Question

If $x^2+y^2=a^2$, then ${\int }_0^a\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx=$

• A. $2\pi a$
• B. $\pi a$
• C. $\frac{1}{2}\pi a$
• D. $\frac{1}{4}\pi a$

Answer 1

The correct option is C $\frac{1}{2}\pi a$

$x^2+y^2=a^2$
Differentiating both sides, we get:
$\frac{dy}{dx}=-\frac{x}{y}$
${\int }_0^a\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$

$={\int }_0^a\sqrt{1+\frac{x^2}{y^2}}dx$

$=a{\int }_0^a\frac{1}{\sqrt{a^2-x^2}}dx$

$=a{\left[{\sin }^{-1}\frac{x}{a}\right]}_0^a$
$=\frac{1}{2}\pi a$