Double integration, multivariable calculus I was given that a triangle has vertices at $(3,10),(3,1),(−2,1)$ and the question asks me to evaluate $\displaystyle\iint x-y~dA$ of the triangle.
Ive calculated $\displaystyle\int_{-2}^3\int_{1}^{(9x+23)/5}x-y~dA$ to be my limits off my integral. Which I think is correct? 
I calculate the limits by letting $x$ be $-2\le x\le3$ and $y$ be $1\le y\le(9x+23)/5$ because the equation of the line through the points $(3,10)$ & $(−2,1)$ is $y=9/5x + 23/5$
By doing the integral of $\displaystyle\int_{1}^{(9x+23)/5}x-y~dy=x/5(9x+23)-1/50(9x+23)^2-x+1/2$ 
Now doing $\displaystyle\int_{-2}^3 x/5(9x+23)-1/50(9x+23)^2-x+1/2~dx=-1242/25-258/25$ and this simplifies to $-60$ 
But if I calculate this I get an answer of $-60.$ I'm finding it difficult to understand how I can have a negative area? I was hoping someone could expand on this, if my limits are correct.
If my limits are wrong, could you not inform me of the answer but give a me hint instead or point out where I've made a mistake?
Thank you in advance   
 A: You don't have a "negative area"
The area of the region is:
$\int_{-2}^3\int_1^{\frac {9x+23}{5}} \ dA$
But that isn't your integral.
your integral is 
$\int_{-2}^3\int_1^{\frac {9x+23}{5}} x-y \ dA$
If you have a negative result is is because $x-y < 0$ over much of the area.
You might think of a volume created if we added the plane $z = x - y$ to the mix.  This will create two volumes, a small tetrahedron above the xy plane and a larger volume below the xy plane.
Update
Here is your region:

The red portion of the region shows where $x-y < 0$
We are summing negative numbers over most of the region, and that will create a negative total.
A: Keep in mind you can often double check double integrals with Stoke's Theorem by translating the problem into line integrals. 
In this case $(x-y)$ is the $z$ component of the curl of $\vec{r}=\frac{x^2}{2}\hat{i}+\frac{y^2}{2}\hat{j}$
So you can break up the problem into 3 line integrals.
Integrate $\vec{r}\cdot \hat{i}$ from (-2,1) to (3,1), then $\vec{r}\cdot\hat{j}$ from (3,1) to (3,10). Finally peform the appropriate line integral along the diagonal, properly parameterized, $\vec{x}=<3-t, \frac{-9}{5}t+10>.$ Where $t \in [0,5]$.
