MHB Volume enclosed by rotating a curve segment

[Guest2]

The volume enclosed by rotating the segment of the curve $y = \frac{1}{2}|x-1|$ between $x = 0$ and $x = 2$ about the $x$-axis is equal to:

Is it this simple? Since $x \ge 0$ it's $\frac{\pi}{2} \int_0^2 (\sqrt{(x-1)^2})^2\, dx = \frac{\pi}{3}.$

 

[Prove It]

The volume enclosed by rotating the segment of the curve $y = \frac{1}{2}|x-1|$ between $x = 0$ and $x = 2$ about the $x$-axis is equal to:

Is it this simple? Since $x \ge 0$ it's $\frac{\pi}{2} \int_0^2 (\sqrt{(x-1)^2})^2\, dx = \frac{\pi}{3}.$

No, it's not that simple, because you are going to have two different functions from the absolute value. For $\displaystyle \begin{align*} 0 \leq x < 1 \end{align*}$ we have $\displaystyle \begin{align*} \frac{1}{2} \, \left| x - 1 \right| = -\frac{1}{2} \, \left( x - 1 \right) \end{align*}$ while for $\displaystyle \begin{align*} 1 \leq x \leq 2 \end{align*}$ we have $\displaystyle \begin{align*} \frac{1}{2} \,\left| x - 1 \right| = \frac{1}{2} \, \left( x - 1 \right) \end{align*}$.

 

[Guest2]

No, it's not that simple, because you are going to have two different functions from the absolute value. For $\displaystyle \begin{align*} 0 \leq x < 1 \end{align*}$ we have $\displaystyle \begin{align*} \frac{1}{2} \, \left| x - 1 \right| = -\frac{1}{2} \, \left( x - 1 \right) \end{align*}$ while for $\displaystyle \begin{align*} 1 \leq x \leq 2 \end{align*}$ we have $\displaystyle \begin{align*} \frac{1}{2} \,\left| x - 1 \right| = \frac{1}{2} \, \left( x - 1 \right) \end{align*}$.

Thank you. So it's \frac{\pi}{4}\int_0^1 (1-x)^2\,{dx}+\frac{\pi}{4}\int_1^2 (x-1)^2\,dx = \frac{\pi}{6}?

 

[Guest2]

Is it by coincidence that I get the same answer from what I was doing earlier as well?

I made a mistake so the answer was off, but now I get $\displaystyle \frac{\pi}{4} \int_0^2 (\sqrt{(x-1)^2})^2\, dx = \frac{\pi}{6}.$

 

[MarkFL]

We know, by looking at the graph, and using the formula for the volume of a right circular cone, that we will have:

$$V=2\cdot\frac{\pi}{3}\left(\frac{1}{2}\right)^2(1)=\frac{\pi}{6}$$

If I we going to do this using the calculus, I would shift the function 1 unit to the left, and then use the even function rule to write:

$$V=2\pi\int_0^1 \frac{1}{4}x^2\,dx=\frac{\pi}{2}\left[\frac{x^3}{3}\right]_0^1=\frac{\pi}{6}$$

 

[Greg]

A slightly different approach:

Thank you. So it's \frac{\pi}{4}\int_0^1 (1-x)^2\,{dx}+\frac{\pi}{4}\int_1^2 (x-1)^2\,dx = \frac{\pi}{6}?

Since you're squaring, there's no difference between the two integrands. Consider what the problem is geometrically: two cones with base radius 1/2 and height 1. Thus, we need to calculate

$$2\cdot\pi\int_0^1\left[\dfrac12(1-x)\right]^2\,dx$$

or (neatly),

$$\dfrac{\pi}{2}\int_0^1(1-x)^2\,dx=\dfrac{\pi}{6}$$