# Thread: Why is that we use the boundaries from -1 to 1 instead of 0 to 1?

1. ## Why is that we use the boundaries from -1 to 1 instead of 0 to 1?

The given curve is rotated about the y-axis. Find the area of the resulting surface.

$\sqrt[3]{x^2}+\sqrt[3]{y^2}=1 , 0\leq y \leq1$

We set up the surface area integral:

$\displaystyle \int_{-1}^{1} 2\pi x ds$

I understand how to find the arc-length(ds) and set up the integral but why wouldn't out boundaries stay from 0 to 1? My reasoning is that $\displaystyle 2\pi x$
is taking in account that our radius from 0 to 1 should be doubled. Shouldn't we only care about the arc-length from 0 to 1?

For instance if we were to find the surface area of $\displaystyle y=1/3 \sqrt[2]{x^3}+$ , $0\leq y \leq12$rotating about the y-axis, our integral would be,

$\displaystyle \int_{0}^{12} 2\pi y ds$

2. ## Re: Why is that we use the boundaries from -1 to 1 instead of 0 to 1?

The figure is defined from 0 to 1 but the it is rotated around the y- axis so the is exactly the same figure from -1 to 0. The entire rotated figure will go from x= -1 to x= 1.

3. ## Re: Why is that we use the boundaries from -1 to 1 instead of 0 to 1?

what if it is rotated around the x axis? i thought saying the radious times one side of the curve would give us the surface area? thank you for helping me

4. ## Re: Why is that we use the boundaries from -1 to 1 instead of 0 to 1?

You know that $0\le y \le 1$. Without it being rotated around anything, let's calculate bounds for $x$:

$\sqrt[3]{x^2} + \sqrt[3]{y^2} = 1 \Longrightarrow x = \pm \left(1-\sqrt[3]{y^2}\right)^{\tfrac{3}{2}}$

So, using the bounds for $y$, we find $x \ge -\left(1-\sqrt[3]{0^2}\right)^{\tfrac{3}{2}} = -1$ and $x \le \left(1-\sqrt[3]{0^2}\right)^{\tfrac{3}{2} = 1$. So, $-1 \le x \le 1$.

5. ## Re: Why is that we use the boundaries from -1 to 1 instead of 0 to 1?

I wonder if the original question meant to restrict x to $\displaystyle 0 \leq x \leq 1$. The function $\displaystyle x^\frac{2}{3} + y^\frac{2}{3} = 1$ looks like an inverted "v" for $\displaystyle 0 \leq y \leq 1$. If this inverted "v" is rotated about the y-axis, it will yield twice the usual surface area if one were to use $\displaystyle -1 \leq x \leq 1$. I believe that the OP is in the right to be skeptical of the bounds $\displaystyle -1 \leq x \leq 1$. See Example 2, p.4 [Stewart].

M><M