Quote:

The volume of the solid obtained by rotating the region enclosed by

$\displaystyle \[x = 5 y, \quad y^3 = x \quad (\mbox{with } y\geq 0)\] $

about the y-axis can be computed using the method of disks or washers via an integral

$\displaystyle \(\displaystyle V = \int_a^b\) part I'm having trouble with, however it is with respect to y; so dy$

with limits of integration $\displaystyle \(a = 0\)$ and $\displaystyle \(b =\sqrt{5}\)$

The volume is $\displaystyle \(V =?\)$

Here's a graph of the function as well: Quote:

I use the flowing formula:

$\displaystyle \int_{0}^{\sqrt{5}}(\pi(outer\, radius)^2-\pi(inner\, radius)^2)dy$

$\displaystyle \int_{0}^{\sqrt{5}}(\pi(y^3)^2-\pi(5y)^2)dy$

This is what I think the integral should be, but it's not.

$\displaystyle \pi\int_{0}^{\sqrt{5}}((y^6)-(25y^2))dy$

$\displaystyle 25\pi\int_{0}^{\sqrt{5}}((y^6)-(y^2))dy$

$\displaystyle \mid_{0}^{\sqrt{5}}25\pi(\frac{y^{7}}{7}-\frac{y^{3}}{3})$