Question

If \(x^{2}+y^{2}=a^{2}\), then \(\int_{0}^{a} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x=\)
(A) \(2 \pi \mathrm{a}\)
(B) \(\pi \mathrm{a}\)
(C) \(\frac{1}{2} \pi \mathrm{a}\)
(D) \(\frac{1}{4} \pi \mathrm{a}\)

Answer 1

Ans: (C)
Hint : \(y_{1}=-\frac{x}{y}\)
\(\int_{0}^{a} \sqrt{1+\frac{x^{2}}{y^{2}}} d x=a \int_{0}^{a} \frac{1}{y} \cdot d x=a \int_{0}^{a} \frac{d x}{\sqrt{a^{2}-x^{2}}}\)
\(=a\left[\sin ^{-1}\left(\frac{x}{a}\right)\right]_{0}^{a}=a \pi / 2\)

Answer 2

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Answer 3

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