Example 3.31. Compute the volume of the body bounded by the surfaces $z=0, x=0, y=x, z=1-y \sqrt{y}$.
Solution. We will construct a drawing showing the body bounded by the specified surfaces (Fig. 51a). The volume of the body \( V = \iint_{D} z \, dx \, dy \), where \( z = 1 - y \sqrt{y} \) is the equation of the surface bounding the body from above, and \( D \) is the region of integration, representing a triangle in the \( xOy \) plane bounded by the lines \( x = 0 \), \( y = x \), and \( y = 1 \), which are the lines of intersection of the planes \( x = 0 \), \( y = x \), and \( z = 1 - y \sqrt{y} \) (Fig. 51b). The vertices of this triangle are the points \((0; 0)\), \((0; 1)\), and \((1; 1)\), obtained by solving the systems of equations formed from the corresponding pairs of side equations. This region can be considered as enclosed between the lines \( y = 0 \) and \( y = 1 \); to the left, the region is bounded by the line \( x = 0 \), and to the right by the line \( x = y \). Therefore,

\[
\begin{aligned}
V & = \iint_{D} (1 - y \sqrt{y}) \, dx \, dy = \int_{0}^{1} dy \int_{0}^{y} (1 - y \sqrt{y}) \, dx = \int_{0}^{1} (1 - y \sqrt{y}) x \bigg|_{0}^{y} \, dy = \\
& = \int_{0}^{1} (1 - y \sqrt{y})_{y} \, dy = \int_{0}^{1} y - y^{\frac{5}{2}} \, dy = \frac{1}{2} y^{2} - \frac{2}{7} y^{\frac{7}{2}} \bigg|_{0}^{1} = \frac{3}{14}
\end{aligned}
\]

Thus, \( V = \frac{3}{14} \) cubic units.

## Section 4

## DIFFERENTIAL EQUATIONS

## 4.1. Basic Concepts

An equation that relates the independent variable, the unknown function, and its derivatives or differentials of various orders is called a differential equation.

The order of a differential equation is the order of the highest derivative appearing in the equation. For example, the equation \( y' \sin x + y \operatorname{tg} x = 1 \) is of the first order; \( \frac{d^{2} y}{d x^{2}} = x^{3} \) is of the second order; \( y''' - 5 x y' + x y = 0 \) is of the third order.

A function \( y = \phi(x) \) that satisfies a differential equation is called a solution of the equation. The graph of a solution is called an integral curve of the equation.

If a function that satisfies a differential equation is given implicitly, i.e., by a relation of the form \( \phi(x, y) = 0 \), then it is called an integral of the equation.

A solution of a differential equation that contains as many independent arbitrary constants as the order of the equation is called the general solution of the equation. For a first-order equation, the general solution has the form:

\[
y = \phi(x, C)
\]

and for a second-order equation, it has the form:

\[
y = \phi(x, C_{1}, C_{2})
\]

Functions obtained from the general solution by assigning different numerical values to the arbitrary constants are called particular solutions of the equation.

Geometrically, the general solution defines a family of curves, and a particular solution defines a specific curve in this family.

To find a particular solution of a differential equation, initial conditions are specified. For a first-order equation, they have the form \( y(x_{0}) = y_{0} \); for a second-order equation, they have the form \( y(x_{0}) = y_{0} \), \( y'(x_{0}) = y'_{0} \). Using these initial conditions, the values of the arbitrary constants in the general solution are determined, resulting in particular solutions that satisfy the given initial conditions.