## Task Condition

Calculate the volumes of solids formed by rotating figures bounded by the graphs of functions. The axis of rotation is $O x$.

$$
x=\sqrt[3]{y-2}, x=1, y=1
$$
## Solution

## Since

the $O x$ 

axis 

is 

the 

axis 

of 

rotation, the volume is found 

by 

the formula:

![](https://cdn.mathpix.com/cropped/2024_05_22_cfec74fe4e2360126fc5g-11.jpg?height=1505&width=1242&top_left_y=1021&top_left_x=541)

$$
V=\pi \cdot \int_{a}^{b} y^{2} d x
$$

Let's find the limits of integration: From the problem statement, we already have: $x_{2}=1$ Now let's find the lower limit:

$$
y=x^{3}+2=1 \Rightarrow x_{1}=-1
$$

We will find the volume of the solid as the difference in volumes of two solids of revolution:

$$
\begin{aligned}
& V=\pi \cdot \int_{-1}^{1}\left(\left(x^{3}+2\right)^{2}-1^{2}\right) d x= \\
& =\pi \cdot \int_{-1}^{1}\left(x^{6}+4 x^{3}+4-1\right) d x= \\
& =\pi \cdot\left(\int_{-1}^{1} x^{6} d x+4 \int_{-1}^{1} x^{3} d x+3 \int_{-1}^{1} d x\right)= \\
& =\pi \cdot\left(\left.\frac{x^{7}}{7}\right|_{-1} ^{1}+\left.4 \cdot \frac{x^{4}}{4}\right|_{-1} ^{1}+\left.3 \cdot x\right|_{-1} ^{1}\right)= \\
& =\pi \cdot\left(\frac{1^{7}}{7}-\frac{(-1)^{7}}{7}+1^{4}-(-1)^{4}+3 \cdot 1-3 \cdot(-1)\right)= \\
& =\pi \cdot\left(\frac{1}{7}+\frac{1}{7}+1-1+3+3\right)= \\
& =\pi \cdot\left(\frac{2}{7}+6\right)=\frac{44}{7} \cdot \pi
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+21-6 \%$

Categories: Kuznetsov's Problem Book Integrals Problem 21 | Integrals | Problems for Checking

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## Problem Kuznetsov Integrals 21-7

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