Question

If $x^2+y^2=a^2$, then $\underset{0}{\overset{a}{\int }}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx=$

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

Answer 1

Answer: (C)

$y_1=-\frac{x}{y}$
$\underset{0}{\overset{a}{\int }}\sqrt{1+\frac{x^2}{y^2}}dx=a\underset{0}{\overset{a}{\int }}\frac{1}{y}\cdot dx$
$=a\underset{0}{\overset{a}{\int }}\frac{dx}{\sqrt{a^2-x^2}}$
$=a{\left[{\sin }^{-1}\left(\frac{x}{a}\right)\right]}_0^a=\frac{a\pi }{2}$