# Use the given transformation to evaluate the integral: \iint_R (3x+12y)dA, where R is the...

## Question:

Use the given transformation to evaluate the integral: {eq}\iint_R (3x+12y)dA {/eq}, where {eq}R {/eq} is the parallelogram with vertices {eq}(-1,2), (1,-2),(4,1) {/eq} and {eq}(2,5); \ x = \frac{1}{3} (u+v), y = \frac {1}{3}(v-2u) {/eq}

## Integral:

We know that {eq}\int \int _Rf(x,y)\,dx\,dy=\int _u\int _vf(u,v)J\,dv\,du {/eq}

{eq}J=\left | \begin{matrix} \frac{\partial x}{\partial u}&\frac{\partial y}{\partial u}\\\frac{\partial x}{\partial v}&\frac{\partial y}{\partial v}\end{matrix} \right | {/eq}

{eq}x = \frac{1}{3} (u+v), y = \frac {1}{3}(v-2u) {/eq}

Find f(u,v) by putting values of x and y in {eq}f(x,y)=3x+12y {/eq}

{eq}x = \frac{1}{3} (u+v), y = \frac {1}{3}(v-2u) {/eq}

To find: J

{eq}J=\left | \begin{matrix} \frac{\partial x}{\partial u}&\frac{\partial y}{\partial u}\\\frac{\partial x}{\partial v}&\frac{\partial y}{\partial v}\end{matrix} \right |\\ =\left | \begin{matrix} \frac{1}{3}&\frac{-2}{3}\\\frac{1}{3}&\frac{1}{3}\end{matrix} \right |\\ =\frac{1}{9}+\frac{2}{9}\\ =\frac{3}{9}\\ =\frac{1}{3} {/eq}

To find: {eq}f(u,v) {/eq}:

{eq}f(x,y)=3x+12y {/eq}

Put {eq}x = \frac{1}{3} (u+v), y = \frac {1}{3}(v-2u) {/eq}

{eq}f(u,v)=3\left [\frac{1}{3} (u+v) \right ]+12\left [ \frac {1}{3}(v-2u) \right ]\\ =u+v+4v-8u\\ =-7u+5v {/eq}

Also, {eq}-3\leq u\leq 3\,,\,0\leq v\leq 9 {/eq}

So,

{eq}\int \int _R 3x+12y\,dx\,dy=\frac{1}{3}\int_{-3}^{3}\int_{0}^{9}-7u+5v\,dv\,du\\ =\frac{1}{3}\int_{-3}^{3}\left [-7uv+\frac{5v^2}{2} \right ]_{0}^{9}\,du\\ =\frac{1}{3}\int_{-3}^{3}\left [ -63u+\frac{405}{2} \right ]\,du\\ =\frac{1}{3}\left [\frac{-63u^2}{2}+\frac{405u}{2} \right ]_{-3}^{3}\\ =\frac{1}{3}\left [ 1215 \right ]\\ =405 {/eq}