Double integration, multivariable calculus

Question

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

Answer 1

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.

Answer 2

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 $t \in [0,5]$.