Exercise 8.1 (Revised) - Chapter 8 - Applications Of Integrals - Ncert Solutions class 12 - Maths

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Chapter 8 - Applications Of Integrals NCERT Solutions Class 12 Maths

Ex 8.1 Question 1.

Find the area of the region bounded by the ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$.

Answer.

Equation of ellipse is $\frac{x^2}{16}+\frac{y^2}{9}=1$ ..(i)

Here $a^2(=16)>b^2(=9)$
From eq. (i), $\frac{y^2}{9}=1-\frac{x^2}{16}=\frac{16-x^2}{16}$
$
\begin{aligned}
& \Rightarrow y^2=\frac{9}{16}\left(16-x^2\right) \\
& \Rightarrow y= \pm \frac{3}{4} \sqrt{16-x^2}
\end{aligned}
$

For arc of ellipse in first quadrant.
Ellipse (i) ia symmetrical about $x$ - axis,
( $\because$ On changing $y \rightarrow-y$ in eq. (i), it remains unchanged)
Ellipse (i) ia symmetrical about $y$-axis,
( $\because$ On changing $x \rightarrow-x$ in eq. (i), it remains unchanged)
Intersections of ellipse (i) with $x$-axis $(y=0)$
Putting $y=0$ in eq. (i), $\frac{x^2}{16}=1 \Rightarrow x^2=16 \Rightarrow x= \pm 4$
Therefore, Intersections of ellipse (i) with $x$-axis are $(0,4)$ and $(0,-4)$.
Intersections of ellipse (i) with $y-\operatorname{axis}(x=0)$
Putting $x=0$ in eq. (i), $\frac{y^2}{9}=1 \Rightarrow y^2=9 \Rightarrow y= \pm 3$
Therefore, Intersections of ellipse (i) with $y$-axis are $(0,3)$ and $(0,-3)$.
Now Area of region bounded by ellipse $(i)=$ Total shaded area

$=4 \times \text { Area OAB of ellipse in first quadrant }$

$\begin{aligned}
& =4\left|\int_0^4 y d x\right|[\because \text { At end B of arc AB of ellipse; } x=0 \text { and at end } \mathrm{A} \text { of } \operatorname{arc} \mathrm{AB} ; x=4] \\
& =4\left|\int_0^4 \frac{3}{4} \sqrt{16-x^2} d x\right|=4\left|\int_0^4 \frac{3}{4} \sqrt{4^2-x^2} d x\right| \\
& =3\left[\frac{x}{2} \sqrt{4^2-x^2}+\frac{4^2}{2} \sin ^{-1} \frac{x}{4}\right]_0^4\left[\because \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right] \\
& =3\left[\frac{4}{2} \sqrt{16-16}+8 \sin ^{-1} 1-\left(0+8 \sin ^{-1} 0\right)\right]=3\left[0+\frac{8 \pi}{2}\right] \\
& =3(4 \pi)=12 \pi \text { sq. units }
\end{aligned}$

Ex 8.1 Question 2.

Find the area of the region bounded by the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$.

Answer.

Equation of ellipse is $\frac{x^2}{4}+\frac{y^2}{9}=1$

Here $a^2(=4) From eq. (i), $\frac{y^2}{9}=1-\frac{x^2}{4}=\frac{4-x^2}{4}$
$
\Rightarrow y^2=\frac{9}{4}\left(4-x^2\right)
$

$
\Rightarrow y= \pm \frac{3}{2} \sqrt{4-x^2} \text {. }
$
for arc of ellipse in first quadrant.
Ellipse (i) ia symmetrical about $x$ - axis,
( $\because$ On changing $y \rightarrow-y$ in eq. (i), it remains unchanged)
Ellipse (i) ia symmetrical about $y$-axis,
( $\because$ On changing $x \rightarrow-x$ in eq. (i), it remains unchanged)
Intersections of ellipse (i) with $x$ - axis $(y=0)$
Putting $y=0$ in eq. (i), $\frac{x^2}{4}=1$
$
\begin{aligned}
& \Rightarrow x^2=4 \\
& \Rightarrow x= \pm 2
\end{aligned}
$

Therefore, Intersections of ellipse (i) with $x$-axis are $(0,2)$ and $(0,-2)$.
Intersections of ellipse (i) with $y-$ axis $(x=0)$
Putting $x=0$ in eq. (i), $\frac{y^2}{9}=1$
$
\Rightarrow y^2=9 \Rightarrow y= \pm 3
$

Therefore, Intersections of ellipse (i) with $y$-axis are $(0,3)$ and $(0,-3)$.
Now Area of region bounded by ellipse $(i)=$ Total shaded area
$=4 \times$ Area OAB of ellipse in first quadrant

$=4\left|\int_0^2 y d x\right|[\because \text { At end B of arc AB of ellipse; } x=0 \text { and at end A of } \operatorname{arc~AB} ; x=2]$

$\begin{aligned}
& =4\left|\int_0^2 \frac{3}{2} \sqrt{4-x^2} d x\right|=4\left|\int_0^4 \frac{3}{2} \sqrt{2^2-x^2} d x\right| \\
& =6\left[\frac{x}{2} \sqrt{2^2-x^2}+\frac{2^2}{2} \sin ^{-1} \frac{x}{2}\right]_0^4\left[\because \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right] \\
& =6\left[\frac{2}{2} \sqrt{4-4}+2 \sin ^{-1} 1-\left(0+2 \sin ^{-1} 0\right)\right] \\
& =6\left[0+2 \cdot \frac{\pi}{2}-0\right]=6 \pi \text { sq. units }
\end{aligned}$

Ex 8.1 Question 3.

Choose the correct answer: 

Area lying in the first quadrant and bounded by the circle $x^2+y^2=4$ and the lines $x=0$ and $x=2$ is
(A) $\pi$
(B) $\frac{\pi}{2}$
(C) $\frac{\pi}{3}$
(D) $\frac{\pi}{4}$

Answer.

Equation of the circle is $x^2+y^2=2^2$ ..(i)
$
\Rightarrow y=\sqrt{2^2-x^2}
$

$\begin{aligned}
& \text { Required area }=\left|\int_0^2 y d x\right|=\left|\int_0^2 \sqrt{2^2-x^2} d x\right| \\
& \left.=\left\lvert\, \frac{x}{2} \sqrt{2^2-x^2}+\frac{2^2}{2} \sin ^{-1} \frac{x}{2}\right.\right)\left._0^2\right|^2 \\
& =\frac{2}{2} \sqrt{4-4}+2 \sin ^{-1} 1-\left(0+2 \sin ^{-1} 0\right) \\
& =0+2 \cdot \frac{\pi}{2}-0-0=\pi \text { sq. units }
\end{aligned}$

Therefore, option (A) is correct.
Ex 8.1 Question 4.

Choose the correct answer:

Area of the region bounded by the curve $y^2=4 x, y$ - axis and the line $y=3$ is:
(A) 2
(B) $\frac{9}{4}$
(C) $\frac{9}{3}$
(D) $\frac{9}{2}$

Answer.

Equation of the curve (parabola) is $y^2=4 x$

$
\begin{aligned}
& \text { Required area }=\text { Area OAM }=\left|\int_0^3 x d y\right|=\left|\int_0^3 \frac{y^2}{4} d y\right| \\
& =\frac{1}{4}\left|\left(\frac{y^3}{3}\right)_0^3\right|=\frac{1}{4}\left|\frac{27}{3}-0\right|=\frac{9}{4} \text { sq. units }
\end{aligned}
$

Therefore, option (B) is correct.