Question

Find the volume of the solid of revolution formed by rotating the bounded region about the x-axis.

f(x)=4-x2, y=0, x=-1, x=1.

Answer 1

Use the disk method:

`\displaystyle\pi\int_(-1)^1(4-x^2)^2\,\mathrm dx=2\pi\int_0^1(16-8x^2+x^4)\,\mathrm dx`

where we take advantage of symmetry to rewrite the limits, and expand the integrand.

`=\displaystyle2\pi\left(16x-\frac83x^3+\frac15x^5\right)\bigg|_0^1=2\pi\left(16-\frac83+\frac15\right)=(406\pi)/(15)`