If x^{2}+y^{2}=a^{2}, then int_{0}^{a} sqrt{1 | vedclass

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(C) Given the equation of the circle $x^{2}+y^{2}=a^{2}$.Differentiating with respect to $x$, we get $2x + 2y \frac{dy}{dx} = 0$, which implies $\frac

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$\frac{1}{2} \pi a$

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(C) Given the equation of the circle $x^{2}+y^{2}=a^{2}$.Differentiating with respect to $x$, we get $2x + 2y \frac{dy}{dx} = 0$, which implies $\frac{dy}{dx} = -\frac{x}{y}$.Now, substitute this into the integral: $\sqrt{1+\left(\frac{dy}{dx}\right)^{2}} = \sqrt{1+\frac{x^{2}}{y^{2}}} = \sqrt{\frac{y^{2}+x^{2}}{y^{2}}} = \sqrt{\frac{a^{2}}{y^{2}}} = \frac{a}{y}$.The integral becomes $\int_{0}^{a} \frac{a}{y} dx$.Since $y = \sqrt{a^{2}-x^{2}}$, the integral is $\int_{0}^{a} \frac{a}{\sqrt{a^{2}-x^{2}}} dx$.Evaluating this, we get $a \left[ \sin^{-1} \left( \frac{x}{a} \right) \right]_{0}^{a} = a \left( \sin^{-1}(1) - \sin^{-1}(0) \right) = a \left( \frac{\pi}{2} - 0 \right) = \frac{\pi a}{2}$.

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

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