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**Reckoner** Using the shell method, you can integrate with respect to $\displaystyle x\colon$

$\displaystyle V=2\pi\int_a^bp(x)h(x)\,dx$

$\displaystyle =2\pi\int_0^1(1-x)\left(5\sqrt x-5x^3\right)\,dx$

(here $\displaystyle p(x)$ is the distance from the axis of revolution at $\displaystyle x,$ and $\displaystyle h(x)$ is the height of the region at $\displaystyle x$). This can be easily integrated.

Alternatively, solve for $\displaystyle x$ in the two equations to get $\displaystyle x=y^2/25$ and $\displaystyle x=y^{1/3}/\sqrt[3]5\text.$ Then we have

$\displaystyle V=\pi\int_a^b\left([R(y)]^2-[r(y)]^2\right)dy$

$\displaystyle =\pi\int_0^5\left[\left(1-\frac{y^2}{25}\right)^2-\left(1-\frac{y^{1/3}}{\sqrt[3]5}\right)^2\right]dy\text.$