Example 1. Calculate the area of the figure bounded by the curve $(2 x+3 y+5)^{2}+(x+2 y-3)^{2}=64$.
Solution. Let $\left\{\begin{array}{l}2 x+3 y+5=u, \\ x+2 y-3=v,\end{array}\right.$ or $\left\{\begin{array}{l}x=2 u-3 v-19, \\ y=-u+2 v+11 .\end{array}\right.$

We have: $\frac{\partial x}{\partial u}=2, \frac{\partial x}{\partial v}=-3, \frac{\partial y}{\partial u}=-1, \frac{\partial y}{\partial v}=2, I(u, v)=\left|\begin{array}{rr}2 & -3 \\ -1 & 2\end{array}\right|=1$.

In the new coordinates $(u, v)$, the curve has the form $u^{2}+v^{2}=64$. This means that the written formulas transform the circle $u^{2}+v^{2} \leqslant 64$ (region $G$) into the given figure $D:(2 x+3 y+5)^{2}+(x+2 y-3)^{2} \leqslant 64$ (ellipse with interior). Since the area of the circle $G$ is $\sigma=\pi R^{2}=64 \pi$, we have

$$
S=\iint_{D} d x d y=\iint|I(u, v)| d u d v=64 \pi
$$

## Answer. $S=64 \pi$.