# Math Help - Calculus 3 - Double Integral

1. ## Calculus 3 - Double Integral

As seen in the picture below that's what I have to integrate. You can see I got $\frac{1}{2}$ as my answer yet they say the correct answer is $\frac{3}{2}$. I don't see how they possibly got that answer. Where did I go wrong?

$=\int_{0}^{3}[\frac{x^2}{2} - \frac{xy}{3}]_{\frac{y}{3}}^{\frac{y}{3}+1} dy$

$=\int_{0}^{3}[\frac{1}{2}(\frac{y}{3}+1)^2 -\frac{y}{3}(\frac{y}{3}+1) - (\frac{1}{3}(\frac{y}{3})^2 - \frac{y}{3}(\frac{y}{3}))] dy$

$=\int_{0}^{3}[-\frac{y^2}{9} + \frac{1}{2}] dy$

$=[-\frac{y^3}{27} + \frac{y}{2}]_{0}^{3} \rightarrow [-\frac{27}{27} + \frac{3}{2}] \rightarrow \frac{1}{2}$

2. Ah I see my error its when I was evaluating the lower limit.