Question

Evaluate the double integral i=∫∫dxyda where d is the triangular region with vertices (0,0),(4,0),(0,4).

Answer 1

$I=\displaystyle\iint_D\mathrm dA=\int_{x=0}^{x=4}\int_{y=0}^{y=4-x}\mathrm dy\,\mathrm dx$
$I=\displaystyle\int_{x=0}^{x=4}(4-x)\,\mathrm dx$
$I=8$

To verify this, we can simply find the area of the triangle using the well-known formula $\dfrac12bh$, where $b$ and $h$ are the base and height, respectively of the triangular region $D$. We have $b=h=4$, so the area is $\dfrac{4^2}2=8$, as expected.