Example 1. Calculate the surface area of $z=\frac{x y}{a}$, located inside the cylinder $x^{2}+y^{2}=a^{2}, a>0$.
Solution. We have $z_{x}^{\prime}=\frac{y}{a}, \quad z_{y}^{\prime}=\frac{x}{a}, \quad \sqrt{1+z_{z}^{\prime 2}+z_{y}^{\prime 2}}=$ $=\frac{1}{a} \sqrt{a^{2}+y^{2}+x^{2}}$. The domain of integration is the circle $x^{2}+y^{2} \leqslant a^{2}$. We will transition to polar coordinates. We have:

$$
\begin{gathered}
S=\int_{D} \frac{1}{a} \sqrt{a^{2}+x^{2}+y^{2}} d x d y=\frac{1}{a} \int_{D} \sqrt{a^{2}+r^{2}} r d r d \varphi= \\
=\frac{1}{a} \int_{0}^{2 \pi} d \varphi \int_{0}^{a}\left(a^{2}+r^{2}\right)^{1 / 2} d\left(a^{2}+r^{2}\right) \cdot \frac{1}{2}= \\
=\left.\left.\frac{1}{a} \cdot \varphi\right|_{0} ^{2 \pi} \cdot \frac{\left(a^{2}+r^{2}\right)^{3 / 2}}{3}\right|_{0} ^{a}=\frac{2 \pi}{3} a^{2}(2 \sqrt{2}-1)
\end{gathered}
$$

$$
\text { Answer. } S=\frac{2 \pi}{3} a^{2}(2 \sqrt{2}-1) \text {. }
$$