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The volume obtained on revolving about $x=a/2$, the area enclosed between the curves $xy^{2} = a^{2}(a-x)$ and $(a-x)y^{2} = a^{2}x$ is ......$?$

I've drawn both curves and both intersect at $x=a/2$, but the line $x=a/2$ lies in middle of both curves. Now, I have no idea how to find the volume of solid. I know the formula, but don't know if area enclosed between both curves is symmetric about $x=a/2$, so that I can find the volume for one curve only. If area is not symmetric then how would I find the generated volume$?$

Mathaddict
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1 Answers1

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We want the volume of the solid generated by revolving the area through an angle of pi.

Calculate half the volume of revolution of the area between the curve $xy^2=a^2(a-x)$ and the line $x=\frac{a}2$, then add half the volume of revolution of the area between the curve $(a-x)y^2=a^2x$ and the line $x=\frac{a}2$

Angela Pretorius
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