Use the given transformation to evaluate the integral.
-$$\begin{array} { l }
\mathrm { u } = \mathrm { x } + \mathrm { y } , \mathrm { v } = \mathrm { x } - \mathrm { y } ; \
\quad \int _ { \mathrm { R } } \int ^ { \mathrm { s } } ( \mathrm { x } + \mathrm { y } ) \sin ( \mathrm { x } - \mathrm { y } ) \mathrm { dx } \mathrm { dy } ,
\end{array}$$
where $R$ is the parallelogram bounded by the lines $y = x , y = x - 9 , y = - x , y = - x + 4$
A) 8(1 - cos 9)
B) 4(1 - cos 9)
C) 8(1 + cos 9)
D) 4(1 + cos 9)

Question

Use the given transformation to evaluate the integral.
-$$\begin{array} { l } 
\mathrm { u } = \mathrm { x } + \mathrm { y } , \mathrm { v } = \mathrm { x } - \mathrm { y } ; \
\quad \int _ { \mathrm { R } } \int ^ { \mathrm { s } } ( \mathrm { x } + \mathrm { y } ) \sin ( \mathrm { x } - \mathrm { y } ) \mathrm { dx } \mathrm { dy } ,
\end{array}$$
where $R$ is the parallelogram bounded by the lines $y = x , y = x - 9 , y = - x , y = - x + 4$
A) 8(1 - cos 9)
B) 4(1 - cos 9)
C) 8(1 + cos 9)
D) 4(1 + cos 9)

Quizplus · Accepted Answer

The given transformation $u = x + y, v = x - y$ simplifies the integral into a form that involves $u$ and $v$, making it easier to evaluate. The Jacobian of the transformation is $|\frac{\partial(x, y)}{\partial(u, v)}| = \frac{1}{2}$, which accounts for the change in area elements from $dx dy$ to $du dv$. The limits of integration for $u$ and $v$ correspond to the given parallelogram's boundaries in the $x-y$ plane, translating into a rectangular region in the $u-v$ plane. The integral simplifies to an expression involving $\sin(v)$ and $u$, which, after integration, yields the result $4(1 - \cos(9))$, taking into account the area scaling factor from the Jacobian.

Use the Given Transformation to Evaluate the Integral Where RRR Is the Parallelogram Bounded by the Lines

Use the Given Transformation to Evaluate the Integral Where $R$ Is the Parallelogram Bounded by the Lines