## Task Condition

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

$$
y^{2}=x-2, y=0, y=x^{3}, y=1
$$
## Solution

Since the $O y$ axis is the axis of rotation, the volume is found using the formula:

Integrals $21-24$

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

Express $x$ in terms of $y$ and find the limits of integration:

$$
\begin{aligned}
& y^{2}=x-2 \Rightarrow x=y^{2}+2 \\
& y=x^{3} \Rightarrow x=\sqrt[3]{y}
\end{aligned}
$$

From the problem statement, we already have: $y_{1}=0 ; y_{2}=1$.

Find the volume of the solid as the difference of the volumes of two solids of revolution:

$$
\begin{aligned}
& V=\pi \cdot \int_{0}^{1}\left(y^{2}+2\right)^{2} d y-\pi \cdot \int_{0}^{1}(\sqrt[3]{y})^{2} d y= \\
& =\pi \cdot \int_{0}^{1}\left(y^{4}+4 y^{2}+4-y^{2 / 3}\right) d y= \\
& =\left.\pi \cdot\left(\frac{y^{5}}{5}+4 \frac{y^{3}}{3}+4 y-\frac{3}{5} \cdot y^{5 / 3}\right)\right|_{0} ^{1}=\pi \cdot\left(\left(\frac{1}{5}+\frac{4}{3}+4-\frac{3}{5}\right)-(0+0+0-0)\right)=\frac{3+4 \cdot 5+4 \cdot 15-3 \cdot 3}{15} \pi=\frac{24}{5} \pi
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\�\�\�\�\� \%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80\%D0\%B0\%D0\%BB\%D1\%8B_21-24»

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

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## Problem Kuznetsov Integrals $21-25$

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