## General Slicing Method

Suppose a solid object extends from x = a to x = b and the cross section of the solid perpendicular to the x-axis has an area given by a function A that is integrable on [a, b]. The volume of the solid is:

So I just wanted to chime in a particularly important analysis you have to remember in order to properly perform the “general Slicing Method”. What it looks like to me is that you have to decide the shape of the cross section, and factor that into calculating the area.

For example, a square with x side length will have an area of x^{2}.

For example, a circle with r radius will have an area of πr^{2}. But if you are taking a cross section of a parabola, then you would actually have an (?approximate) area of (1/2)*πr^{2}.

To further elaborate, notice in the video above that he squares what is in-between the brackets. That’s because he is taking the cross-sectional area of a **square**.

## Disk Method about the X Axis

Let f be continuous with f(x) ≥ 0 on the interval [a, b]. If the region R bounded by the graph of f, the x-axis, and the lines x = a and x = b is revolved about the x-axis, the volume of the resulting solid of revolution is:

Area of circle = πr^{2}.

You get r from measuring the distance of f(x) to the x-axis.

dx is the thickness or width of the disk.

∫ for interval [a, b] is a summation of all the disks between [a, b].

## Washer Method about the x-axis

Let f and g be continuous functions with f(x) ≥ g(x) ≥ 0 on [a, b]. Let R be the region bounded by the curves y = f(x) and y = g(x), and the lines x = a and x = b. When R is revolved about the x-axis, the volume of the resulting solid of revolution is

We aren’t measuring the area between two curves in a 2D fashion. No, this is quite a bit more.

We are revolving that area between the two curves around an axis. At least, that’s why I understand.

The area of a circle is πr^{2}. Given that we are revolving around the x-axis (that’s my assumption atm), the length from f(x) to the x-axis is not the diameter. No. It’s the radius. So that’s why f(x) = r, in general.

But we are dealing with a washer. Imagine another f(x), the same, but smaller, and you subtract it from the bigger f(x). That is g(x). g(x) is responsible for the hole of the washer. That’s why you have f(x) – g(x).

f(x) – g(x) becomes π(f(x)^{2} – g(x)^{2}) because you are calculating the area of the circle____? You are substituting for πr^{2}, the area of the circle.

Times dx for the thickness of each slice. And you summate the slices by integrating between the interval [a, b].

I suppose that’s how you get the volume of a washer.

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