# Volume of the solid generated by revolution of the given curve.

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$$?$$

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 $$x=\frac{a}2$$