Exercise 9.9-Additional Problems - Chapter 9 - Applications of Integration - 12th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Additional Questions
Question 1.
Find the volume of the solid that results when the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad(\mathrm{a}>\mathrm{b}>0)$ is revolved about the minor axis.
Solution:

Volume of the solid is obtained by revolving the right side of the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ about the $y$-axis.
Limits for $y$ is obtained by putting $x=0 \Rightarrow y^2=b^2 \Rightarrow y=\pm b$.
From the given curve $x^2=\frac{a^2}{b^2}\left(b^2-y^2\right)$
$\therefore$ Volume is given by
$
\mathrm{V}=\int_c^d \pi x^2 d y=\int_{-b}^b \pi \frac{a^2}{b^2}\left(b^2-y^2\right) d y=2 \pi \frac{a^2}{b^2}\left(b^2 y-\frac{y^3}{3}\right)_0^b-a
$
- $=2 \pi \frac{a^2}{b^2}\left(b^3-\frac{b^3}{3}\right)=\frac{4 \pi}{3} a^2 b$ cubic units.

Question 2.
Find the volume of the solid generated when the region enclosed by $\mathrm{y}=\sqrt{x}, \mathrm{y}=2$ and $\mathrm{x}=0$ is revolved about the $y$ - axis.
Solution:

Since the solid is generated by revolving about the $\mathrm{y}$-axis, rewrite $\mathrm{y}=\sqrt{x}$ as $\mathrm{x}=\mathrm{y}^2$.
Taking the limits for $\mathrm{y}, \mathrm{y}=0$ and $\mathrm{y}=2$ (Putting $\mathrm{x}=0$ in $\mathrm{x}=\mathrm{y}^2$, we get $\mathrm{y}=0$ )
$
\text { Volume is given by } \begin{aligned}
\mathrm{V} & =\int_c^d \pi x^2 d y \\
& =\int_0^2 \pi y^4 d y \\
& =\left[\frac{\pi y^5}{5}\right]_0^2=\frac{32 \pi}{5} \text { cubic units. }
\end{aligned}
$