consider the following integral and transformation. ∬_r (20x + 15y) da, where r is the parallelogram with…

Explanation:

Step1: Find the Jacobian

First, find the partial - derivatives. Given $x=\frac{1}{5}(u + v)$ and $y=\frac{1}{5}(v-4u)$. $\frac{\partial x}{\partial u}=\frac{1}{5}$, $\frac{\partial x}{\partial v}=\frac{1}{5}$, $\frac{\partial y}{\partial u}=-\frac{4}{5}$, $\frac{\partial y}{\partial v}=\frac{1}{5}$. The Jacobian $J=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{vmatrix}=\frac{1}{5}\times\frac{1}{5}-(-\frac{4}{5})\times\frac{1}{5}=\frac{1 + 4}{25}=\frac{1}{5}$.

Step2: Transform the region

The transformation $x=\frac{1}{5}(u + v)$ and $y=\frac{1}{5}(v-4u)$ maps the parallelogram $R$ in the $xy$ - plane to a rectangle in the $uv$ - plane. We can find the limits of integration for $u$ and $v$ by substituting the vertices of the parallelogram into the transformation equations. After substitution, we find that the region in the $uv$ - plane is $- 10\leq u\leq10$ and $0\leq v\leq20$. So $A = 20$.

Step3: Evaluate the double - integral

The original integral $\iint_R(20x + 15y)dA$ becomes $\int_{-10}^{10}\int_{0}^{20}\frac{1}{5}(7v-8u)dvdu$. First, integrate with respect to $v$: $\int_{-10}^{10}\left[\frac{1}{5}\left(\frac{7v^{2}}{2}-8uv\right)\big|{v = 0}^{v = 20}\right]du=\int{-10}^{10}\frac{1}{5}\left(\frac{7\times20^{2}}{2}-8u\times20\right)du$. $=\int_{-10}^{10}\frac{1}{5}(1400 - 160u)du=\int_{-10}^{10}(280-32u)du$. Integrate with respect to $u$: $\left(280u-16u^{2}\right)\big|_{-10}^{10}=(280\times10-16\times10^{2})-(280\times(-10)-16\times(-10)^{2})$. $=(2800 - 1600)-(-2800 - 1600)=1200 + 4400=5600$.

Answer:

$A = 20$ The value of the integral is $5600$