• Jun 1st 2010, 10:19 AM
WahooMan
Find the volume of the solid of revolution formed by revolving the planar region, $f(x)=x^3$ and $f(x)=x$, about the y-axis. You may choose either the shell or the disk method.

Here is what I did:

v = pi*[integral from 0 to 1 of (y^2 - (cube root(y))^2)dy]
= pi*[integral from 0 to 1 of (y^2 - y^(2/3))dy]
= pi*(1/3y^3 - 3/5y^5/3)
= pi*(1/3 - 3/5)
= pi*(5/15 - 9/15)
= -4pi/15

I'm pretty sure I can't have a negative volume, can I? What did I do wrong?

Also, how do you use the math tags to write a definite integral?
• Jun 1st 2010, 10:29 AM
Unknown008
If you rotate it about the x axis, the area formed by the curve f(x) = x^3 is greater than the area under the curve f(x) = x for y between 0 and 1. Hence you are getting a negative answer. :)

v = pi*[integral from 0 to 1 of ((cube root(y))^2 - y^2)dy]
• Jun 1st 2010, 10:36 AM
bigli
Solution
Use shell method and $\int_0^1\! 2\pi x(x-x^3)\, dx=\frac{4\pi}{15}$.
• Jun 1st 2010, 11:02 AM
WahooMan
Quote:

Originally Posted by Unknown008
If you rotate it about the x axis, the area formed by the curve f(x) = x^3 is greater than the area under the curve f(x) = x for y between 0 and 1. Hence you are getting a negative answer. :)

v = pi*[integral from 0 to 1 of ((cube root(y))^2 - y^2)dy]

Ah, thank you so much. This stuff can be so tricky..

Now, I have to revolve it around the x-axis, and I'm trying to use the shell method, but I'm getting a negative answer.

v = 2pi*[integral from 0 to 1 of (y(y - cube root(y))dy)]
= 2pi*[integral from 0 to 1 of (y^2 - y^4/3)dy]
= 2pi*(1/3y^3 - 3/7y^7/3)
= 2pi*(1/3 - 3/7)
= -0.598

Where did I go wrong?