## Task Condition

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

$$
y=\sqrt{x-1}, y=0, y=1, x=0.5
$$
## 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:

![](https://cdn.mathpix.com/cropped/2024_05_22_cfec74fe4e2360126fc5g-42.jpg?height=846&width=1111&top_left_y=979&top_left_x=815)

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

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:

$$
\begin{aligned}
& V=\pi \cdot \int_{0}^{1}\left(y^{2}+1\right)^{2} d y-\pi \cdot \int_{0}^{1}\left(\frac{1}{2}\right)^{2} d y= \\
& V=\pi \cdot \int_{0}^{1}\left(y^{4}+2 y^{2}+1\right) d y-\pi \cdot \int_{0}^{1} \frac{1}{4} d y= \\
& =\left.\pi \cdot\left(\frac{y^{5}}{5}+\frac{2 y^{3}}{3}+y\right)\right|_{0} ^{1}-\left.\pi \cdot \frac{y}{4}\right|_{0} ^{1}= \\
& =\pi \cdot\left(\frac{1}{5}+\frac{2}{3}+1\right)-\pi \cdot \frac{1}{4}=\pi \cdot \frac{97}{60}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+21-21$ »

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

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- Last edited on this page: 05:23, 9 July 2010.
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