Question

The area in the first quadrant enclosed by the $x$-axis, the line $y=\frac{1}{\sqrt{3}}x$ and circle $x^2+y^2=4$ is given by.

• A. ${\int }_0^{\sqrt{3}}(x+\sqrt{4-x^2})dx$
• B. ${\int }_0^2(x+\sqrt{4-x^2})dx$
• C. ${\int }_0^2(\frac{x}{\sqrt{3}}+\sqrt{4-x^2})dx$
• D. ${\int }_0^{\sqrt{3}}\frac{x}{\sqrt{3}}dx+{\int }_{\sqrt{3}}^2\sqrt{4-x^2}dx$

Answer 1

Answer: (D)

${\int }_0^{\sqrt{3}}\frac{x}{\sqrt{3}}dx+{\int }_{\sqrt{3}}^2\sqrt{4-x^2}dx$

$y=\frac{1}{\sqrt{3}}x$ and $x^2+y^2=4$

$x^2+\frac{1}{3}{x}^2=4$

$\frac{4}{3}{x}^2=4$

$x=\pm \sqrt{3}$

Since area is in $1^{st}$ quadrant, so, $x=\sqrt{3}$ and $y=1$

So,

Area $={\int }_0^{\sqrt{3}}\frac{1}{\sqrt{3}}xdx+{\int }_{\sqrt{3}}^2\sqrt{4-x^2}dx$

Hence, this is the answer.