Outros tipos de volumes "cortes"

Previously, we learned how to calculate volumes of solids made by revolving a flat region around an axis. In this section, we'll learn how to find volumes of solids with general cross-sections perpendicular to some closed region.

Suppose we start with some arbitrary region bounded by two functions $f(x)$ and $g(x)$, from $x=a$ to $x=b$. Also assume that the cross-sections are perpendicular to the $x$-axis.

Let's also suppose that at any value of $a\le x\le b$, the cross-sectional area of the solid perpendicular to the plane of the region is equal to $A(x)$. If we partition the interval $[a,b]$ into $n$ equal pieces $[x_{i-1},x_i]$ with $$x_i-x_{i-1}=\Delta x$$ and $$a=x_0<x_1<\cdots<x_{n-1}<x_n=b,$$ then each ``slice'' of the solid at $x=x_i$ is a prism with base area $A(x_i)$ and height $\Delta x$. So, the volume of each slice is $$S_i=A(x_i)\Delta x,$$ meaning that the volume of the final solid can be estimated by $$V\approx \sum_{i=1}^n S_i=\sum_{i=1}^n A(x_i)\Delta x.$$ And as $n\to\infty$, the sum approaches the integral $$V=\lim_{n\to\infty} \sum_{i=1}^n A(x_i)\Delta x=\int_a^b A(x)\, dx.$$ Similarly, if the cross-sections are perpendicular to the $y$-axis, then the volume is gotten by $$V=\int_c^d A(y)\, dy,$$ where the initial region ranges from $y=c$ to $y=d$.

Great! Now that we have a way of finding the volume of such figures, let's look at an example:

Find the volume of a cylinder of radius $r$ and height $h$.

To solve this problem, let's place the cylinder sideways on the coordinate grid, such that its center axis is the $x$-axis, and the left side of the cylinder is at $x=0$. (This is merely for convenience.) We can think of this solid as a solid with a rectangular base and vertical cross-sections of circles:

Since the cross-sections are circles, we find that at any $x$, $$A(x)=\pi\cdot r^2.$$ Therefore, the volume of the cylinder becomes $$V=\int_0^h \pi\cdot r^2\, dx=\pi\cdot r^2\int_0^h \, dx=\pi r^2h,$$ which is what we expected.

Next, let's try a slightly more difficult question:

Suppose we have a region bounded by the curve $f(x)=\sqrt{1-x^2}$ and the $x$-axis on $[-1,1]$, and let's find the volume of the solid whose base is the mentioned region and whose vertical cross-sections are squares perpendicular to the $x$-axis.

Since the area of a square is the square of its side length, we get that $$A=y^2=\left(\sqrt{1-x^2}\right)^2=1-x^2.$$

We apply the formula proven above: $$V=\int_{-1}^1 A(x)\, dx=\int_{-1}^1 (1-x^2)\, dx=\left(x-\frac{x^3}{3}\right)\bigg|_{-1}^1=\frac{4}{3}.$$

Now you try on your own!

First, we observe that the cross-sections are perpendicular to the $y$-axis, meaning that we need to use the formula $$V=\int_c^d A(y)\, dy.$$ So, we draw change the diagram as follows:

Note that this time, we need to represent all points in terms of $y$. Since our original function was $$y=\sqrt{1-x^2},$$ we can rearrange this to get $$x=\sqrt{1-y^2},$$ as indicated in the graph. Now, this tells us that the length of each square is $$2x=2\sqrt{1-y^2}.$$ So, our area function is $$A(y)=\left(2\sqrt{1-y^2}\right)^2=4-4y^2.$$ Therefore, since the graph clearly ranges from $y=0$ to $y=1$, by the formula above $$V=\int_0^1 A(y)\, dy=\int_0^1 (4-4y^2)\, dy=\left(4y-\frac{4y^3}{3}\right)\bigg|_0^1=\frac{8}{3}.$$

We end with one final remark: recall the disk and washer methods, used to find volumes of solids gotten by revolving a region about an axis. For example, we found that when the took the region bounded by functions $f(x)$ the $x$-axis in the interval $[a,b)$, and we revolved that region about the $x$-axis, the volume of the resulting solid was equal to $$V=\int_a^b \pi f(x)^2\, dx.$$ Do you see a connection between that method and the one described in this section?

In fact, the one described here is a generalization of the disk and washer methods! To illustrate, suppose we revolve the region bounded by $f(x)=x^2$ and the $x$-axis on the interval $[0,1]$ about the $x$-axis.

The solid gotten above would be identical to the solid with base $R$ (denoted in the diagram above) and cross-sections of circles perpendicular to the $x$-axis. Since the ``radius'' of these cross-sections at any value of $x$ is simply $y=f(x)$, we have $$V=\int_0^1 A(x)\, dx=\int_0^1 \pi f(x)^2\, dx,$$ which is consistent with the former method.

What does this mean? It means that this method (of calculating volumes of solids through their general cross-sections) is stronger and more applicable than the previous methods shown. So, if all else fails, keep this method in mind!