Question

If $x^{2}+y^{2}=a^{2}$, then $\int\limits_{0}^{a} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x=$

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

Answer 1

Answer: (C)

$y_{1}=-\frac{x}{y}$
$\int\limits_{0}^{a} \sqrt{1+\frac{x^{2}}{y^{2}}} d x=a \int\limits_{0}^{a} \frac{1}{y} \cdot d x$
$=a \int\limits_{0}^{a} \frac{d x}{\sqrt{a^{2}-x^{2}}}$
$=a\left[\sin ^{-1}\left(\frac{x}{a}\right)\right]_{0}^{a}=\frac{a \pi}{2}$