Question

Find the volume V of the solid obtained by rotating the region bounded by the given curves about the specified line. Only 1 try remaining for this problem or else I get a zero. Please help if you know the answer or how to solve.

Answer 1

Using the shell method, the volume is

$\displaystyle 2\pi \int_0^1 (2-x) \cdot 8x^3 \, dx = 16\pi \int_0^1 (2x^3 - x^4) \, dx$

Each cylindrical shell has radius $2-x$ (the horizontal distance from the axis of revolution to the curve $y=8x^3$); has height $8x^3$ (the vertical distance between a point on the $x$-axis in $0\le x\le1$ and the curve $y=8x^3$).

Compute the integral.

$\displaystyle 16 \pi \int_0^1 (2x^3 - x^4) \, dx = 16\pi \left(\frac{x^4}2 - \frac{x^5}5\right) \bigg|_{x=0}^{x=1} \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = 16\pi \left(\frac12 - \frac15\right) \\\\ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ = \frac{24}5\pi = \boxed{4.8\pi}$