## Task Condition

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

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

Since the $O y$ axis is the axis of rotation, the volume

Integrals $21-25$ are found using the formula:

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

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

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

![](https://cdn.mathpix.com/cropped/2024_05_22_cfec74fe4e2360126fc5g-49.jpg?height=1210&width=1194&top_left_y=1094&top_left_x=728)

$$
y=x^{2}=x^{3} \Rightarrow x_{1}=0 ; x_{2}=1
$$

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

$$
V=\pi \cdot \int_{0}^{1}(\sqrt[3]{y})^{2} d y-\pi \cdot \int_{0}^{1}(\sqrt{y})^{2} d y=\pi \cdot \int_{0}^{1}\left(y^{2 / 3}-y\right) d y=
$$

$$
=\left.\pi \cdot\left(\frac{3}{5} \cdot y^{5 / 3}-\frac{y^{2}}{2}\right)\right|_{0} ^{1}=\pi \cdot\left(\frac{3}{5}-\frac{1}{2}\right)-(0-0)=\frac{\pi}{10}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{\%} 0 \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD}$ $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+21-25$ » Categories: Kuznetsov's Problem Book Integrals Problem 21 | Integrals

- Last edited: 05:52, 1 June 2010.
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Created by GeeTeatoo

## Problem Kuznetsov Integrals 21-26

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