Question

Use the given transformation to evaluate the integral. 8xy dA R , where R is the region in the first quadrant bounded by the lines y = 2 3 x and y = 3 2 x and the hyperbolas xy = 2 3 and xy = 3 2 ; x = u/v, y = v

Answer 1

$\displaystyle\iint_R8xy\,\mathrm dA$

Assuming $R$ is the region bounded by the curves,

$\begin{cases}y=\frac23x\\y=\frac32x\\xy=\frac23\\xy=\frac32\end{cases}$

then substituting $x=\dfrac uv$ and $y=v$ changes the boundaries of $R$ in the $u$-$v$ plane to

$\begin{cases}v=\frac23\frac uv\\v=\frac32\frac uv\\\frac uvv=\frac23\\\frac uvv=\frac32\end{cases}\iff\begin{cases}v=\sqrt{\frac23u}\\v=\sqrt{\frac32u}\\u=\frac23\\v=\frac32\end{cases}$

so that the region $R$ is given by all the points in the set $\right\{(u,v)~:~\sqrt{\dfrac23u}\le v\le\sqrt{\dfrac32u},\,\dfrac23\le u\le\dfrac32\right\}$.

Now the area element of the integral over $R$ in the $u$-$v$ plane is given by the determinant of the following Jacobian matrix:

$J=\dfrac{\partial(x,y)}{\partial(u,v)}=\begin{bmatrix}x_u&x_v\\y_u&y_v\end{bmatrix}$
$\det J=\begin{vmatrix}\frac1v&-\frac u{v^2}\\0&1\end{vmatrix}=\dfrac1v$
$|\det J|=\mathrm dA=\mathrm dx\,\mathrm dy=\dfrac1v\,\mathrm du\,\mathrm dv$

So the integral is equivalent to

$\displaystyle\iint_R8xy\,\mathrm dA=\int_{u=2/3}^{u=3/2}\int_{v=\sqrt{2u/3}}^{v=\sqrt{3u/2}}8\frac uv v\frac1v\,\mathrm dv\,\mathrm du$
$=\displaystyle8\int_{u=2/3}^{u=3/2}\int_{v=\sqrt{2u/3}}^{v=\sqrt{3u/2}}\frac uv\,\mathrm dv\,\mathrm du$
$=\displaystyle8\int_{u=2/3}^{u=3/2}u\ln v\bigg|_{v=\sqrt{2u/3}}^{v=\sqrt{3u/2}}\,\mathrm du$
$=\displaystyle8\ln\frac32\int_{u=2/3}^{u=3/2}u\,\mathrm du$
$=\displaystyle4\ln\frac32u^2\bigg|_{u=2/3}^{u=3/2}$
$=\dfrac{65}{18}\ln\dfrac32$