# Evaluate the following integrate. \int\int_R \sqrt \frac {x - y}{x + y + 1} dA. Where R is the...

## Question:

Evaluate the following integrate.

{eq}\int\int_R \sqrt \frac {x - y}{x + y + 1} dA. {/eq}

Where R is the square with vetices (0, 0), (1, -1) (2, 0) and (1, 1).

## Double integral over flat regions:

Let {eq}f(x,y) {/eq} be a continuous function in a certain region {eq}R \subset \mathbb{R}^{2} {/eq}. The integral of {eq}f {/eq} over the region can be denoted as follow:

{eq}I=\iint_{R} f(x, y) d y d x {/eq}

If the region is a square {eq}[a, b] \times[c, d] {/eq}, then

{eq}I=\int_{a}^{b} \int_{c}^{d} f(x, y) d y d x {/eq}

This double integral evaluates two simple integral simultaneously,

{eq}I=\int_{a}^{b}\left\{\int_{c}^{d} f(x, y) d y\right\} d x {/eq}

where the variable is consider a constant in the internal integral.

Since {eq}g(x)=\int_{c}^{d} f(x, y) d y {/eq} is a function of {eq}x {/eq}, finally we evaluate the integral {eq}I=\int_{a}^{b} g(x) d x {/eq}.

To evaluate {eq}f(x, y)=\sqrt{\frac{x-y}{x+y+1}} {/eq} in the given region it is necessary taking into account the following change of variable:

{eq}\begin{aligned} &u=x-y\\ &v=x+y \end{aligned} {/eq}

then, {eq}0 \leq u, v \leq 2 {/eq} and

{eq}\iint_{R} f(x, y) d y d x=\int_{0}^{2} \int_{0}^{2} f(u, v) J d u d v {/eq}

where

{eq}J=\left|\begin{array}{ll} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{array}\right|=\left|\begin{array}{cc} \frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{array}\right|=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} {/eq}

Thus,

{eq}\begin{aligned} \iint_{R} f(x, y) d y d x &=\frac{1}{2} \int_{0}^{2} \int_{0}^{2} \sqrt{\frac{u}{v+1}} d u d v \\ &=\frac{1}{2} \int_{0}^{2} \sqrt{u} d u \int_{0}^{2} \frac{d v}{\sqrt{v+1}} \\ &=\frac{1}{2} \frac{2}{3} \left[u^{3 / 2}\right]_{0}^{2} 2\left[\sqrt{v+1}\right]_{0}^{2}\\ &=\frac{4\sqrt{2}\left(\sqrt{3}-1\right)}{3} \end{aligned} {/eq}