Find volume of solid generated by revolving parabola y²= 4ax, cut of by latus rectum, about tangent at vertex

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$4 π a^{3}$

$\frac{4 {πa}^{2}}{5}$

$\frac{4 π}{5}$

$\frac{4 π a^{3}}{5}$

answer is A.

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Detailed Solution

$\begin{aligned} V = 2 \int_{y_{1}}^{y_{2}} π x^{2} d y = 2 π \int_{0}^{2 a} {(\frac{y^{2}}{4 a})}^{2} d y \\ = \frac{2 π}{16 a^{2}} {(\frac{y^{5}}{5})}_{0}^{2 a} = \frac{4 π a^{3}}{5} \end{aligned}$

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