let (d) be the region (0leq yleq x) and (0leq xleq 4). evaluate the integral by making the change of…

Explanation:

Step1: Find the Jacobian

First, find the partial - derivatives. Given $x = u + v$ and $y=u - v$, we have $\frac{\partial x}{\partial u}=1$, $\frac{\partial x}{\partial v}=1$, $\frac{\partial y}{\partial u}=1$, $\frac{\partial y}{\partial v}=-1$. The Jacobian $J=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix}1&1\1& - 1\end{vmatrix}=-1 - 1=-2$, and $|J| = 2$. Also, $x + y=(u + v)+(u - v)=2u$.

Step2: Find the new limits of integration

From $0\leq y\leq x$ and $0\leq x\leq4$. Substitute $x = u + v$ and $y = u - v$. $y\leq x$ gives $u - v\leq u + v$, which simplifies to $v\geq0$. $y\geq0$ gives $u - v\geq0$ or $u\geq v$. $x\leq4$ gives $u + v\leq4$. $x\geq0$ and $y\geq0$ also imply non - negative values for $u$ and $v$. The new region in the $uv$ - plane is $0\leq v\leq u$ and $0\leq u\leq4$.

Step3: Evaluate the integral using the change of variables

The double - integral $\iint_{D}(x + y)dxdy$ becomes $\iint_{D^{*}}2u|J|dudv=\int_{0}^{4}\int_{0}^{u}2u\times2dvdu$. First, integrate with respect to $v$: $\int_{0}^{4}2u\times2\left[v\right]{0}^{u}du=\int{0}^{4}4u\times udu=\int_{0}^{4}4u^{2}du$. Then, integrate with respect to $u$: $4\times\frac{u^{3}}{3}\big|_{0}^{4}=\frac{4\times4^{3}}{3}=\frac{256}{3}$.

Step4: Evaluate the integral directly

The original integral $\iint_{D}(x + y)dxdy=\int_{0}^{4}\int_{0}^{x}(x + y)dydx$. First, integrate with respect to $y$: $\int_{0}^{4}\left[xy+\frac{y^{2}}{2}\right]{0}^{x}dx=\int{0}^{4}\left(x\cdot x+\frac{x^{2}}{2}-0\right)dx=\int_{0}^{4}\left(x^{2}+\frac{x^{2}}{2}\right)dx=\int_{0}^{4}\frac{3x^{2}}{2}dx$. Then, integrate with respect to $x$: $\frac{3}{2}\times\frac{x^{3}}{3}\big|_{0}^{4}=\frac{1}{2}\times4^{3}=\frac{64}{2}\times2=\frac{256}{3}$.

Answer:

$\frac{256}{3}$