I am wondering why I get an incorrect answer when trying to find the volume of a rotated function about the x-axis when using the Average Value Theorem.
I want to find the volume of $y=\sqrt{x-2}$ as it is revolved around the $x-axis$ on $ x\in[2,6]$
Classic Calculus (area with disks)
$\int_2^6\left(\pi x - 2\pi\right)dx$
Why can't I do...
$4\pi\left(\frac14\int_2^6\sqrt{x-2}dx\right)^2$
Basically my plan is to create a cylinder with r = average radius. It is using the volume for a cylinder $V=\pi r^2h$. The version with variables looks like...
$\pi(a-b)\left(\frac1{b-a}\int_a^bf(x)dx\right)^2$
The problem is that the computation of the average value of $y$ assigns equal weights to all of the $y$ values, but unlike the situation for areas, they don’t contribute equally to the volume.
The contribution of a particular $y$ value to the total area under a curve is directly proportional to $y$, and thus also directly proportional to $\Delta y=y-\bar y$, so replacing $y$ by $\bar y$, the unweighted average, is appropriate. For a volume of rotation, however, you have $dV=\pi y^2\,dx$, and $$\pi(y+\Delta y)^2-\pi\bar y^2 = \pi(2\bar y\Delta y+\Delta y^2),$$ which is nonlinear. You could, of course, use an average value to compute a volume of rotation, but it would have to be an appropriately-weighted one.