17-22 use the given transformation to evaluate the integral. 17. $iint_r (x - 3y) da$, where $r$ is the…

Explanation:

Step1: Find the Jacobian

The Jacobian $J$ of the transformation $x = 2u + v,y=u + 2v$ is given by the determinant $J=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{vmatrix}$. We have $\frac{\partial x}{\partial u}=2,\frac{\partial x}{\partial v}=1,\frac{\partial y}{\partial u}=1,\frac{\partial y}{\partial v}=2$. So $J=\begin{vmatrix}2&1\1&2\end{vmatrix}=2\times2 - 1\times1=3$.

Step2: Express $x - 3y$ in terms of $u$ and $v$

Substitute $x = 2u + v$ and $y=u + 2v$ into $x - 3y$: $x-3y=(2u + v)-3(u + 2v)=2u + v-3u - 6v=-u - 5v$.

Step3: Find the new - region in the $uv$ - plane

For the point $(0,0)$: Set $x = 0,y = 0$, then $\begin{cases}2u + v=0\u + 2v=0\end{cases}$, solving gives $u = 0,v = 0$. For the point $(2,1)$: Set $x = 2,y = 1$, then $\begin{cases}2u + v=2\u + 2v=1\end{cases}$. Multiply the second - equation by 2, we get $2u+4v = 2$. Subtract the first equation from it: $(2u + 4v)-(2u + v)=2 - 2$, $3v=0$, so $v = 0$ and $u = 1$. For the point $(1,2)$: Set $x = 1,y = 2$, then $\begin{cases}2u + v=1\u + 2v=2\end{cases}$. Multiply the first equation by 2, we get $4u+2v = 2$. Subtract the second equation from it: $(4u + 2v)-(u + 2v)=2 - 2$, $3u=0$, so $u = 0$ and $v = 1$. The region $S$ in the $uv$ - plane is the triangular region with vertices $(0,0),(1,0),(0,1)$. The double - integral becomes $\iint_{R}(x - 3y)dA=\iint_{S}(-u - 5v)|J|dudv$. Since $|J| = 3$, the integral is $3\int_{0}^{1}\int_{0}^{1 - u}(-u - 5v)dvdu$.

Step4: Evaluate the double - integral

First, integrate with respect to $v$: $\int_{0}^{1 - u}(-u - 5v)dv=\left[-uv-\frac{5v^{2}}{2}\right]{0}^{1 - u}=-u(1 - u)-\frac{5(1 - u)^{2}}{2}=-u+u^{2}-\frac{5(1 - 2u+u^{2})}{2}=-u+u^{2}-\frac{5}{2}+5u-\frac{5u^{2}}{2}=-\frac{3u^{2}}{2}+4u-\frac{5}{2}$. Then integrate with respect to $u$: $3\int{0}^{1}\left(-\frac{3u^{2}}{2}+4u-\frac{5}{2}\right)du=3\left[-\frac{3u^{3}}{6}+\frac{4u^{2}}{2}-\frac{5u}{2}\right]_{0}^{1}=3\left(-\frac{1}{2}+2-\frac{5}{2}\right)=3\times(-1)=-3$.

Answer:

$-3$