A curve y = $\frac{1}{2 \sqrt{x}}$ is allowed to revolve around x axis. The volume of solid of revolution for $3 \leq x \leq 4$ is

x (\sqrt{3} + \frac{2}{\sqrt{3}})

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\frac{π}{2} l n (\frac{8}{3})

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\frac{π}{4} l n (\frac{4}{3})

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\frac{π}{2} (\frac{1}{2 \sqrt{2}} - \frac{1}{\sqrt{3}})

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Solution

The correct option is C $\frac{π}{4} l n (\frac{4}{3})$
Volume of solid = $\int_{a}^{b} π y^{2} d x$
Given $y = \frac{1}{2 \sqrt{x}}$
Volume of the solid = $\int_{3}^{4} \frac{π}{4 x} d x = \frac{π}{4} (l n x)_{3}^{4}$
$= \frac{π}{4} l n (\frac{4}{3})$

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