Use Methods of Variable Transformation to find the value of $\iint _{D} e^{( y-2x)} dxdy,\ D=\{( x,y) |5\leqslant 3y-x\leqslant 10,-4y\leqslant 2x\leqslant y\}$
Here is my thinking:
\begin{array}{l} Method\ 1:\ Jacobian\\ \\ Let\ u=3y-x,\ v=2x\\ \begin{cases} x=\frac{1}{2} v\\ y=\frac{1}{3}\left( u+\frac{1}{2} v\right) \end{cases} \ \ \ \ \ \ \ \ \ \ \\ \therefore 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} 0 & \frac{1}{2}\\ \frac{1}{3} & \frac{1}{6} \end{vmatrix} =-\frac{1}{6}\\ \iint _{D} e^{( y-2x)} dxdy=\int _{5}^{10}\int _{-\frac{4}{3}\left( u+\frac{1}{2} v\right)}^{\frac{1}{3}\left( u+\frac{1}{2} v\right)} \ e^{\frac{1}{3}\left( u+\frac{1}{2} v\right) -v} \cdotp |J|dvdu\\ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =........( Please\ let\ me\ omit\ the\ hard\ process)\\ \\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =-\frac{3}{20} \cdotp e^{\frac{1}{2} v}\left( e^{\frac{40}{3}} -e^{\frac{20}{3}}\right)\left( e^{\frac{10}{9}} -e^{-\frac{5}{18}}\right) \ ( werid\ result)\\ \\ \\ Method\ 2:\ without\ Variable\ Transformation\\ 5\leqslant 3y-x\leqslant 10\Longrightarrow \frac{5+x}{3} \leqslant y\leqslant \frac{10+x}{3}\\ -4y\leqslant 2x\leqslant y\Longrightarrow \frac{y}{2} \leqslant x\leqslant -2y\\ \\ \iint _{D} e^{( y-2x)} dxdy\\ \\ =\int _{\frac{5+x}{3}}^{\frac{10+x}{3}} e^{y}\int _{-2y}^{\frac{y}{2}} e^{-2x} dxdy\ \ \ \ \ \ \ \ \ ( Let\ u=-2x,\ du=-2dx)\\ \\ =\int _{\frac{5+x}{3}}^{\frac{10+x}{3}} e^{y}\int _{-y}^{4y} e^{u} dudy\\ \\ =\int _{\frac{5+x}{3}}^{\frac{10+x}{3}}\left( e^{5y} -1\right) dy\\ \\ \\ =e^{\frac{50+5x}{3}} -e^{\frac{25+5x}{3}} -\frac{5}{3} \end{array} I am unsure if the process of Method 1 is correct, especially regarding the range of v and u. Additionally, the results obtained from the two methods do not match (I apologize for not having the correct answer). Therefore, I would appreciate your advice on this matter. Thank you.
To begin, the first problem I see with your integrals is that despite them being definite integrals, you obtain results with variables. They should not be present at all. A definite integral definitely has a defined value (when the integral is nice).
Since this fundamental mistake was made, I decided to walk through the entire process of computing this integral, both for my sake as well as anyone else who might wish to read this answer.
So just for reference here are some images depicting the domain of integration.
$$ D=\left\{(x,y)|(5 \le 3y-x\le 10)\wedge(-4y \le 2x \le y)\right\} $$
Perhaps this is too many images, so a moderator is welcome to modify the presentation of these images. However, we see that there are four points of interest which define a trapezoidal domain of integration.
$$ I_D = \iint_D e^{y-2x} dx dy $$
Firstly, it is worth noting that $I_D \ge 0$. This is because the integrand is positive of the domain of integration and the axes are not reversed. It is also the case that computing the antiderivative of this is relatively trivial, however if we choose to calculate the integral as presented with respect to x the integral splits into three separate pairs
$$ I_D = \int_{x=-4}^{x=-2} \int_{y=-\frac{1}{2}x}^{y=\frac{x+10}{3}} e^{y-2x} dy dx + \int_{x=-2}^{x=1} \int_{y=\frac{x+5}{3}}^{y=\frac{x+10}{3}} e^{y-2x} dy dx + \int_{x=1}^{x=2} \int_{y=2x}^{y=\frac{x+10}{3}} e^{y-2x} dy dx $$
This is obviously annoying to calculate. It would be nice if we could specify a coordinate transformation that nicely captures $D$ as a simple square. To this end we define an affine transformation
$$ \vec{r} = x\hat{x}+y\hat{y} = \begin{bmatrix} x\\y \end{bmatrix} = \begin{bmatrix} -2\\1\end{bmatrix} + u\left(\begin{bmatrix} 3\\1\end{bmatrix}\right) + v\left(\begin{bmatrix} -2\\1\end{bmatrix}\right) \\ \Rightarrow \left\{\begin{matrix} x=&-2+3u-2v \\ y=&1+u+v \end{matrix}\right. \\ \Rightarrow \begin{bmatrix}x\\y\\1\end{bmatrix} = \begin{bmatrix}3&-2&-2\\1&1&1\\0&0&1\end{bmatrix} \begin{bmatrix}u\\v\\1\end{bmatrix} $$
Inverting the matrix yields the inverse affine transformation
$$ \begin{bmatrix}u\\v\\1\end{bmatrix} = \begin{bmatrix}3&-2&-2\\1&1&1\\0&0&1\end{bmatrix}^{-1} \begin{bmatrix}x\\y\\1\end{bmatrix} = \frac{1}{5} \begin{bmatrix}1&2&0\\-1&3&-5\\0&0&5\end{bmatrix} \begin{bmatrix}x\\y\\1\end{bmatrix}\\ \Rightarrow \left\{\begin{matrix}u=\frac{x+2y}{5}\\v=\frac{-x+3y-5}{5}\end{matrix}\right. $$
This yields the jacobian matrix and corresponding determinant.
$$ J = \left|\begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{matrix}\right|= \begin{bmatrix} 3&-2\\ 1&1 \end{bmatrix} |J|=5 $$
The domain can be expressed in terms of the $(u,v)$ coordinates. The first two boundaries become
$$ 5\le 3x-y \le 10 \rightarrow 5 \le 3(u+v+1)-(3u-2v-2) \le 10 \\ \rightarrow 5 \le 5v+5 \le 10 \\ \rightarrow 0 \le v \le 1 \\ \ \\ -4y \le 2x \le y \rightarrow -4(u+v+1) \le 2(3u-2v-2) \le u+v+1 \\ \rightarrow -4u-4v-4 \le 6u-4v-4 \le u+v+1 \\ \rightarrow 0 \le 10u \le 5u+5v+5 \\ \rightarrow (0 \le 2u)\wedge( 2u \le u+v+1) \\ \rightarrow (0/2 \le u)\wedge( u \le v+1) \\ \rightarrow 0 \le u \le v+1 $$
$$ D= \left\{ (u,v) | ( 0 \le v \le 1 ) \wedge ( 0 \le u \le v+1 ) \right\} $$
We can see the domain in the new coordinates is much nicer.
This all yields the integral $I_D$
$$ I_D = \iint_D e^{y-2x} dx dy \int_{v=0}^{v=1}\int_{u=0}^{u=v+1} e^{u+v+1-2(3u-2v-2)} \left|5\right| du dv \\ = 5 \int_{v=0}^{v=1}\int_{u=0}^{u=v+1} e^{-5u+5v+5} du dv \\ = 5e^5 \int_{v=0}^{v=1} e^{5v} \int_{u=0}^{u=v+1} e^{-5u} du dv \\ = 5e^5 \int_{v=0}^{v=1} e^{5v} \left[\frac{-1}{5}e^{-5u}\right]_{u=0}^{u=v+1} dv \\ = -e^5 \int_{v=0}^{v=1} e^{5v} \left(e^{-5v-5}-1\right) dv \\ = -e^5 \int_{v=0}^{v=1} e^{-5}-e^{5v} dv \\ = -e^5 \left[ e^{-5}v-\frac{1}{5}e^{5v} \right]_{v=0}^{v=1} \\ = -e^5\left( \left( e^{-5}-\frac{1}{5}e^{5} \right) - \left( -\frac{1}{5} \right)\right) \\ = \frac{e^{10}}{5} - \frac{e^5}{5} - 1 \approx 4374.61 $$
I took the liberty of plugging the original integral into Wolfram Alpha, and obtained the same result.