Question

Use the given transformation to evaluate the integral.

\(\iint_{R}{{}}(4x+8y)dA,\) where R is the parallelogram with vertices \((-1,3),(1,-3),(3,-1)\) and \((1,5);\)\(x=\frac{1}{4}(u+v),\;y=\frac{1}{4}(v-3u)\) \((1,5);\)

Short answer

The required integral is \(\mathop{\iint }_{R}(4x+8y)dA=192\)

Step-by-step solution

Formula

Formula for the Jacobian

\(\frac{{\partial (x,y)}}{{\partial (u,v)}} = \left| {\begin{array}{*{20}{l}}{\frac{{\partial x}}{{\partial u}}}&{\frac{{\partial x}}{{\partial v}}}\\{\partial y}&{\partial y}\end{array}} \right|\)

Find partial derivatives and using Jacobian formula

Firstly, we need to find partial derivatives

\(\begin{aligned}{}&\begin{aligned}\frac{{\partial x}}{{\partial u}} &= \frac{1}{4}\\\frac{{\partial x}}{{\partial v}} &= \frac{1}{4}\end{aligned}\\{}&\begin{aligned}\frac{{\partial y}}{{\partial u}} &= \frac{{ - 3}}{4}\\\frac{{\partial y}}{{\partial v}} &= \frac{1}{4}\end{aligned}\end{aligned}\)

And by plugging them in the formula for the Jacobian we get:

\(\begin{aligned}{\frac{{\partial (x,y)}}{{\partial (u,v)}}}&{\: = \left| {\begin{aligned}{\frac{1}{4}}&{\frac{1}{4}}\\{\frac{{ - 3}}{4}}&{\frac{1}{4}}\end{aligned}} \right|}\\{}&{\: = \frac{1}{4}}\end{aligned}\)

Simplify the integral equation

Now, we know the substitution we are using is:

\(\begin{array}{*{20}{c}}{}&{u = x - y}\\{}&{v = 3x + y}\end{array}\)

With \(-4\le u\le 4\) and \(0\le v\le 8\)

Therefore,

\(\begin{aligned}\mathop{\iint }_{R}(4x+8y)dA&=\int _{-4}^{4}\int _{0}^{8}\left( \frac{-5u+3v}{4} \right)dvdu \\&=\int _{-4}^{4}\left( \frac{-5uv}{4}+\frac{3{{v}^{2}}}{8} \right)_{0}^{8}du \\&=\int _{-4}^{4}(-10u+24)du \\&=\left( \frac{-10{{u}^{2}}}{2}+24u \right)_{-4}^{4} \\&=192 \\\end{aligned}\)

Hence, the required integral is \(\mathop{\iint }_{R}(4x+8y)dA=192\)