# Solids of Rotation.

• December 1st 2010, 01:03 PM
Zanderist
Solids of Rotation.
My current object with this problem is just to get to the integral. Once I have that I can solve for volume myself.

The Question is:

Quote:

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

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

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

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

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

The volume is $$$V =?$$$
Here's a graph of the function as well:
http://www4d.wolframalpha.com/Calcul...44&w=367&h=301

My Attempt:
Quote:

I use the flowing formula:

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

$\int_{0}^{\sqrt{5}}(\pi(y^3)^2-\pi(5y)^2)dy$
This is what I think the integral should be, but it's not.

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

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

$\mid_{0}^{\sqrt{5}}25\pi(\frac{y^{7}}{7}-\frac{y^{3}}{3})$
• December 1st 2010, 01:27 PM
Opalg
As far as I can see, your only mistake is to have the inner and outer radii the wrong way round. For $0\leqslant y\leqslant\sqrt5$, $5y$ is bigger than $y^3$, so $5y$ should be the outer radius. Remember that you are taking horizontal disks, so the larger radius is the one that is further from the y-axis.

Edit. Just seen another mistake: you have multiplied the $y^6$ by 25, as well as the $y^2$. Obviously you shouldn't have done that.
• December 1st 2010, 01:45 PM
Zanderist
Quote:

Originally Posted by Opalg
Edit. Just seen another mistake: you have multiplied the $y^6$ by 25, as well as the $y^2$. Obviously you shouldn't have done that.

I do not see this. Or at least do not understand what your trying to point out.

However, I'm thinking you mean when I pulled out the 25 of the integral.

Okay I got the volume now, that's this problem solved.