Question

Find the volume of the solid formed by revolving the region bounded by the graphs of y = x^3, x = 2, and y = 1 about the y-axis.

Answer 1

If you're using the shell method, the setup would be

$\displaystyle2\pi\int_1^2x(x^3-1)\,\mathrm dx$

Reasoning: consider just one shell (see the image below). The height of the shell is determined by the vertical distance between the curve $y=x^3$ and the line $y=1$. The radius of each shell is determined by the value of $x$, which can only range from 1 to 2 within the region of interest, and the height is determined exactly by $x^3-1$ at this $x$. Multiply by $2\pi$ to get the surface area of this infinitesimally thin cylindrical shell, then integrate along the interval [1, 2] to get the volume.

We have

$\displaystyle2\pi\int_1^2x(x^3-1)\,\mathrm dx=2\pi\int_1^2(x^4-x)\,\mathrm dx=2\pi\left(\dfrac{x^5}5-\dfrac{x^2}2\right)\bigg|_1^2=\dfrac{47\pi}5$