## Task Condition

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

$$
y=\ln x, x=2, y=0
$$
## Solution

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

$$
y=\ln x \Rightarrow x=e^{y}
$$

From the problem statement, we already have: $y_{1}=0$. Now let's find the upper limit:

![](https://cdn.mathpix.com/cropped/2024_05_22_cfec74fe4e2360126fc5g-44.jpg?height=1497&width=1131&top_left_y=979&top_left_x=811)

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

$$
V=\pi \cdot \int_{0}^{\ln 2}(2)^{2} d y-\pi \cdot \int_{0}^{\ln 2}\left(e^{y}\right)^{2} d y=
$$

$$
\begin{aligned}
& =\left.\pi \cdot 4 y\right|_{0} ^{\ln 2}-\left.\pi \cdot \frac{e^{2 y}}{2}\right|_{0} ^{\ln 2}= \\
& \left.=\pi \cdot(4 \ln 2-0)-\frac{1}{2}\left(e^{2 \ln 2}-e^{0}\right)\right)= \\
& =\pi \cdot\left(4 \ln 2-\frac{1}{2}(4-1)\right)=\pi \cdot\left(4 \ln 2-\frac{3}{2}\right)
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 84 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_21-22$ »

Categories: Kuznetsov's Problem Book Integrals Problem 21 | Integrals

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## Problem Kuznetsov Integrals $21-23$

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