Exercise 9.9 - Chapter 9 Applications of Integration 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $9.9$
Question 1.
Find, by integration, the volume of the solid generated by revolving about the $\mathrm{x}$-axis, the region enclosed by $y=2 x^2, y=0$ and $x=1$
Solution:

$
\begin{aligned}
\mathrm{V} & =\int_a^b \pi y^2 d x \\
& =\int_0^1 \pi\left(2 x^2\right)^2 d x=4 \pi \int_0^1 x^4 d x \\
\mathrm{~V} & =4 \pi\left[\frac{x^5}{5}\right]_0^1=\frac{4 \pi}{5} \text { cubic units. }
\end{aligned}
$
Question 2.
Find, by integration, the volume of the solid generated by revolving about the $\mathrm{x}$-axis, the region enclosed by $y=e^{-2 x} y=0, x=0$ and $x=1$
Solution:
$
\begin{aligned}
\mathrm{V} & =\int_a^b \pi y^2 d x \\
& =\int_0^1 \pi\left(e^{-2 x}\right)^2 d x \\
& =\pi \int_0^1 e^{-4 x} d x=\pi\left[-\frac{e^{-4 x}}{4}\right]_0^1 \\
& =\frac{\pi}{4}\left[-e^{-4}+e^0\right]=\frac{\pi}{4}\left[1-e^{-4}\right] \text { cubic units. }
\end{aligned}
$

Question 3.
Find, by integration, the volume of the solid generated by revolving about the y-axis, the region enclosed by $x^2=1+y$ and $y=3$.
Solution:

$
\begin{aligned}
\mathrm{V} & =\int_a^b \pi x^2 d y \\
& =\int_{-1}^3 \pi(1+y) d y \\
& =\pi\left[y+\frac{y^2}{2}\right]_{-1}^3=\pi\left[\left(3+\frac{9}{2}\right)-\left(-1+\frac{1}{2}\right)\right] \\
\mathrm{V} & =\pi\left[\frac{15}{2}+\frac{1}{2}\right]=\pi\left[\frac{16}{2}\right]=8 \pi \text { cubic units. }
\end{aligned}
$
Question 4.
The region enclosed between the graphs of $y=x$ and $y=x^2$ is denoted by $R$, Find the volume generated when $\mathrm{R}$ is rotated through $360^{\circ}$ about $\mathrm{x}$ - axis.
Solution:

To find points of intersection, solving $y=x^2$ and $y=x$, we get $(0,0)$ and $(1,1)$
$
\begin{aligned}
& \text { Volume }=\pi \int_0^1\left[(f(x))^2-(g(x))^2\right] d x \\
& =\pi \int_0^1\left[(x)^2-\left(x^2\right)^2\right] d x=\pi \int_0^1\left(x^2-x^4\right) d x \\
& =\pi\left[\frac{x^3}{3}-\frac{x^5}{5}\right]_0^1=\pi\left[\frac{1}{3}-\frac{1}{5}\right]=\frac{2 \pi}{15} \text { cubic units. }
\end{aligned}
$

Question 5.
Find, by integration, the volume of the container which is in the shape of a right circular conical frustum as shown in the Figure.
Solution:

$
\begin{aligned}
\text { Volume of the frustum } & =\int_0^h \pi\left[\frac{(\mathrm{R}-r)\left(\frac{\mathrm{R} h}{\mathrm{R}-r}-y\right)}{h}\right]^2 \\
\text { Here } r=1 ; \mathrm{R}=2 ; h & =2 \\
& =\int_0^2 \pi\left[\frac{(4-y)}{2}\right]^2 d y=\frac{\pi}{4} \int_0^2(4-y)^2 d y \\
& =\frac{\pi}{4} \int_0^2\left(16+y^2-8 y\right) d y=\frac{\pi}{4}\left[16 y+\frac{y^3}{3}-\frac{8 y^2}{2}\right]_0^2 \\
& =\frac{\pi}{4}\left[32+\frac{8}{3}-16\right]=\frac{\pi}{4}\left[16+\frac{8}{3}\right] \\
& =\frac{\pi}{4}\left[\frac{56}{3}\right]=\frac{14}{3} \pi \mathrm{m}^3 .
\end{aligned}
$
Question 6.
A watermelon has an ellipsoid shape which can be obtained by revolving an ellipse with major-axis 20 $\mathrm{cm}$ and minor-axis $10 \mathrm{~cm}$ about its major-axis. Find its volume using integration.
Solution:
From the given data $\mathrm{a}=10 \mathrm{~cm}$ and $\mathrm{b}=5 \mathrm{~cm}$
Equation of the Ellipse

$\begin{aligned}
& \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\
& \frac{x^2}{100}+\frac{y^2}{25}=1 \\
& \frac{y^2}{25}=1-\frac{x^2}{100} \\
& \text { Volume }=\int_a^b=\frac{25}{100}\left(100-y^2 d x=\frac{1}{4}\left(100-x^2\right)\right. \\
& =\int_{-10}^{10} \frac{\pi}{4}\left(100-x^2\right) d x=\frac{2 \pi}{4} \int_0^{10}\left(100-x^2\right) d x \\
& =\frac{\pi}{2}\left[100 x-\frac{x^3}{3}\right]_0^{10}=\frac{\pi}{2}\left[1000-\frac{1000}{3}\right] \\
& =\frac{\pi}{2}\left[\frac{2000}{3}\right]=\frac{1000}{3} \pi \mathrm{cm}^3 .
\end{aligned}$