## Limit for Double Integrals When Boundaries are Functions

Evaluation of double integrals is often simpple when the region of integration is rectangular eg
$0\leq x\leq 1, 2 \leq y\leq 3$
. It is not so simple when one or more boundaries of the region of integration is given by an equation in the variables.
Suppose we are to evaluate
$\int_R xy dR$
over the region
$R$
satisfying
$x+2y \leq 4, x-3y \geq6, x \geq0$
. We can evaluate this integral by considering the limits. For the region
$R$

$x/3 -2 \leq y\leq2-x/2$
and
$0 \leq x\leq 24/5$

Not that
$\frac{24}{5}$
is the
$x$
intecpt of the top and bottom boundaries of
$R$
.
We can write the integral as
\begin{equation} \begin{aligned} \int^{24/5}_0 \int^{2-x/2}_{x/3-2} xy dy dx &= \int^{24/5}_0 [xy^2/2]^{2-x/2}_{x/3-2} dx \\ &= \int^{24/5}_0 ((x(2-x/2)^2 - x(x/3-2)^2) dx \\ &= \int^{24/5}_0 (-2x^2/3+5x^3/36) dx \\ &= [-2x^3/9 5x^4/144]^{24/5}_0 \\ &= 663/125 \end{aligned} \end{equation}

Notice that on the left hand boundary
$x=0$
, a constant, and on the right hand boundary,
$x=24/5$
.
This is why we integrated with respect to
$y$
first, then
$x$
. 