Question

Which of the following is/are true?
I. The area of the circle $4x^2+4y^2=9$ which is interior to the parabola $x^2=4y$, 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 $4x^2+4y^2=9$ and the given parabola is $x^2=4y$. The two curves meet at
$4(4y)+4y^2=9\Rightarrow 4y^2+16y-9=0$
$\Rightarrow y=\frac{-16\pm \sqrt{256+144}}{2\times 4}$
$=\frac{-16\pm \sqrt{400}}{8}=\frac{-16\pm 20}{8}=\frac{1}{2},-\frac{9}{2}$

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)
$=2\underset{0}{\overset{\sqrt{2}}{\int }}\left(y_2-y_1\right)dx=2\left\{\underset{0}{\overset{\sqrt{2}}{\int }}\sqrt{\frac{9-4x^2}{4}}dx-\underset{0}{\overset{\sqrt{2}}{\int }}\frac{x^2}{4}dx\right\}$
$=2\underset{0}{\overset{\sqrt{2}}{\int }}\sqrt{{\left(\frac{3}{2}\right)}^2-x^2}dx-\frac{2}{4}{\left[\frac{x^3}{3}\right]}_0^{\sqrt{2}}$
$=2{\left[\frac{x}{2}\sqrt{{\left(\frac{3}{2}\right)}^2-x^2}+\frac{(3/2{)}^2}{2}{\sin }^{-1}\left(\frac{x}{3/2}\right)\right]}_0^{\sqrt{2}}-\frac{1}{6}\left[(\sqrt{2}{)}^3-0\right]$
$={\left[x\sqrt{\frac{9}{4}-x^2}+\frac{9}{4}{\sin }^{-1}\left(\frac{2x}{3}\right)\right]}_0^{\sqrt{2}}-\frac{2\sqrt{2}}{6}$
$=\left[\sqrt{2}\sqrt{\frac{9}{4}-2}+\frac{9}{4}{\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)\right]-\{0-0\}-\frac{2\sqrt{2}}{6}$
$=\frac{\sqrt{2}}{2}+\frac{9}{4}{\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)-\frac{2\sqrt{2}}{6}=\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{3}\right)+\frac{9}{4}{\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)$
$=\frac{\sqrt{2}}{6}+\frac{9}{4}{\sin }^{-1}\left(\frac{2\sqrt{2}}{3}\right)$ 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 $2x=1$ or $x=\frac{1}{2}$.
Required area (shown in shaded region)
$=2\left[\underset{0}{\overset{1/2}{\int }}{y}_{1}dx+\underset{1/2}{\overset{1}{\int }}{y}_{2}dx\right]$
$=2\left[\underset{0}{\overset{1/2}{\int }}\sqrt{1-(x-1{)}^2}dx+\underset{1/2}{\overset{1}{\int }}\sqrt{1-x^2}dx\right]$
$=2{\left[\frac{x-1}{2}\sqrt{1-(x-1{)}^2}+\frac{1}{2}{\sin }^{-1}\frac{x-1}{1}\right]}_0^{1/2}$
$+2{\left[\frac{x}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}x\right]}_{1/2}^1$
$=2\left[\frac{\frac{1}{2}-1}{2}\sqrt{1-\frac{1}{4}}+\frac{1}{2}{\sin }^{-1}\left(-\frac{1}{2}\right)-\left(\frac{-1}{2}\right)0-\frac{1}{2}{\sin }^{-1}(-1)\right]$
$+2\left[0+\frac{1}{2}{\sin }^{-1}(1)-\frac{1}{4}\sqrt{1-\frac{1}{4}}-\frac{1}{2}{\sin }^{-1}\frac{1}{2}\right]$
$=2\left[-\frac{1}{4}\cdot \frac{\sqrt{3}}{2}-\frac{1}{2}\cdot \frac{\pi }{6}+0+\frac{1}{2}\cdot \frac{\pi }{2}\right]+\frac{\pi }{2}-\frac{1}{2}\cdot \frac{\sqrt{3}}{2}-\frac{\pi }{6}$
$=-\frac{\sqrt{3}}{4}-\frac{\pi }{6}+\frac{\pi }{2}+\frac{\pi }{2}-\frac{\sqrt{3}}{4}-\frac{\pi }{6}=\left(\frac{2\pi }{3}-\frac{\sqrt{3}}{2}\right)$ 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 $(OABDO)$ - Area $(OCDO)$
$=[$ Area under $y=x^2+2$ between $x=0$ and $x=3$ ]
- [Area under $y=x$ between $x=0$ and $x=3$ ]
$=\underset{0}{\overset{3}{\int }}\left(x^2+2\right)dx-\underset{0}{\overset{3}{\int }}xdx$
$={\left[\frac{x^3}{3}+2x\right]}_0^3-{\left[\frac{x^2}{2}\right]}_0^3=\left[\frac{3^3}{3}+6-0\right]-\left[\frac{3^2}{2}-0\right]$
$=9+6-\frac{9}{2}=\frac{21}{2}$ sq units