# Thread: frustrated - area by slice

1. ## frustrated - area by slice

I need to find the area of the ring defined by $y=x^3$, y=0, and x=1, with the center of the ring being x=2.

Part of the problem I am having is that while I understand the concept of what I need to do, I am getting confused between how I figure out which way the slices go (parallel to x axis, or parallel to y axis), whether I need to set up the integral with respect to x or with respect to y. For example, I have $y=x^3$, do I plug that into the integral, or do I need to solve for x (getting $x=y^\frac{1}{3}$ and then plug that into the integral? I'm not sure how to keep that all straight.

I am getting:

$\pi\int_0^1(2-y^\frac{1}{3})^2-(2-1)^2$ which becomes $\pi\int_0^1(3-2y^\frac{1}{3}+y^\frac{2}{3}$

Take the integral and get:

$\pi\[_0^1(3y-\frac{3}{2}y^\frac{4}{3}+\frac{3}{5}y^\frac{5}{3})$ (haven't quite got the syntax of the left bracket and intervals correct) , or $\pi(\frac{30}{10}-\frac{15}{10}+\frac{6}{10})$

The answer I get is $\frac{21\pi}{10}$ and the book has $\frac{3\pi}{5}$

I'm pretty confident my setup is incorrect, but I've tried several different setups and not gotten to the correct one. I'm not sure what else to try at this point.

I would appreciate some help with the setup and also some pointers on how to set up these problems in general.

Thanks!

2. ## Re: frustrated - area by slice

If some idiot (namely me) were to square a binomial correctly, things would go much better.

3. ## Re: frustrated - area by slice

$(2-y^\frac{1}{3})^2 = 4 - 4y^\frac{1}{3} + y^\frac{2}{3}$

You got that, right?