# Evaluate the integral over the region D described below. Double integral over D of 6y dA, D is...

## Question:

Evaluate the integral over the region {eq}D {/eq} described below.

{eq}\iint_{D} 6y \, \mathrm{d}A, \; D {/eq} is the region bounded by {eq}\; y = \frac{x}{2}, \; y = 0, \; x = 4 {/eq}.

## Double Integrals

The integral {eq}\displaystyle \iint_{\mathcal{D}}f(x,y) dA, {/eq} is evaluated as an iterative integral

{eq}\displaystyle \int_a^b\int_{u(x)}^{v(x)}f(x,y) dydx, {/eq}

if the region of integration is given as a y-simple region, {eq}\displaystyle \mathcal{D}=\{(x,y)|\, a\leq x\leq b, u(x)\leq y\leq v(x)\}. {/eq}

If the region is given as a x-simple region, {eq}\displaystyle \mathcal{D}=\{(x,y)|\, c\leq y\leq d, w(y)\leq x\leq h(y)\}, {/eq}

the integral is written as {eq}\displaystyle \int_c^d\int_{w(y)}^{h(y)}f(x,y) dxdy, {/eq}

## Answer and Explanation:

The region of integration for {eq}\displaystyle \int_{\mathcal{D}}6y\ dA {/eq}

is bounded above by the line {eq}\displaystyle y=\frac{x}{2}, {/eq} below, by the x-axis,

for {eq}\displaystyle 0\leq x\leq 4. {/eq}

So, the region is described as a y-simple region, whose integral is written iteratively as

{eq}\displaystyle \begin{align*}\int_{\mathcal{D}}6y\ dA &=\int_0^4 \int_0^{x/2} 6y \ dy dx\\ &= \int_0^4 3y^2 \bigg\vert_{y=0}^{y=x/2} \ dx\\ &= \int_0^4\frac{3}{4}x^2\ dx\\ &= \frac{1}{4} x^3 \bigg\vert_0^4\\ &= \frac{1}{4} 4^3 =\boxed{16}. \end{align*} {/eq}