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If $x^{2}+y^{2}=a^{2}$, then $\int_{0}^{a} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x=$
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Verified Answer
The correct answer is:
$\frac{1}{2} \pi \mathrm{a}$
Hint:
$y_{1}=-\frac{x}{y}$
$\int_{0}^{a} \sqrt{1+\frac{x^{2}}{y^{2}}} d x=a \int_{0}^{a} \frac{1}{y} \cdot d x=a \int_{0}^{a} \frac{d x}{\sqrt{a^{2}-x^{2}}}$
$=a\left[\sin ^{-1}\left(\frac{x}{a}\right)\right]_{0}^{]}=a \pi / 2$
$y_{1}=-\frac{x}{y}$
$\int_{0}^{a} \sqrt{1+\frac{x^{2}}{y^{2}}} d x=a \int_{0}^{a} \frac{1}{y} \cdot d x=a \int_{0}^{a} \frac{d x}{\sqrt{a^{2}-x^{2}}}$
$=a\left[\sin ^{-1}\left(\frac{x}{a}\right)\right]_{0}^{]}=a \pi / 2$
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