 = 25-(50/3)x(+25/9)x^2
(5-5/3x)^3 = 125-250/3+(125/9)x^2-(125/3)x+(250/9)x^2-(125/27)x^3 \rightarrow 125-375/3x+375/9x^2-125/27x^3
\int^{3}_{0}75x-50x^2+25/3x^3-25x^250/3x^3-25/9x^4-50x+50x2-50x^3+50/27x^4 dx
Which simplifies to
25x - 25x^2 + 25x^3 -25/27x^4 which after integration gives me
25/2x^2-25/3x^3+25/4x^4 -25/(27*5) * x^5
And after plugging in I get 348.75 which is wrong.
Is there a simpler way of doing this because that was really painful to integrate.

 

[cragar]

instead of multiply those out like x(x+1) for example just integrate it by parts.
and also
(x+2)^2
instead of multiplying this out the integral is just
\frac{(x+2)^3}{3}
and where is you dz for your integral .

 

[SammyS]

The limits of integration look fine, so the error must be doing the integrations.