# Evaluate the integral by making an appropriate change of variables. Let \iint_R 3...

## Question:

Evaluate the integral by making an appropriate change of variables.

Let {eq}\iint_R 3 \cos\Big(3\Big(y - \frac{x}{y} + x\Big)\Big) dA {/eq} where R is the trapezoidal region with vertices {eq}(7, 0) {/eq}, {eq}(8, 0) {/eq}, {eq}(0, 8) {/eq}, and {eq}(0, 7) {/eq}.

## Transformations To Solve Double Integrals:

If a function F(x,y) is to be integrated over a region R bounded by some given curves in XY coordinate system, then the integration can be simplified under som transformation {eq}\displaystyle x=g(u,v) {/eq} and {eq}\displaystyle y=h(u,v) {/eq} which transform region R to some region Q such that the integral is transformed to some simpler integral given by,

$$\displaystyle \iint_RF(x,y) \ dA=\iint_QF(g(u,v),h(u,v))\left|\frac{\partial(x,y)}{\partial(u,v)}\right| \ d\bar{A}$$

where,{eq}\displaystyle \left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\frac{\partial x}{\partial u}\frac{\partial y}{\partial v}-\frac{\partial x}{\partial v}\frac{\partial y}{\partial u} \quad \text{ and } \quad \displaystyle d\bar{A}=\ du \ dv {/eq}

To ensure that the transformation simplifies the calculation, the choice of transformation is the key. Else we may end up with an integral which would be even more difficult to handle.

## Answer and Explanation:

Please refer the figure given below

Let us consider the transformation {eq}\displaystyle u=3(x-y) \text{ and }v=y+x {/eq}. Under this transformation the x and y values in terms of u and v will be given by,

{eq}\displaystyle \Rightarrow x=\frac{u+3v}{6}\text{ and }y=\frac{3v-u}{6} {/eq}

The region R is given bounded by the four given points forming a trapezoid as shown in the above figure, which will have four straight lines with the equations as shown below (under the considered transformations)

{eq}\displaystyle y=-x+7\Rightarrow \frac{3v-u}{6}=-\frac{3v+u}{6}+7\Rightarrow v=7\\ \displaystyle y=-x+8\Rightarrow \frac{3v-u}{6}=-\frac{3v+u}{6}+8\Rightarrow v=8\\ \displaystyle y=0\Rightarrow \frac{3v-u}{6}=0\Rightarrow u=3v\\ \displaystyle x=0\Rightarrow \frac{3v+u}{6}=0\Rightarrow u=-3v\\ {/eq}

Next the Jacobian will be

{eq}\displaystyle \left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\frac{1}{6}\frac{1}{2}-\frac{1}{2}\frac{-1}{6}=\frac{1}{6} {/eq}

So the given integral will be transformed to

{eq}\displaystyle \begin{align} \iint_R3\cos\left( 3\left( \frac{y-x}{y+x}\right)\right) \ dA&=\iint_R3\cos\left( 3\left( \frac{x-y}{y+x}\right)\right) \ dA& \because \cos(-a)=\cos(a)\\ &=\int_{7}^{8}\int_{-3v}^{3v}3\cos\left( \frac{u}{v}\right)\left|\frac{1}{6}\right| \ du \ dv\\ &=\frac{1}{2}\int_{7}^{8}\left[ \sin\left( \frac{u}{v}\right)(v)\right]_{-3v}^{3v} \ du \ dv\\ &=\frac{1}{2}\int_{7}^{8}\left[v(\sin(3))-v\sin(-3)\right] \ dv\\ &=\frac{1}{2}\int_{7}^{8}\left[2v\sin(3)\right] \ dv&\because \sin(-a)=-\sin(a)\\ &=\sin(3)\int_{7}^{8}v \ dv\\ &=\sin(3)\left[\frac{v^2}{2}\right]_{7}^{8}\\ &=\frac{\sin(3)}{2}[64-49]\\ &=\frac{15\sin(3)}{2}\approx 1.0584 \end{align} {/eq}

#### Learn more about this topic:

Double Integration: Method, Formulas & Examples

from AP Calculus AB & BC: Help and Review

Chapter 12 / Lesson 15
43K