Evaluate $\iint_D\sin(xy)dA$ where $D$ is bounded by $y=\frac 1x, y=\frac2x, y=x, y=2x$ in the first quadrant. Evaluate $\iint_D\sin(xy)dA$ where $D$ is bounded by $y=\frac 1x, y=\frac2x, y=x, y=2x$ in the first quadrant.
By subbing numbers into the equation, I see that $1\leq x\leq 2, 1\leq y\leq 2.$
Without solving the equation, can someone tell me if this is correct? The final answer I got was $-\frac14\sin(4)+\frac 54\sin(2)-\sin(1)$

Okay, here's what I got. Evaluating this double integral seems to be much harder on usual cartesian coordinates, since the region is pretty ugly.

Courtesy of Mathematica. If we perform the coordinate change
$$
\begin{cases}
xy & = u \\
\frac{y}{x} & = v,
\end{cases}
$$
we get the much neater region that is a square.

What we have to do now is find the Jacobian and integrate. We have
$$dA = dx \, dy = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \, du \, dv.$$
From our definitions we find
$$
\begin{align}
x & = \sqrt{\frac{u}{v}} \\
y & = \sqrt{uv}.
\end{align}
$$
Then
$$
\begin{align}
\mathcal{J} & = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \\
 & = \begin{vmatrix} \frac{1}{2} \frac{1}{\sqrt{uv}} & - \frac{1}{2} \frac{\sqrt{u}}{v} \\ \frac{1}{2} \sqrt{\frac{v}{u}} & \frac{1}{2} \sqrt{\frac{u}{v}} \end{vmatrix} \\
 & = \frac{1}{4} \left[ \frac{1}{v} + \sqrt{v} \right].
\end{align}
$$
This is nonzero in the region considered, thus a valid coordinate change. Rewriting the integral we get
$$I = \iint\limits_{D} \sin (xy) \, dA = \int_1^2 \hspace{-5pt} \int_1^2 \sin (u) \frac{1}{4} \left[ \frac{1}{v} + \sqrt{v} \right] \, du \, dv.$$
Evaluating this I found
$$I = \frac{1}{4} \left( (\cos (1) - \cos(2)) \left( \ln (2) + \frac{2}{3} \left( 2^{3/2} -1 \right) \right) \right).$$
Using Mathematica I numerically evaluated this, and $I \approx 0,457206$. This seems correct to me.
Note: We must take the absolute value of the Jacobian, but in this case it is positive so the absolute value is unnecessary.
