# Double integral problem [closed]

Evaluate $$I = \iint_D x y \ dx dy$$ where $D$ is the triangular region with vertices $(0, 0)$, $(4, 0)$, $(0, 3)$.

Appreciate any and all help!

-

## closed as off-topic by 900 sit-ups a day, Thomas, martini, studiosus, user91500Jul 8 at 15:28

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 900 sit-ups a day, Thomas, martini, studiosus, user91500
If this question can be reworded to fit the rules in the help center, please edit the question.

Do you have an attempt...? –  sidht Dec 7 '12 at 5:39

$x$ ranges from $0$ to $4$, and for each value of $x$, $y$ ranges from $0$ to $-3x/4 + 3$, so $$I = \int_0^4 dx \ x \int_0^{-3x/4 + 3} dy \ y.$$ I'll let you evaluate.
You can solve this question with Stokes' theorem. If you have the vector field $\mathbf{F} (x,y)$, you know that $$\int_{D} (\mathbf{\nabla}\times \mathbf{F})_z\,dx\,dy =\int_{\partial D} \mathbf{F}\cdot d\mathbf{r}.$$
With $\mathbf{F}=- \tfrac12 x y^2 \mathbf{e}_x$, you find that \begin{align} \int_{D} xy\,dx\,dy& = \int_{\partial D} \mathbf{F}\cdot d\mathbf{r} = \underbrace{\int_0^4 F_x(x,0)\,dx}_{=0} + \underbrace{\int_0^3 F_y(4,y)\,dy}_{=0} - \int_0^1 [4 F_x(\mathbf{r}) + \underbrace{3 F_y(\mathbf{r})}_{=0}]\,dt\\ &=\int_0^1 2 [4(1-t)][3(1-t)]^2\,dt = 72 \int_0^1(1-t)^3\,dt \end{align} where we parameterized the last segment via $\mathbf{r}(t) = \begin{pmatrix}4(1-t)\\ 3(1-t)\end{pmatrix}$, $t\in[0,1]$.