The-loop-of-curve-2ay-2-x-x-a-2-revolves-about-straight-line-y-a-Find-the-volume-of-the-solid-generated-

Question Number 130208 by liberty last updated on 23/Jan/21

The loop of curve {2 ay}^{2} = x {(x - a)}^{2}

revolves about straight line

y = a . Find the volume of the solid

generated .

Answered by benjo_mathlover last updated on 23/Jan/21

The given curve is {2 ay}^{2} = x {(x - a)}^{2} \dots (1)

the curve (1) is symetrical about the x - axis

and the loop lies between x = 0 and x = a .

Differentiating (1) w . r . t x, we get

4 ay (dy / dx) = 2 x (x - a) + {(x - a)}^{2} = {3 x}^{2} - 4 ax + a^{2}

(dy / dx) = 0 when {3 x}^{2} - 4 ax + a^{2} = 0 or when

x = \frac{a}{3} which gives from (1), y = \frac{a \sqrt{2}}{3 \sqrt{3}} .

Area of the loop A = 2 \int_{0}^{a} y dx = 2 \int_{0}^{a} \frac{(x - a) \sqrt{x}}{\sqrt{2 a}} dx

A = \sqrt{\frac{2}{a}} \int_{0}^{a} (x^{3 / 2} - {ax}^{1 / 2}) dx = \frac{4}{15} \sqrt{2} a^{2}

∴ by Pappus theorem the required volume

= 2 π a \times A = \frac{8}{15} \sqrt{2} π a^{3} . \dots \dots \dots \dots \dots \dots . ◊