Menu Close

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

Tinku Tara 0 Comments
Pin




Question Number 130208 by liberty last updated on 23/Jan/21
The loop of curve 2ay^2 =x(x−a)^2 revolves about straight line y=a. Find the volume of the solid generated.
$$\:\mathrm{The}\:\mathrm{loop}\:\mathrm{of}\:\mathrm{curve}\:\mathrm{2ay}^{\mathrm{2}} =\mathrm{x}\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{2}} \\ $$$$\mathrm{revolves}\:\mathrm{about}\:\mathrm{straight}\:\mathrm{line}\: \\ $$$$\mathrm{y}=\mathrm{a}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{solid} \\ $$$$\mathrm{generated}. \\ $$
Answered by benjo_mathlover last updated on 23/Jan/21
The given curve is 2ay^2 =x(x−a)^2 ...(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 4ay (dy/dx) = 2x(x−a)+(x−a)^2 =3x^2 −4ax+a^2 (dy/dx)=0 when 3x^2 −4ax+a^2 =0 or when x=(a/3) which gives from (1) , y=((a(√2))/(3(√3))) . Area of the loop A=2∫_0 ^( a) y dx = 2∫_0 ^( a) (((x−a)(√x))/( (√(2a)))) dx A= (√(2/a)) ∫_0 ^( a) (x^(3/2) −ax^(1/2) ) dx = (4/(15))(√2) a^2 ∴ by Pappus theorem the required volume = 2πa × A = (8/(15)) (√2) πa^3 . ...................◊
$$\:\mathrm{The}\:\mathrm{given}\:\mathrm{curve}\:\mathrm{is}\:\mathrm{2ay}^{\mathrm{2}} =\mathrm{x}\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{2}} …\left(\mathrm{1}\right) \\ $$$$\:\mathrm{the}\:\mathrm{curve}\:\left(\mathrm{1}\right)\:\mathrm{is}\:\mathrm{symetrical}\:\mathrm{about}\:\mathrm{the}\:\mathrm{x}−\mathrm{axis} \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{loop}\:\mathrm{lies}\:\mathrm{between}\:\mathrm{x}=\mathrm{0}\:\mathrm{and}\:\mathrm{x}=\mathrm{a}. \\ $$$$\mathrm{Differentiating}\:\left(\mathrm{1}\right)\:\mathrm{w}.\mathrm{r}.\mathrm{t}\:\mathrm{x}\:,\:\mathrm{we}\:\mathrm{get}\: \\ $$$$\mathrm{4ay}\:\left(\mathrm{dy}/\mathrm{dx}\right)\:=\:\mathrm{2x}\left(\mathrm{x}−\mathrm{a}\right)+\left(\mathrm{x}−\mathrm{a}\right)^{\mathrm{2}} =\mathrm{3x}^{\mathrm{2}} −\mathrm{4ax}+\mathrm{a}^{\mathrm{2}} \\ $$$$\left(\mathrm{dy}/\mathrm{dx}\right)=\mathrm{0}\:\mathrm{when}\:\mathrm{3x}^{\mathrm{2}} −\mathrm{4ax}+\mathrm{a}^{\mathrm{2}} =\mathrm{0}\:\mathrm{or}\:\mathrm{when} \\ $$$$\mathrm{x}=\frac{\mathrm{a}}{\mathrm{3}}\:\mathrm{which}\:\mathrm{gives}\:\mathrm{from}\:\left(\mathrm{1}\right)\:,\:\mathrm{y}=\frac{\mathrm{a}\sqrt{\mathrm{2}}}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$$$\mathrm{Area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{loop}\:\mathrm{A}=\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{a}} \mathrm{y}\:\mathrm{dx}\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{a}} \frac{\left(\mathrm{x}−\mathrm{a}\right)\sqrt{\mathrm{x}}}{\:\sqrt{\mathrm{2a}}}\:\mathrm{dx} \\ $$$$\mathrm{A}=\:\sqrt{\frac{\mathrm{2}}{\mathrm{a}}}\:\int_{\mathrm{0}} ^{\:\mathrm{a}} \left(\mathrm{x}^{\mathrm{3}/\mathrm{2}} −\mathrm{ax}^{\mathrm{1}/\mathrm{2}} \right)\:\mathrm{dx}\:=\:\frac{\mathrm{4}}{\mathrm{15}}\sqrt{\mathrm{2}}\:\mathrm{a}^{\mathrm{2}} \\ $$$$\therefore\:\mathrm{by}\:\mathrm{Pappus}\:\mathrm{theorem}\:\mathrm{the}\:\mathrm{required}\:\mathrm{volume} \\ $$$$=\:\mathrm{2}\pi\mathrm{a}\:×\:\mathrm{A}\:=\:\frac{\mathrm{8}}{\mathrm{15}}\:\sqrt{\mathrm{2}}\:\pi\mathrm{a}^{\mathrm{3}} \:.\:……………….\lozenge \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *