How do you find the volume of the solid obtained by rotating the region bounded by the curves Y=2x and Y=x^2 rotated around the x-axis?

May 23, 2015

Remembering that the area between $f \left(x\right)$ and $g \left(x\right)$, with $f \left(x\right)$ over $g \left(x\right)$ from ${x}_{1}$ to ${x}_{2}$ is given by:

${\int}_{{x}_{1}}^{{x}_{2}} \left[f \left(x\right) - g \left(x\right)\right] \mathrm{dx}$,

and since $y = 2 x$ is over the function $y = {x}^{2}$, and the intercept point is given by the system:

$y = 2 x$

$y = {x}^{2}$

$2 x = {x}^{2} \Rightarrow x \left(x - 2\right) = 0 \Rightarrow$

${x}_{1} = 0$ and ${x}_{2} = 2$,

then:

${\int}_{0}^{1} \left(2 x - {x}^{2}\right) \mathrm{dx} = {\left[{x}^{2} - {x}^{3} / 3\right]}_{0}^{2} =$

$= \left[{2}^{2} - {2}^{3} / 3 - \left({0}^{2} - {0}^{3} / 3\right)\right] = 4 - \frac{8}{3} = \frac{4}{3}$.