# Volume between identity function and parabola

I have two functions \begin{align*}y_1=& \ x \\ y_2 =& \ x^2, [0, \infty) \end{align*}

And I am supposed to "rotate the region between $$y_1$$ and $$y_2$$ from $$x=0$$ to $$x=1$$ about the x-axis" to then find the volume of that region.

I understand at this point doing this gets a curved object with circular ends, and my job is to integrate the difference of the areas to find volume. My problem is finding the radius of the circle. In my notes I'm given that volume $$v$$ is $$\int_0^1 \pi(x)^2-\pi(x^2)^2 dx \\ v= 2*\pi/15$$ where the first quantity, $$\pi(x)^2$$, is the volume of $$y_1$$ rotated about the x axis and likewise the second quantity for $$y_2$$. It looks like here that the radius of $$y_1=x$$ and the radius of $$y_2=x^2$$, but why? I'm having a difficult time visualizing the rotation here. I tried to derive the rotation of $$y_2$$ to see if that would help, so I figured $$y_2^{-1}$$ would represent the rotated curve (and it indeed looks like it does), but after reading some stack threads it looks like the inverse isn't always a rotation about the x axis and is not always a function.

Can someone explain the physical process going on here, or provide a neat animation for it?

I've included a quick sketch of the region to be rotated about the $$x$$-axis, coloured in blue. We can think of rotating the blue region strip-by-strip to form many small volume elements and then adding the volumes of all elements to find the total volume of the solid formed. One such volume element is shown. Note that cross-sections of the solid are annular regions with outer radius determined by $$y_{1} = x$$ and inner radius determined by $$y_{2} = x^{2}.$$

$$\hskip1in$$ Divide the interval $$[0,1]$$ into subintervals using nodes $$x_{0} = 0 < x_{1} < x_{2} < \cdots < x_{n}= 1$$, and set $$\Delta x_{i} = x_{i} - x_{i-1}$$ for $$1 \leq i \leq n$$. We can calculate the element of volume formed by rotating the $$i$$th strip about the $$x$$-axis (using the right value, $$x_{i}$$, of the subinterval as the $$x$$-coordinate) as

\begin{align*} \Delta V_{i} &= \text{Area of cross section}\times \text{Width of strip} \\ &= \mathopen{}\left(\pi x_{i}^{2} - \pi (x_{i}^{2})^{2} \right)\mathclose{}\times \Delta x_{i} \\ &= \pi (x_{i}^{2} - x_{i}^{4})\Delta x_{i}. \end{align*} The total volume can be approximated by \begin{align*} V &\approx \sum_{i=1}^{n}\Delta V_{i} \\ &= \sum_{i=1}^{n} \pi (x_{i}^{2} - x_{i}^{4})\Delta x_{i}. \end{align*} As $$\Delta x_{i} \to 0$$ (or equivalently, as $$n \to \infty$$), the error in the approximation becomes smaller and we obtain \begin{align*} V &= \lim_{n \to \infty}\sum_{i=1}^{n}\pi(x_{i}^{2} -x_{i}^{4})\Delta x_{i} \\ &= \int_{0}^{1} \pi (x^{2} - x^{4})\, \mathrm{d}x \\ &= \frac{2\pi }{15}. \end{align*} An animation for a similar problem to this one can be found here: https://www.youtube.com/watch?v=3oAjcLD34kc.

Visualize a horizontal circle with its center being at the y axis, with one of the flat faces of the circle pointing towards the x axis such that the circle is "coming out" of the xy-plane.

If you were to look at this circle like you do any other graph, it would look like just a line, only once you looked at it from an angle would you see that it is circular, no?

Now imagine a line coming out of the center of the circle and going to one of its points, representing the radius. Imagine this line going all around the circle, but stop at the point when it is going right through parallel to the x axis. At this exact moment, if we went back to our xy graph, we would just see a line coming out of the y axis and going straight to the end of the "circle" (that would look just like a line again now), this here is the radius, and we can assign an x value to it by simply going straight down from the point at the end of our line.

Now take this visualization, look at the shape that y=x makes from 0 to 1, and imagine each of those points representing the end of a line that spins around the y axis just like our initial one did, that would create the shape you are taking the area of.