# Find the volume of the solid generated by rotating the region bounded

1. Feb 5, 2013

### jorgegalvan93

1. The problem statement, all variables and given/known data

Find the volume of the solid generated by rotating the region bounded by the x-axis, the curve y=3x^4, and the lines x=1 and x= -1. The axis of rotation is the y-axis.

2. Relevant equations

Washers method: V=∏∫ [(R)^2 - (r)^2]dr
x = (y/3)^(1/4)

3. The attempt at a solution

So for the following problem, I used washers method to solve for the volume of the solid. R representing the outer radius, and r representing the inner radius (hole).
Because my washer is perpendicular do the y-axis, I will set my expression with respect to y… dy
I will use symmetry, so my R is 1, and r is a vertical distance so it is x, which is equal to (y/3)^(1/4)

V=∏∫ [(1)^2 - ((y/3)^(1/4))^2]dy , where the limits of integration are a = 0 and b = 3
V = 2∏∫ [(1) - √(y/3)]dy (step where problem arises)
V = 2∏[(y - 2/3(y/3)^(3/2)] from 0 to 3
V = 2∏[3 - 2/3(3/3)^(3/2)]
V = 2∏[3 - 2/3(1)]
V = 2∏[3 -(2/3)] … V = 2∏(7/3)….

V = 14∏/3

My professor said that on the second step performed, I should have done a substitution… u = y/3. I don't think that's correct though, because I would still have a 1.
Can anyone give me a hand with this problem?

Thank you!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 6, 2013

### SammyS

Staff Emeritus
You are combining some aspects of the washer method and some aspects of the shell method.

The washer method should look like
$\displaystyle \pi\int_{a}^{b}\,(f(y))^2-(g(y))^2\,dy\ .$​

3. Feb 6, 2013

### jorgegalvan93

That's the method I used. Where f(y) = 1 and g(y) = (y/3)^(1/4)
I have a 2 next to ∏ because I am using the symmetry of the function. I thought it would be much simpler that way. But I suppose I can find the volume without using symmetry. By doing that, f(y) = 2 and g(y) = 2x???
I'm confused as to what g(y) would have to be in this instance.

4. Feb 6, 2013

### SammyS

Staff Emeritus
Do not multiply ny 2. Only use the right half or the left half of the region. Either will "sweep out" the entire volume that you're interested in.

Also, I agree with your professor. The 1 is not the issue.

I integrating $\displaystyle \ \ \sqrt{\frac{y}{3}}\,, \$ you have that the anti-derivative of that is $\displaystyle \ \ \frac{2}{3}\left(\frac{y}{3}\right)^{3/2}\ .$

Now, if you check that result by taking the derivative, you get $\displaystyle \ \ \frac{2}{3}\frac{3}{2}\left(\frac{y}{3}\right)^{1/2}\left(\frac{1}{3}\right) \ .\$ The last 1/3 coming from the chain rule.