Exercise 9.10-Additional Problems - Chapter 9 Applications of Integration 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Choose the correct or the most suitable answer from the given four alternatives:
Question 1.
The area bounded by the line $y=x$, the $x$-axis, the ordinates $x=1, x=2$ is
(a) $\frac{3}{2}$
(b) $\frac{5}{2}$
(c) $\frac{1}{2}$
(d) $\frac{7}{2}$
Solution:
(a) $\frac{3}{2}$
Hint:
$
\begin{aligned}
\text { Area } & =\int_1^2 y d x=\int_1^2 x d x=\left[\frac{x^2}{2}\right]_1^2 \\
& =\frac{1}{2}\left[2^2-1\right]=\frac{1}{2}[4-1]=\frac{3}{2} \text { sq. units. }
\end{aligned}
$

Question 2.
The area of the region bounded by the graph of $y=\sin x$ and $y=\cos x$ between $x=0$ and $x=\frac{\pi}{4}$ is
(a) $\sqrt{2}$
(b) $\sqrt{2}-1$
(c) $2 \sqrt{2}-2$
(d) $2 \sqrt{2}+2$
Solution:
(b) $\sqrt{2}-1$

Hint:
$
\begin{aligned}
\text { Area } & =\int_0^{\pi / 4}(\cos x-\sin x) d x \\
& =[\sin x+\cos x]_0^{\pi / 4}=\sin \frac{\pi}{4}+\cos \frac{\pi}{4}-(\sin (0)+\cos (0)) \\
& =\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)-(0+1)=\left(\frac{2}{\sqrt{2}}-1\right)=(\sqrt{2}-1) \text { sq. units. }
\end{aligned}
$

Question 3.
The area between the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and its auxiliary circle is .......
(a) $\pi \mathrm{b}(\mathrm{a}-\mathrm{b})$
(b) $2 \pi a(a-b)$
(c) $\pi a(a-b)$
(d) $2 \pi \mathrm{b}(\mathrm{a}-\mathrm{b})$
Solution:
(c) $\pi a(a-b)$
Hint:
Area of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1=\pi a b$
Area of the auxillary line $x^2+y^2=a^2=\pi a^2$
$\therefore$ Area between them $=\pi a^2-\pi a b=\pi a(a-b)$ sq. units.

Question 4.
The area bounded by the parabola $y^2=x$ and its latus rectum is ........
(a) $\frac{4}{3}$
(b) $\frac{1}{6}$
(c) $\frac{2}{3}$
(d) $\frac{8}{3}$
Solution:
(b) $\frac{1}{6}$

Hint:
$
\begin{aligned}
y^2=x \Rightarrow 4 a=1 \Rightarrow a & =\frac{1}{4} \\
\text { Required area } & =2 \int_0^{\frac{1}{4}} y d x \quad\left[\because y^2=x \Rightarrow y=\sqrt{x}\right] \\
& =2 \int_0^{\frac{1}{4}} \sqrt{x} d x=2\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^{\frac{1}{4}}=\frac{4}{3}\left[x^{\frac{3}{2}}\right]_0^{\frac{1}{4}} \\
& =\frac{4}{3}\left[\left(\frac{1}{4}\right)^{\frac{3}{2}}-0^{\frac{3}{2}}\right]=\frac{4}{3}\left[\frac{1}{2^3}\right]=\frac{4}{3} \times \frac{1}{8}=\frac{1}{6} \text { sq.units. }
\end{aligned}
$

Question 5.
The volume of the solid obtained by revolving $\frac{x^2}{9}+\frac{y^2}{16}=1$ about the minor axis is ......
(a) $48 \pi$
(b) $64 \pi$
(c) $32 \pi$
(d) $128 \pi$
Solution:
(b) $64 \pi$
Hint:
The volume of the solid obtained by revolving $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ about the minor axis is $\frac{4 \pi}{3} a^2 b$ cubic units.
Given equation of the given ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Here $a^2=16, a=4, b^2=9, b=3$
$\therefore$ Required volume $=\frac{4 \pi}{3} \times 16 \times 3=64 \pi$ cu. units.

Question 6.
The volume, when the curve $\mathrm{y}=\sqrt{3+x^2}$ from $\mathrm{x}=0$ to $\mathrm{x}=4$ is rotated about $\mathrm{x}$-axis is
(a) $100 \pi$
(b) $\frac{100}{9} \pi$
(c) $\frac{100}{3} \pi$
$(d) \frac{100}{3}$
Solution:
(c) $\frac{100}{3} \pi$
Hint:
$
\begin{aligned}
& \text { Volume of the solid rotated about } x \text {-axis }=\int_a^b \pi y^2 d x \\
& \qquad \begin{aligned}
y & =\sqrt{3+x^2} \Rightarrow y^2=3+x^2 \\
\text { Required volume } \mathrm{V} & =\pi \int_0^4\left(3+x^2\right) d x \\
& =\pi\left[3 x+\frac{x^3}{3}\right]_0^4=\pi\left[12+\frac{64}{3}\right]=\frac{100 \pi}{3} \text { cu. units. }
\end{aligned}
\end{aligned}
$

Question 7.
The volume generated when the region bounded by $y=x, y=1, x=0$ is rotated about $y$-axis is
(a) $\frac{\pi}{4}$
(b) $\frac{\pi}{2}$
(c) $\frac{\pi}{3}$
(d) $\frac{2 \pi}{3}$
Solution:
(c) $\frac{\pi}{3}$
Hint:
Volume of the solid rotated about $y$-axis $=\int_0^1 x^2 d x$ $$ \pi \int_0^1 x^2 d x=\pi\left[\frac{x^3}{3}\right]_0^1=\frac{\pi}{3} \text { cu. units. } $$
Question 8.
Volume of solid obtained by revolving the area of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ about major and minor axes are in the ratio
(a) $b^2: a^2$
(b) $a^2: b^2$
(c) $a: b$
(d) $b: a$
Solution:
(d) $b: a$

Hint:
The volume of the solid obtained by revolving the area of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ about major axis $=\frac{4 \pi}{3} a^2 b$ cu. units.
The volume of the solid obtained by revolving the area of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ about minor axis $=\frac{4 \pi}{3} a b^2 \mathrm{cu}$. units.
Ratio of the volume $=\frac{\frac{4}{3} a b^2}{\frac{4}{3} a^2 b}=\frac{b}{a}$ Ratio is $b: a$

Question 9.
The volume generated by rotating the triangle with vertices at $(0,0),(3,0)$ and $(3,3)$ about $\mathrm{x}$-axis is
(a) $18 \pi$
(b) $2 \pi$
(c) $36 \pi$
(d) $9 \pi$
Solution:
(d) $9 \pi$
Hint:

Required volume $=$ Volume of the cone with $r=3 ; h=3$
Required volume $=\frac{1}{3} \pi r^2 h=\frac{1}{3} \pi 3^2 \times 3=9 \pi$ cubic units.

Question 10.
The length of the arc of the curve $x^{2 / 3}+y^{2 / 3}=4$ is ......
(a) 48
(b) 24
(c) 12
(d) 96
Solution:
(a) 48
Hint:
Length of the arc of the curve $=6 \mathrm{a}$
$
\begin{aligned}
& =x^{2 / 3}+y^{2 / 3}=a^{2 / 3} \\
\text { Given, } a^{2 / 3} & =4 \Rightarrow a^2=64 \Rightarrow a=8 \\
\therefore \text { Required length } & =6 a=6 \times 8=48 \text { units. } \\
\therefore \text { Required length }=6 a & =6 \times 8=48 \text { units. }
\end{aligned}
$

Question 11.
The surface area of the solid of revolution of the region bounded by $\mathrm{y}=2 \mathrm{x}, \mathrm{x}=0$ and $\mathrm{x}=2$ about $\mathrm{x}$-axis is
(a) $8 \sqrt{5} \pi$
(b) $2 \sqrt{5} \pi$
(c) $\sqrt{5} \pi$
$(d) 4 \sqrt{5} \pi$
Solution:
(a) $8 \sqrt{5} \pi$
Hint:
$
\begin{aligned}
& \text { Required surface area }=\int_a^b 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^2 d x} \\
& \text { Given, } y=2 x \\
& \frac{d y}{d x}=2 \Rightarrow\left(\frac{d y}{d x}\right)^2=4 \\
& \sqrt{1+\left(\frac{d y}{d x}\right)^2}=\sqrt{1+4}=\sqrt{5} \\
& \therefore \text { Required surface area } \\
& \int_0^2 2 \pi 2 x \sqrt{5}=4 \pi \sqrt{5} \int_0^2 x d x \\
&=4 \pi \sqrt{5}\left[\frac{x^2}{2}\right]_0^2=8 \sqrt{5} \pi .
\end{aligned}
$

Question 12.
The curved surface area of a sphere of radius 5 , intercepted between two parallel planes of distance 2 and 4 from the centre is .....
(a) $20 \pi$
(b) $40 \pi$
(c) $10 \pi$
(d) $30 \pi$
Solution:
(a) $20 \pi$
Hint:
The curved surface area of a sphere of radius $r$ intercepted between two parallel planes at a distance $a$ and $b$ from the centre of the sphere is $2 \pi r(b-a)$
Given radius, $\mathrm{r}=5 ; \mathrm{a}=2 ; \mathrm{b}=4$
Required surface area $=2 \pi \mathrm{r}(\mathrm{b}-\mathrm{a})$
$=2 \pi \times 5 \times(4-2)=20 \pi$ sq. units