Question

find the area of the loop of the curve 3ay2=x(x-a)2
it answer is
8a2 /15root3
so please give me explanation

Answer 1

Answer:

Given curve is

$3a\:y^2=x(x-a)^2$

since the curve is symmetrical about x-axis, the area of the loop of the curve is twice the area of the loop of the curve in the I quadrant.

Therefore,

Loop of the curve

$=2\int\limits^a_0{y}\:dx$

$=2\int\limits^a_0\frac{-\sqrt{x}(x-a)}{\sqrt{3a}}\:dx$

$=\frac{-2}{\sqrt{3a}}\int\limits^a_0\sqrt{x}(x-a)\:dx$

$=\frac{-2}{\sqrt{3a}}\int\limits^a_0[x^\frac{3}{2}-a\:x\frac{1}{2}]\:dx$

$=\frac{-2}{\sqrt{3a}}[\frac{x^\frac{5}{2}}{\frac{5}{2}}-\frac{a\:x\frac{3}{2}}{\frac{3}{2}}]^a_0$

$=\frac{-2}{\sqrt{3a}}[\frac{2x^\frac{5}{2}}{5}-\frac{2a\:x\frac{3}{2}}{3}]^a_0$

$=\frac{-4}{\sqrt{3a}}[\frac{x^\frac{5}{2}}{5}-\frac{a\:x\frac{3}{2}}{3}]^a_0$

$=\frac{-4}{\sqrt{3a}}[(\frac{a^\frac{5}{2}}{5}-\frac{a\:a\frac{3}{2}}{3})-(0)]$

$=\frac{-4}{\sqrt{3a}}[(\frac{a^\frac{5}{2}}{5}-\frac{a\frac{5}{2}}{3})-(0)]$

$=\frac{-4}{\sqrt{3a}}[\frac{3a^\frac{5}{2}-5a^\frac{5}{2}}{15}]$

$=\frac{-4}{\sqrt{3a}}[\frac{-2a^\frac{5}{2}}{15}]$

$=\frac{8}{\sqrt{3a}}[\frac{a^\frac{5}{2}}{15}]$

$=\frac{8}{\sqrt{3}\sqrt{a}}[\frac{a^2\:\sqrt{a}}{15}]$

$=\frac{8}{\sqrt{3}}[\frac{a^2}{15}]$

$=\frac{8a^2}{15\sqrt{3}}$