1. ## 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:

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:

My Attempt:
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.

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

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

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

2. As far as I can see, your only mistake is to have the inner and outer radii the wrong way round. For $\displaystyle 0\leqslant y\leqslant\sqrt5$, $\displaystyle 5y$ is bigger than $\displaystyle y^3$, so $\displaystyle 5y$ 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 $\displaystyle y^6$ by 25, as well as the $\displaystyle y^2$. Obviously you shouldn't have done that.

3. Originally Posted by Opalg
Edit. Just seen another mistake: you have multiplied the $\displaystyle y^6$ by 25, as well as the $\displaystyle y^2$. 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.