Question

Which of the following is/are true?
I. The area of the circle $4 x^2+4 y^2=9$ which is interior to the parabola $x^2=4 y$, is $\frac{\sqrt{2}}{6}+\frac{9}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$ sq units.
II. The area bounded by the curves $(x-1)^2+y^2=1$ and $x^2+y^2=1$ is $\left(\frac{2 \pi}{3}+\frac{\sqrt{3}}{2}\right)$ sq units.
III. The area of the region bounded by the curves $y=x^2+2, y=x, x=0$ and $x=3$ is 21 sq units.

• A. II and III are true
• B. Only I is true
• C. I and III are true
• D. All are true

Answer 1

Answer: (B)

I. Given circle is $4 x^2+4 y^2=9$ and the given parabola is $x^2=4 y$. The two curves meet at
$ 4(4 y)+4 y^2=9 \Rightarrow 4 y^2+16 y-9=0$
$ \Rightarrow y=\frac{-16 \pm \sqrt{256+144}}{2 \times 4}$
$ =\frac{-16 \pm \sqrt{400}}{8}=\frac{-16 \pm 20}{8}=\frac{1}{2},-\frac{9}{2} $

But $y>0$, therefore the two curves meet when $y=\frac{1}{2}$
i.e., when $x^2=4 \times \frac{1}{2}=2$, i.e., when $x=\pm \sqrt{2}$.
$\therefore$ Required area (shown in shaded region)
$=2 \int\limits_0^{\sqrt{2}}\left(y_2-y_1\right) d x=2\left\{\int\limits_0^{\sqrt{2}} \sqrt{\frac{9-4 x^2}{4}} d x-\int\limits_0^{\sqrt{2}} \frac{x^2}{4} d x\right\}$
$ =2 \int\limits_0^{\sqrt{2}} \sqrt{\left(\frac{3}{2}\right)^2-x^2} d x-\frac{2}{4}\left[\frac{x^3}{3}\right]_0^{\sqrt{2}}$
$=2\left[\frac{x}{2} \sqrt{\left(\frac{3}{2}\right)^2-x^2}+\frac{(3 / 2)^2}{2} \sin ^{-1}\left(\frac{x}{3 / 2}\right)\right]_0^{\sqrt{2}}-\frac{1}{6}\left[(\sqrt{2})^3-0\right]$
$=\left[x \sqrt{\frac{9}{4}-x^2}+\frac{9}{4} \sin ^{-1}\left(\frac{2 x}{3}\right)\right]_0^{\sqrt{2}}-\frac{2 \sqrt{2}}{6}$
$=\left[\sqrt{2} \sqrt{\frac{9}{4}-2}+\frac{9}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\right]-\{0-0\}-\frac{2 \sqrt{2}}{6}$
$=\frac{\sqrt{2}}{2}+\frac{9}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)-\frac{2 \sqrt{2}}{6}=\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{3}\right)+\frac{9}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$
$=\frac{\sqrt{2}}{6}+\frac{9}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$ sq units
II. Given curve is
$ (x-1)^2+y^2 =1 $ ...(i)
$\therefore y = \sqrt{1-(x-1)^2}$

which represents a circle with centre $(1,0)$ and radius 1 and curve $ x^2+y^2=1$...(ii)
$\therefore y=\sqrt{1-x^2}$
which represents a circle with centre $(0,0)$ and radius 1 . Both the curves are circles and meet where $(x-1)^2=x^2$ i.e., where $2 x=1$ or $x=\frac{1}{2}$.
Required area (shown in shaded region)
$ =2\left[\int\limits_0^{1 / 2} y_1 d x+\int\limits_{1 / 2}^1 y_2 d x\right] $
$=2\left[\int\limits_0^{1 / 2} \sqrt{1-(x-1)^2} d x+\int\limits_{1 / 2}^1 \sqrt{1-x^2} d x\right] $
$ =2\left[\frac{x-1}{2} \sqrt{1-(x-1)^2}+\frac{1}{2} \sin ^{-1} \frac{x-1}{1}\right]_0^{1 / 2}$
$ +2\left[\frac{x}{2} \sqrt{1-x^2}+\frac{1}{2} \sin ^{-1} x\right]_{1 / 2}^1$
$ =2\left[\frac{\frac{1}{2}-1}{2} \sqrt{1-\frac{1}{4}}+\frac{1}{2} \sin ^{-1}\left(-\frac{1}{2}\right)-\left(\frac{-1}{2}\right) 0-\frac{1}{2} \sin ^{-1}(-1)\right] $
$ +2\left[0+\frac{1}{2} \sin ^{-1}(1)-\frac{1}{4} \sqrt{1-\frac{1}{4}}-\frac{1}{2} \sin ^{-1} \frac{1}{2}\right]$
$ =2\left[-\frac{1}{4} \cdot \frac{\sqrt{3}}{2}-\frac{1}{2} \cdot \frac{\pi}{6}+0+\frac{1}{2} \cdot \frac{\pi}{2}\right]+\frac{\pi}{2}-\frac{1}{2} \cdot \frac{\sqrt{3}}{2}-\frac{\pi}{6}$
$ =-\frac{\sqrt{3}}{4}-\frac{\pi}{6}+\frac{\pi}{2}+\frac{\pi}{2}-\frac{\sqrt{3}}{4}-\frac{\pi}{6}=\left(\frac{2 \pi}{3}-\frac{\sqrt{3}}{2}\right) $ sq unit
III. Given curve $y=x^2+2$ represents a parabola and it is symmetrical about $Y$-axis having vertex $(0,2)$.

The given region bounded by $y=x^2+2, y=x, x=0$ and $x=3$, is represented by the shaded area.
The point of intersection of the curve $y=x^2+2$ and the line $x=3$ is $(3,11)$
Required area (shown in shaded region)
$=$ Area $(O A B D O)$ - Area $(O C D O)$
$=\left[\right.$ Area under $y=x^2+2$ between $x=0$ and $x=3$ ]
- [Area under $y=x$ between $x=0$ and $x=3$ ]
$=\int\limits_0^3\left(x^2+2\right) d x-\int\limits_0^3 x d x$
$=\left[\frac{x^3}{3}+2 x\right]_0^3-\left[\frac{x^2}{2}\right]_0^3=\left[\frac{3^3}{3}+6-0\right]-\left[\frac{3^2}{2}-0\right]$
$=9+6-\frac{9}{2}=\frac{21}{2}$ sq units