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}`