How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region y = 16 - x^2 and 0 ≤ x ≤ 4 rotated about the x-axis?

Oct 28, 2016

The volume is 546.1

Explanation:

The volume of a shell is surface area multiplied by thickness
The surface area is $\pi {y}^{2}$
And the volume is =surface area x thickness

$\mathrm{dV} = \pi {y}^{2} \mathrm{dx}$
$y = 16 - {x}^{2}$
so ${y}^{2} = 256 - 32 {x}^{2} + {x}^{4}$

The volume is $V = \pi {\int}_{0}^{4} \left(256 - 32 {x}^{2} + {x}^{4}\right) \mathrm{dx}$

$V = \pi {\left(256 x - \frac{32 {x}^{3}}{3} + {x}^{5} / 5\right)}_{0}^{4}$
$V = \pi \left(256 \cdot 4 - 32 \cdot \frac{64}{3} + \frac{1024}{5}\right)$
$V = \pi \left(1024 - \frac{2048}{3} + \frac{1024}{5}\right) = 546.1$