Question

Use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region about the y-axis. y = 8x − x2 x = 0 y = 16

Answer 1

Completing the square gives

`y=8x-x^2=16-(x-4)^2`

and

`16=16-(x-4)^2\implies(x-4)^2=0\implies x=4`

tells us the parabola intersect the line
`y=16` at one point, (4, 16).

Then the volume of the solid obtained by revolving shells about
`x=0` is

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

`=\pi\frac{(x-4)^3}3\bigg|_(x=0)^(x=4)=\boxed{\frac{64\pi}3}`