# Evaluate the given integral by making an appropriate change of variables: \iint_R 2...

## Question:

Evaluate the given integral by making an appropriate change of variables: {eq}\iint_R 2 \frac{x-4y}{5x-y} dA {/eq}, where {eq}R {/eq} is the parallelogram enclosed by the lines {eq}x-4y=0, x-4y=6, 5x-y=5, 5x-y=10 {/eq}

## Evaluate a Double Integral :

The Jacobian of {eq}x {/eq} and {eq}y {/eq} with respect to {eq}u {/eq} and {eq}v, {/eq} where {eq}x=g(u,v) {/eq} and {eq}y=h(u,v) {/eq} is denoted by

{eq}\displaystyle j=\frac{\partial (x,y)}{\partial (u,v)}=\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}{/eq}

We use the Jacobian to change variables in a double integral by the formula

{eq}\displaystyle \int \int _{\mathbb{R}}f(x,y)dy\: dx=\int \int _{\mathbb{S}}f(g(u,v),h(u,v))\left |j \right |du\: dv {/eq}

## Answer and Explanation:

Consider the integral

{eq}\displaystyle \iint_R 2\left( \frac{x-4y}{5x-y} \right) dA\\ \displaystyle R:\; \; x-4y=0, \; x-4y=6,\; 5x-y=5, \; 5x-y=10 {/eq}

Let us assume that The substitution :

{eq}\displaystyle u = x - 4y\\ \displaystyle v = 5x - y {/eq}

Solving these two equations we get

{eq}\displaystyle x=\frac{1}{19}(4v-u) ,\; \; y=\frac{1}{19}(v-5u)\\ \displaystyle j=\frac{\partial (x,y)}{\partial (u,v)}=\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix}=\displaystyle \begin{vmatrix} -\frac{1}{19} & \frac{4}{19}\\ -\frac{5}{19} & \frac{1}{19} \end{vmatrix}=-\frac{1}{(19)^{2}}+\frac{20}{(19)^{2}}=\frac{19}{(19)^{2}}=\frac{1}{19} {/eq}

Therefore, the limit of integration is as follows

{eq}\displaystyle 0\leq u\leq 6,\quad 5\leq v\leq 10 {/eq}

Therefore, the integral become as follows

{eq}\displaystyle \int \int _{\mathbb{R}}f(x,y)dy\: dx=\int \int _{\mathbb{S}}f(g(u,v),h(u,v))\left |j \right |du\: dv\\ \displaystyle \iint_R 2 \left( \frac{x-4y}{5x-y} \right) dA=\int_{v=5}^{10}\int_{u=0}^{6} 2\cdot \frac{u}{v}\cdot \frac{1}{19}du\: dv\\ \displaystyle =\int _5^{10}\left(\frac{36}{19v}\right)dv\\ \displaystyle =\frac{36}{19}\left[\ln \left(v\right)\right]_5^{10}\\ \displaystyle \boxed{ \iint_R 2 \left( \frac{x-4y}{5x-y} \right) dA =\frac{36}{19}\left(\ln \left(10\right)-\ln \left(5\right)\right) } {/eq}

#### Learn more about this topic:

from GRE Math: Study Guide & Test Prep

Chapter 15 / Lesson 4