3. use the transformation given by $x = \\frac{1}{4}(u + v), y = \\frac{1}{4}(v - 3u)$ to compute the double…

Explanation:

Step1: Find the Jacobian

First, find the partial - derivatives. $\frac{\partial x}{\partial u}=\frac{1}{4}$, $\frac{\partial x}{\partial v}=\frac{1}{4}$, $\frac{\partial y}{\partial u}=-\frac{3}{4}$, $\frac{\partial y}{\partial v}=\frac{1}{4}$. The Jacobian $J$ of the transformation $x = \frac{1}{4}(u + v)$ and $y=\frac{1}{4}(v - 3u)$ is given by the determinant $J=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix}\frac{1}{4}&\frac{1}{4}\-\frac{3}{4}&\frac{1}{4}\end{vmatrix}=\frac{1}{4}\times\frac{1}{4}-\left(-\frac{3}{4}\right)\times\frac{1}{4}=\frac{1 + 3}{16}=\frac{1}{4}$.

Step2: Express $4x + 8y$ in terms of $u$ and $v$

Substitute $x=\frac{1}{4}(u + v)$ and $y=\frac{1}{4}(v - 3u)$ into $4x + 8y$. [ \begin{align*} 4x+8y&=4\times\frac{1}{4}(u + v)+8\times\frac{1}{4}(v - 3u)\ &=(u + v)+2(v - 3u)\ &=u + v+2v-6u\ &=- 5u + 3v \end{align*} ]

Step3: Find the limits of integration

We use the transformation $x=\frac{1}{4}(u + v)$ and $y=\frac{1}{4}(v - 3u)$ to find the limits of integration for $u$ and $v$. For the line segment between $(-1,3)$ and $(1, - 3)$: [ \begin{cases} -1=\frac{1}{4}(u + v)\ 3=\frac{1}{4}(v - 3u) \end{cases}\text{ and }\begin{cases} 1=\frac{1}{4}(u + v)\ -3=\frac{1}{4}(v - 3u) \end{cases} ] Solving these systems gives the appropriate bounds. The region $R$ in the $uv$ - plane is a rectangle with $-2\leq u\leq2$ and $2\leq v\leq6$.

Step4: Compute the double - integral

The double - integral $\iint_R(4x + 8y)dA$ in the $uv$ - plane is $\iint_{S}(-5u + 3v)|J|dudv$, where $S$ is the rectangle in the $uv$ - plane and $|J|=\frac{1}{4}$. [ \begin{align*} \iint_{S}(-5u + 3v)\times\frac{1}{4}dudv&=\frac{1}{4}\int_{2}^{6}\int_{-2}^{2}(-5u + 3v)dudv\ &=\frac{1}{4}\int_{2}^{6}\left[-\frac{5}{2}u^{2}+3vu\right]{u = - 2}^{u = 2}dv\ &=\frac{1}{4}\int{2}^{6}\left(-\frac{5}{2}(4 - 4)+3v(2+2)\right)dv\ &=\frac{1}{4}\int_{2}^{6}12vdv\ &=\frac{1}{4}\times12\times\left[\frac{1}{2}v^{2}\right]_{2}^{6}\ & = 3\times\frac{1}{2}(36 - 4)\ &=48 \end{align*} ]

Answer:

$48$