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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!

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closed as off-topic by 900 sit-ups a day, Thomas, martini, studiosus, user91500 Jul 8 at 15:28

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

2 Answers 2

$x$ 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.

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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 $\mathbf{r}(t) = \begin{pmatrix}4(1-t)\\ 3(1-t)\end{pmatrix}$, $t\in[0,1]$.

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