Example. Calculate the volume of the solid formed by rotating the region bounded by the graphs of the functions

$$
y=x^{3} \quad y=\sqrt{x}
$$

around the $O X$ axis.
## Solution.

1. Define the region $D$:

a) find the abscissas $a$ and $b$ of the points of intersection of the graphs. For this, solve the equation

$$
x^{3}=\sqrt{x}
$$

We get

$$
a=0, \quad b=1
$$

b) on the interval $[0,1] \sqrt{x} \geq x^{3}$. Therefore, $u(x)=x^{3}$ and $v(x)=\sqrt{x}$.

2. Calculate the volume using formula (1):

$$
V=\pi \int_{0}^{1}\left[(\sqrt{x})^{2}-\left(x^{3}\right)^{2}\right] d x=\pi \int_{0}^{1}\left(x-x^{6}\right) d x=\frac{5 \pi}{14} .
$$

Answer. $5 \pi / 14$ (units of length) $)^{3}$.

Conditions of the problems. Calculate the volumes of solids formed by rotating around the $O X$ axis the regions bounded by the graphs of the given functions.

1. $y=-x^{2}+1, \quad y=0$. 2. $y=\sin (\pi x / 2), y=x$.
2. $y=x^{2}$,

$y=\sqrt{x}$.
4. $y=x^{2}, y=2 x$.
5. $y=\cos x$,

$y=\sin x$

$x=0$

$(0 \leq x \leq \pi / 4)$.
6. $y=\sin ^{2} x$,

$y=0$,

$x=\pi / 2$

$(0 \leq x \leq \pi / 2)$.
7. $y=e^{x}$,

$y=1$,

$x=1$.
8. $y=\ln x$,

$y=0, \quad x=e$.
9. $y=\frac{2}{x}$,

$y=1$,

$x=1$.
10. $y=\cos ^{2} x$

$y=0$

$(-\pi / 2 \leq x \leq \pi / 2)$.

Answers. 1. $16 \pi / 15$. 2. $\left(5 \pi^{2}-48\right) /(6 \pi)$. 3. $3 \pi / 10$. 4. $16 \pi / 15$. 5. $\pi^{2} / 4-1 / 2$. 6. $3 \pi^{2} / 16$. 7. $\pi\left(e^{2}-4 e+5\right) / 2$. 8. $\pi(e-2)$. 9. $\pi(3-4 \ln 2)$. 10. $3 \pi^{2} / 8$.

## Chapter 9

## CURVILINEAR INTEGRALS

In the study of the topic CURVILINEAR INTEGRALS, you will become familiar with the concepts of curvilinear integrals of the first kind (along the arc length) and the second kind (along coordinates) of functions of two and three variables and learn how to compute them along various planar and spatial curves given parametrically, in Cartesian and polar coordinates, reducing curvilinear integrals to definite integrals.

Using the package SOLVER.CA, you can compute derivatives of functions defining the curve, solve systems of equations determining the boundary values of the parameter, compute the definite integrals you obtain, and verify the correctness of the results you have obtained.

## 9.1. Curvilinear Integrals of the First Kind

Problem Statement. Compute the curvilinear integral

$$
\int_{L} f(x, y, z) d l
$$

where $L$ is a part of a smooth curve given parametrically

$$
\left\{\begin{array}{l}
x=x(t) \\
y=y(t), \\
z=z(t)
\end{array} \quad t_{1} \leq t \leq t_{2}\right.
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_5878ce417e2ca061000bg-203.jpg?height=43&width=549&top_left_y=1794&top_left_x=76)

Plan of Solution. The curvilinear integral of the first kind along the curve $L$ is defined by the formula

$$
\int_{L} f(x, y, z) d l=\int_{t_{1}}^{t_{2}} f(x(t), y(t), z(t)) \sqrt{x^{\prime}(t)^{2}+y^{\prime}(t)^{2}+z^{\prime}(t)^{2}} d t
$$

Note that the curvilinear integral of the first kind does not depend on the direction of traversal of the curve and always $t_{1} \leq t_{2}$.

1. Compute $x^{\prime}(t), y^{\prime}(t), z^{\prime}(t)$ and $d l=\sqrt{x^{\prime}(t)^{2}+y^{\prime}(t)^{2}+z^{\prime}(t)^{2}} d t$.
2. Compute the curvilinear integral using formula (1) and write down the answer.

Remark 1. If the boundary points of the curve $L M\left(x_{1}, y_{1}, z_{1}\right)$ and $N\left(x_{2}, y_{2}, z_{2}\right)$ are given in Cartesian coordinates, then $t_{1}$ and $t_{2}$ are determined by solving the systems of equations

$$
\left\{\begin{array} { l } 
{ x _ { 1 } = x ( t _ { 1 } ) , } \\
{ y _ { 1 } = y ( t _ { 1 } ) , } \\
{ z _ { 1 } = z ( t _ { 1 } ) ; }
\end{array} \quad \left\{\begin{array}{l}
x_{2}=x\left(t_{2}\right) \\
y_{2}=y\left(t_{2}\right) \\
z_{2}=z\left(t_{2}\right)
\end{array}\right.\right.
$$

Remark 2. If the curve is given as the intersection of two surfaces:

$$
\left\{\begin{array}{l}
F_{1}(x, y, z)=0 \\
F_{2}(x, y, z)=0
\end{array}\right.
$$

then it must be parameterized.

Remark 3. If a planar curve is given by the equation $y=y(x)$ $(a \leq x \leq b)$, then the differential of the arc length is $d l=\sqrt{1+y^{\prime}(x)^{2}} d x$ and formula (1) has the form

$$
\int_{L} f(x, y) d l=\int_{a}^{b} f(x, y(x)) \sqrt{1+y^{\prime}(x)^{2}} d x
$$

If a planar curve is given in polar coordinates $(x=\varrho \cos \varphi$, $y=\varrho \sin \varphi)$ by the equation $\varrho=\varrho(\varphi)(a \leq \varphi \leq b)$, then the differential of the arc length is

$$
d l=\sqrt{\varrho(\varphi)^{2}+\varrho^{\prime}(\varphi)^{2}} d \varphi
$$

and formula (1) has the form

$$
\int_{L} f(x, y) d l=\int_{a}^{b} f(\varrho(\varphi) \cos \varphi, \varrho(\varphi) \sin \varphi) \sqrt{\varrho(\varphi)^{2}+\varrho^{\prime}(\varphi)^{2}} d \varphi
$$