1.14. The plate is bounded by the parabola $y^{2}=x$ and its chord passing through the focus perpendicular to the axis of the parabola. Find the mass of the plate if the surface density at each point is inversely proportional to the distance from the point to the directrix of the parabola.
Solution. The plate is shown in Fig. 1.38 with the focus and directrix indicated (for more details, see Part I, Chapter 3).

According to the problem, \( p(x, y) = \frac{K}{\left(x + \frac{1}{4}\right)} \), therefore

$$
M(\Omega) = \iint_{\Omega} p(x, y) d x d y = K \int_{0}^{1 / 4} d x \int_{-\sqrt{x}}^{\sqrt{x}} \frac{d y}{x + \frac{1}{4}} = 2 K \int_{0}^{1 / 4} \frac{\sqrt{x} d x}{x + \frac{1}{4}}
$$

To compute the last integral, we make the substitution \(\sqrt{x} = t\), then

$$
\begin{gathered}
M(\Omega) = 2 K \int_{0}^{1 / 4} \frac{\sqrt{x} d x}{x + \frac{1}{4}} = 4 K \int_{0}^{1 / 2} \frac{t^{2} d t}{t^{2} + \frac{1}{4}} = 4 K \left( \frac{1}{2} - \frac{1}{4} \int_{0}^{1 / 2} \frac{d t}{t^{2} + \frac{1}{4}} \right) = \\
= 2 K \left( 1 - \left. \operatorname{arctg} 2 t \right|_{0}^{1 / 2} \right) = 2 K \left( 1 - \frac{\pi}{4} \right)
\end{gathered}
$$