Use the given transformation to evaluate the integral. ,where R is the square with vertices (0,0),(4,6),(6, ),(10,2)and

Question

Use the given transformation to evaluate the integral.  ,where R is the square with vertices (0,0),(4,6),(6,  ),(10,2)and

Quizplus · Accepted Answer

We will use the transformation $x=u+2v+4$ and $y=2u+3v+6$. This maps the vertices of the square R to the vertices $(4,6)$, $(10,14)$, $(14,17)$, and $(8,9)$ in the $uv$-plane. Also, we have the Jacobian of this transformation is $J= \begin{vmatrix} 1 & 2\2 & 3 \end{vmatrix}= -1$. Then the integral becomes 
\begin{align*}
\iint_R 3y-x\,dA &= \int_0^4\int_{u}^{6-u/2}(3(2u+3v+6)-(u+2v+4))(-1)dv\,du \
&= \int_0^4\int_{u}^{6-u/2}(-u+6v+16)\,dv\,du \
&= \int_0^4\left[\frac{-uv+3v^2+16v}{3}ight]_{v=u}^{v=6-u/2}\,du \
&= \int_0^4 \frac{(2u-6)(u-4)}{3}\,du = \frac{32}{3}.
\end{align*} 
Therefore, the answer is $312$.

Use the Given Transformation to Evaluate the Integral