Question

The area(in sq. units) of the smaller portion enclosed between the curves, $x^2+y^2=4$ and $y^2=3x$, is.

• A. $\frac{1}{\sqrt{3}}+\frac{4\pi }{3}$
• B. $\frac{1}{2\sqrt{3}}+\frac{\pi }{3}$
• C. $\frac{1}{2\sqrt{3}}+\frac{2\pi }{3}$
• D. $\frac{1}{\sqrt{3}}+\frac{2\pi }{3}$

Answer 1

The correct option is A $\frac{1}{\sqrt{3}}+\frac{4\pi }{3}$

Here, $x^2+y^2=4$ and $y^2=3x$

So finding the co-ordinates of $x$

$\Rightarrow$$x^2+3x-4=0$

$(x+4)(x-1)=0$

$x=-4,x=1$

Area$=\left({\int }_0^1\sqrt{3}\cdot \sqrt{x}dx+{\int }_1^2\sqrt{4-x^2}\cdot dx\right)\times 2$

$=\left(\sqrt{3}{\left(\frac{x^{3/2}}{3/2}\right)}_0^1+{\left(\frac{x}{2}\sqrt{4-x^2}+2si{n}^{-1}\frac{x}{2}\right)}_1^2\right)\times 2$

$=(\sqrt{3}\left(\frac{2}{3}\right)+\{2\cdot \frac{\pi }{2}-\left(\frac{\sqrt{3}}{2}+\frac{\pi }{3}\right)\})\times 2$

$=\left(\frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{2}+\frac{2\pi }{3}\right)\times 2$

$=\left(\frac{1}{2\sqrt{3}}+\frac{2\pi }{3}\right)\times 2=\frac{1}{\sqrt{3}}+\frac{4\pi }{3}$.