Question

Area enclosed between the curves $\left|y\right|=1-x^2$ and $x^2+y^2=1$ is

• A. $\frac{3\pi -14}{3}$ sq.units
• B. $\frac{\pi -8}{3}$ sq.units
• C. $\frac{2\pi -8}{3}$ sq.units
• D. None of these

Answer 1

The correct option is A $\frac{3\pi -14}{3}$ sq.units

We have,

$x^2+y^2=1$

$y_1=\sqrt{1-x^2}$ ……. (1)

$\left|y\right|=1-x^2$

$y=1-x^2$ ……. (2)

So, area of first quadrant

$={\int }_0^1(y_1-y_2)dx$

$={\int }_0^1(\sqrt{1-x^2}-(1-x^2))dx$

$={\int }_0^1\left(\sqrt{1-x^2}\right)dx-{\int }_0^1(1-x^2)dx$

We know that

$\int \sqrt{a^2-x^2}dx=\frac{1}{2}a\sqrt{a^2-x^2}+\frac{1}{2}{a}^2{\sin }^{-1}\left(\frac{x}{a}\right)+C$

Therefore,

$={[\frac{1}{2}\sqrt{1-x^2}+\frac{1}{2}{\sin }^{-1}\left(x\right)]}_0^1-{(x-\frac{x^3}{3})}_0^1$

$=[(\frac{1}{2}\sqrt{1-1}+\frac{1}{2}{\sin }^{-1}\left(1\right))-(\frac{1}{2}\sqrt{1-0}+\frac{1}{2}{\sin }^{-1}\left(0\right))]-[(1-\frac{1}{3})-0]$

$=[(0+\frac{1}{2}{\sin }^{-1}\left(\sin \frac{\pi }{2}\right))-(\frac{1}{2}+\frac{1}{2}{\sin }^{-1}\left(\sin 0\right))]-\frac{2}{3}$

$=\frac{\pi }{4}-\frac{1}{2}-\frac{2}{3}$

$=\frac{\pi }{4}-\left(\frac{3+4}{6}\right)$

$=\frac{\pi }{4}-\frac{7}{6}$

$=\frac{3\pi -14}{12}$

Therefore,

Area of all quadrants

$=4\times \left(\frac{3\pi -14}{12}\right)$

$=\frac{3\pi -14}{3}$

Hence, this is the answer.