# 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=x^(1/2), y=0, and x=4 rotated about the x axis?

Aug 18, 2015

This solid to-be-revolved looks like:

graph{(y - sqrtx)(y)sqrt(4-(x-0))/sqrt(4-(x-0)) <= 0 [-5, 5, 0, 5]}

If you want to do it with the shell method, convert your functions to their inverses.

You get:
$y = {x}^{2}$
$x = 0$
$y = 4$

Now your domain is your range and your interval is $\left[0 , 2\right]$.

graph{(y - x^2)sqrt(4-(y-0))/sqrt(4-(y-0)) >= 0 [0, 8, -1, 5]}

$V = 2 \pi \int x f \left(x\right) \mathrm{dx}$

$f \left(x\right) = {x}^{2} \in \left[0 , 2\right]$, bounded from above by $y = 4$, thus $f \left(x\right) = 4 - {x}^{2}$

$V = 2 \pi {\int}_{0}^{2} x \cdot \left(4 - {x}^{2}\right) \mathrm{dx}$

$= 2 \pi \left[2 {x}^{2} - {x}^{4} / 4\right] {|}_{0}^{2}$

$= 2 \pi \left[\left(2 {\left(2\right)}^{2} - {\left(2\right)}^{4} / 4\right) - \cancel{\left(2 {\left(0\right)}^{2} - {\left(0\right)}^{4} / 4\right)}\right]$

= color(blue)(8pi "u") ~~ 25.1327 "u"

Note: if you evaluated this using $f \left(x\right) = {x}^{2}$ instead, you would coincidentally get the same answer, but for the wrong work, so if you did that on a test, you wouldn't get full credit.