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

The transformation is $x=\frac{1}{4}(u + v)$ and $y=\frac{1}{4}(v-3u)$. The partial - derivatives are $\frac{\partial x}{\partial u}=\frac{1}{4}$, $\frac{\partial x}{\partial v}=\frac{1}{4}$, $\frac{\partial y}{\partial u}=-\frac{3}{4}$, $\frac{\partial y}{\partial v}=\frac{1}{4}$. The Jacobian $J$ of the transformation 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 $4x + 8y$: [4x+8y=4\times\frac{1}{4}(u + v)+8\times\frac{1}{4}(v-3u)=u + v+2v-6u=- 5u + 3v]

Step3: Find the new limits of integration

We use the transformation $x=\frac{1}{4}(u + v)$ and $y=\frac{1}{4}(v-3u)$ to find the new vertices. For the point $(-1,3)$: [\begin{cases}-1=\frac{1}{4}(u + v)\3=\frac{1}{4}(v-3u)\end{cases}] Multiply the first equation by 4: $-4=u + v$, multiply the second by 4: $12=v-3u$. Subtract the first new - equation from the second: $12-(-4)=v-3u-(u + v)$, $16=-4u$, $u=-4$, and $v = 0$. For the point $(1,-3)$: [\begin{cases}1=\frac{1}{4}(u + v)\-3=\frac{1}{4}(v-3u)\end{cases}] Multiply by 4: $4=u + v$, $-12=v-3u$. Subtract: $4-(-12)=u + v-(v-3u)$, $16 = 4u$, $u = 4$, $v = 0$. For the point $(3,-1)$: [\begin{cases}3=\frac{1}{4}(u + v)\-1=\frac{1}{4}(v-3u)\end{cases}] Multiply by 4: $12=u + v$, $-4=v-3u$. Subtract: $12-(-4)=u + v-(v-3u)$, $16 = 4u$, $u = 4$, $v = 8$. For the point $(1,5)$: [\begin{cases}1=\frac{1}{4}(u + v)\5=\frac{1}{4}(v-3u)\end{cases}] Multiply by 4: $4=u + v$, $20=v-3u$. Subtract: $20 - 4=v-3u-(u + v)$, $16=-4u$, $u=-4$, $v = 8$. The region in the $uv$ - plane is a rectangle with $-4\leq u\leq4$ and $0\leq v\leq8$.

Step4: Compute the double - integral

The double - integral $\iint_R(4x + 8y)dA$ in the $uv$ - plane is: [\iint_R(4x + 8y)dA=\int_{0}^{8}\int_{-4}^{4}(-5u + 3v)\left|J\right|dudv] Since $\left|J\right|=\frac{1}{4}$, we have: [\int_{0}^{8}\int_{-4}^{4}\frac{1}{4}(-5u + 3v)dudv=\frac{1}{4}\int_{0}^{8}\left(-\frac{5}{2}u^{2}+3vu\right)\big|{u=-4}^{u = 4}dv] [=\frac{1}{4}\int{0}^{8}\left(-\frac{5}{2}(4^{2}-(-4)^{2})+3v(4-(-4))\right)dv=\frac{1}{4}\int_{0}^{8}(24v)dv] [=\frac{1}{4}\times12v^{2}\big|_{0}^{8}=\frac{1}{4}\times12\times64 = 192]

Answer:

$192$