WBJEE · Maths · Definite Integration
If \(x^{2}+y^{2}=a^{2}\), then \(\int_{0}^{a} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x=\)
- A \(2 \pi a\)
- B \(\pi a\)
- C \(\frac{1}{2} \pi \mathrm{a}\)
- D \(\frac{1}{4} \pi \mathrm{a}\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{2} \pi \mathrm{a}\)
Step-by-step Solution
Detailed explanation
Hint: \(y_{1}=-\frac{x}{y}\) \(\int_{0}^{a} \sqrt{1+\frac{x^{2}}{y^{2}}} d x=a \int_{0}^{a} \frac{1}{y} \cdot d x=a \int_{0}^{a} \frac{d x}{\sqrt{a^{2}-x^{2}}}\) \(=a\left[\sin ^{-1}\left(\frac{x}{a}\right)\right]_{0}^{]}=a \pi / 2\)
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