Determining Volumes by Slicing

Volumes by Slices

When we think about volume from an intuitive point of view, we typically think of it as the amount of "space" an item occupies. Unfortunately assigning a number that measures this amount of space can prove difficult for all but the simplest geometric shapes. Calculus provides a new tool that can greatly extend our ability to calculate volume. In order to understand the ideas involved it helps to think about the volume of a cylinder.

The volume of a cylinder is calculated using the formula $V=\pi r^{2}h$ . The base of the cylinder is a circle whose area is given by $A=\pi r^{2}$ . Notice that the volume of a cylinder is derived by taking the area of its base and multiplying by the height $h$ . For more complicated shapes, we could think of approximating the volume by taking the area of some cross section at some height $x$ and multiplying by some small change in height $\Delta x$ then adding up the heights of all of these approximations from the bottom to the top of the object. This would appear to be a Riemann sum. Keeping this in mind, we can develop a more general formula for the volume of solids in $\mathbb {R} ^{3}$ (3 dimensional space).

Formal Definition

Formally the ideas above suggest that we can calculate the volume of a solid by calculating the integral of the cross-sectional area along some dimension. In the above example of a cylinder, every cross section is given by the same circle, so the cross-sectional area is therefore a constant function, and the dimension of integration was vertical (although it could have been any one we desired). Generally, if $S$ is a solid that lies in $\mathbb {R} ^{3}$ between $x=a$ and $x=b$ , let $A(x)$ denote the area of a cross section taken in the plane perpendicular to the $x$-axis, and passing through the point $x$ .

If the function $A(x)$ is continuous on $[a,b]$ , then the volume $V_{S}$ of the solid $S$ is given by:

$V_{S}=\int \limits _{a}^{b}A(x)dx$

Examples

Example 1: A right cylinder

Figure 1

Now we will calculate the volume of a right cylinder using our new ideas about how to calculate volume. Since we already know the formula for the volume of a cylinder this will give us a "sanity check" that our formulas make sense. First, we choose a dimension along which to integrate. In this case, it will greatly simplify the calculations to integrate along the height of the cylinder, so this is the direction we will choose. Thus we will call the vertical direction $x$ (see Figure 1). Now we find the function, $A(x)$ , which will describe the cross-sectional area of our cylinder at a height of $x$ . The cross-sectional area of a cylinder is simply a circle. Now simply recall that the area of a circle is $\pi r^{2}$ , and so $A(x)=\pi r^{2}$ . Before performing the computation, we must choose our bounds of integration. In this case, we simply define $x=0$ to be the base of the cylinder, and so we will integrate from $x=0$ to $x=h$ , where $h$ is the height of the cylinder. Finally, we integrate:

${\begin{aligned}V_{\mathrm {cylinder} }&=\int \limits _{a}^{b}A(x)dx\\&=\int \limits _{0}^{h}\pi r^{2}dx\\&=\pi r^{2}\int \limits _{0}^{h}dx\\&=\pi r^{2}x{\bigg |}_{x=0}^{h}\\&=\pi r^{2}(h-0)\\&=\pi r^{2}h\end{aligned}}$

This is exactly the familiar formula for the volume of a cylinder.

Example 2: A right circular cone

Figure 2: The cross-section of a right circular cone by a plane perpendicular to the axis of the cone is a circle.

For our next example we will look at an example where the cross sectional area is not constant. Consider a right circular cone. Once again the cross sections are simply circles. But now the radius varies from the base of the cone to the tip. Once again we choose $x$ to be the vertical direction, with the base at $x=0$ and the tip at $x=h$ , and we will let $R$ denote the radius of the base. While we know the cross sections are just circles we cannot calculate the area of the cross sections unless we find some way to determine the radius of the circle at height $x$ .

Figure 3: Cross-section of the right circular cone by a plane perpendicular to the base and passing through the tip.

Luckily in this case it is possible to use some of what we know from geometry. We can imagine cutting the cone perpendicular to the base through some diameter of the circle all the way to the tip of the cone. If we then look at the flat side we just created, we will see simply a triangle, whose geometry we understand well. The right triangle from the tip to the base at height $x$ is similar to the right triangle from the tip to the base at height $h$ . This tells us that ${\frac {r}{h-x}}={\frac {R}{h}}$ . So that we see that the radius of the circle at height $x$ is $r(x)={\frac {R}{h}}(h-x)$ . Now using the familiar formula for the area of a circle we see that $A(x)=\pi {\frac {R^{2}}{h^{2}}}(h-x)^{2}$ .

Now we are ready to integrate.

${\begin{aligned}V_{\mathrm {cone} }&=\int \limits _{a}^{b}A(x)dx\\&=\int \limits _{0}^{h}\pi {\frac {R^{2}}{h^{2}}}(h-x)^{2}dx\\&=\pi {\frac {R^{2}}{h^{2}}}\int \limits _{0}^{h}(h-x)^{2}dx\end{aligned}}$

By u-substitution we may let $u=h-x$ , then $du=-dx$ and our integral becomes

${\begin{aligned}&&=\pi {\frac {R^{2}}{h^{2}}}\left(-\int \limits _{h}^{0}u^{2}du\right)\\&&=\pi {\frac {R^{2}}{h^{2}}}\left(-{\frac {u^{3}}{3}}{\bigg |}_{h}^{0}\right)\\&&=\pi {\frac {R^{2}}{h^{2}}}\left(-0+{\frac {h^{3}}{3}}\right)\\&&={\frac {\pi }{3}}R^{2}h\end{aligned}}$

Example 3: A sphere

Figure 4: Determining the radius of the cross-section of the sphere at a distance $|x|$ from the sphere's center.

In a similar fashion, we can use our definition to prove the well known formula for the volume of a sphere. First, we must find our cross-sectional area function, $A(x)$ . Consider a sphere of radius $R$ which is centered at the origin in $\mathbb {R} ^{3}$ . If we again integrate vertically then $x$ will vary from $-R$ to $R$ . In order to find the area of a particular cross section it helps to draw a right triangle whose points lie at the center of the sphere, the center of the circular cross section, and at a point along the circumference of the cross section. As shown in the diagram the side lengths of this triangle will be $R$ , $|x|$ , and $r$ . Where $r$ is the radius of the circular cross section. Then by the Pythagorean theorem $r={\sqrt {R^{2}-|x|^{2}}}$ and find that $A(x)=\pi (R^{2}-|x|^{2})$ . It is slightly helpful to notice that $|x|^{2}=x^{2}$ so we do not need to keep the absolute value.

So we have that

${\begin{aligned}V_{\mathrm {sphere} }&=\int \limits _{a}^{b}A(x)dx\\&=\int \limits _{-R}^{R}\pi (R^{2}-x^{2})dx\\&=\pi \int \limits _{-R}^{R}R^{2}dx-\pi \int \limits _{-R}^{R}x^{2}dx\\&=\pi R^{2}x{\Bigg |}_{-R}^{R}-\pi {\frac {x^{3}}{3}}{\Bigg |}_{-R}^{R}\\&=\pi R^{2}(R-(-R))-\pi \left({\frac {R^{3}}{3}}-{\frac {(-R)^{3}}{3}}\right)\\&=2\pi R^{3}-{\frac {2\pi }{3}}R^{3}={\frac {4\pi }{3}}R^{3}\end{aligned}}$

Volume of Solids of Revolution

In this section we cover solids of revolution and how to calculate their volume. A solid of revolution is a solid formed by revolving a 2-dimensional region around an axis. For example, revolving the semi-circular region bounded by the curve $y={\sqrt {1-x^{2}}}$ and the line $y=0$ around the $x$-axis produces a sphere. There are two main methods of calculating the volume of a solid of revolution using calculus: the disk method and the shell method.

Disk Method

Figure 1: A solid of revolution is generated by revolving this region around the x-axis.

Figure 2: Approximation to the generating region in Figure 1.

Consider the solid formed by revolving the region bounded by the curve $y=f(x)$ , which is continuous on $[a,b]$ , and the lines $x=a$ , $x=b$ and $y=0$ around the $x$-axis. We could imagine approximating the volume by approximating $f(x)$ with the stepwise function $g(x)$ shown in figure 2, which uses a right-handed approximation to the function. Now when the region is revolved, the region under each step sweeps out a cylinder, whose volume we know how to calculate, i.e.

$V_{\rm {cylinder}}=\pi r^{2}h$

where $r$ is the radius of the cylinder and $h$ is the cylinder's height. This process is reminiscent of the Riemann process we used to calculate areas earlier. Let's try to write the volume as a Riemann sum and from that equate the volume to an integral by taking the limit as the subdivisions get infinitely small.

Consider the volume of one of the cylinders in the approximation, say the $k$-th one from the left. The cylinder's radius is the height of the step function, and the thickness is the length of the subdivision. With $n$ subdivisions and a length of $b-a$ for the total length of the region, each subdivision has width

$\Delta x={\frac {b-a}{n}}$

Since we are using a right-handed approximation, the $k$-th sample point will be

$x_{k}=k\Delta x$

So the volume of the $k$-th cylinder is

$V_{k}=\pi f(x_{k})^{2}\Delta x$

Summing all of the cylinders in the region from $a$ to $b$ , we have

$V_{\rm {approx}}=\sum _{k=1}^{n}\pi f(x_{k})^{2}\Delta x$

Taking the limit as $n$ approaches infinity gives us the exact volume

$V=\lim _{n\to \infty }\sum _{k=1}^{n}\pi f(x_{k})^{2}\Delta x$

which is equivalent to the integral

$V=\int \limits _{a}^{b}\pi f(x)^{2}dx$

Example: Volume of a Sphere

Let's calculate the volume of a sphere using the disk method. Our generating region will be the region bounded by the curve $f(x)={\sqrt {r^{2}-x^{2}}}$ and the line $y=0$ . Our limits of integration will be the $x$-values where the curve intersects the line $y=0$ , namely, $x=\pm r$ . We have

${\begin{aligned}V_{\rm {sphere}}&=\int \limits _{-r}^{r}\pi (r^{2}-x^{2})dx\\&=\pi \left(\int \limits _{-r}^{r}r^{2}dx-\int \limits _{-r}^{r}x^{2}dx\right)\\&=\pi \left(r^{2}x{\bigg |}_{-r}^{r}-{\frac {x^{3}}{3}}{\bigg |}_{-r}^{r}\right)\\&=\pi {\Big (}r^{2}{\bigl (}r-(-r){\bigr )}-{\tfrac {1}{3}}{\bigl (}r^{3}-(-r)^{3}{\bigr )}{\Big )}\\&=\pi \left(2r^{3}-{\frac {2r^{3}}{3}}\right)\\&=\pi {\frac {6r^{3}-2r^{3}}{3}}\\&={\frac {4\pi }{3}}r^{3}\end{aligned}}$

Washer Method

Figure 3: A solid of revolution containing an irregularly shaped hole through its center is generated by revolving this region around the x-axis.

Figure 4: Approximation to the generating region in Figure 3.

The washer method is an extension of the disk method to solids of revolution formed by revolving an area bounded between two curves around the $x$-axis. Consider the solid of revolution formed by revolving the region in figure 3 around the $x$-axis. The curve $f(x)$ is the same as that in figure 1, but now our solid has an irregularly shaped hole through its center whose volume is that of the solid formed by revolving the curve $g(x)$ around the $x$-axis. Our approximating region has the same upper boundary, $f_{\rm {step}}(x)$ as in figure 2, but now we extend only down to $g_{\rm {step}}(x)$ rather than all the way down to the $x$-axis. Revolving each block around the $x$-axis forms a washer-shaped solid with outer radius $f_{\rm {step}}(x)$ and inner radius $g_{\rm {step}}(x)$ . The volume of the $k$-th hollow cylinder is

${\begin{aligned}V_{k}&=\pi \cdot f(x_{k})^{2}\Delta x-\pi \cdot g(x_{k})^{2}\Delta x\\&=\pi {\bigl (}f(x_{k})^{2}-g(x_{k})^{2}{\bigr )}\Delta x\end{aligned}}$

where $\Delta x={\frac {b-a}{n}}$ and $x_{k}=k\Delta x$ . The volume of the entire approximating solid is

$V_{\rm {approx}}=\sum _{k=1}^{n}\pi {\bigl (}f(x_{k})^{2}-g(x_{k})^{2}{\bigr )}\Delta x$

Taking the limit as $n$ approaches infinity gives the volume

${\begin{aligned}V&=\lim _{n\to \infty }\sum _{k=1}^{n}\pi {\bigl (}f(x_{k})^{2}-g(x_{k})^{2}{\bigr )}\Delta x\\&=\int \limits _{a}^{b}\pi {\bigl (}f(x)^{2}-g(x)^{2}{\bigr )}dx\end{aligned}}$

Resources

Volume by Slices

• Determine Volume of Solids by Slices by James Sousa, Math is Power 4U
• Ex 1: Volume of a Solid with Known Cross Section Using Integration - Volume by Slices by James Sousa, Math is Power 4U
• Ex 2: Volume of a Solid with Known Cross Section Using Integration - Volume by Slices by James Sousa, Math is Power 4U
• Ex 3: Volume of a Solid with Known Cross Section Using Integration - Volume by Slices by James Sousa, Math is Power 4U
• Ex 4: Volume of a Solid with Known Cross Section Using Integration - Volume by Slices by James Sousa, Math is Power 4U
• Ex 5: Volume of a Solid with Known Cross Section Using Integration - Volume by Slices by James Sousa, Math is Power 4U

• Volume with Cross Sections: Introduction by Khan Academy
• Volume with Cross Sections: Semicircle by Khan Academy
• Volume with Cross Sections: Triangle by Khan Academy

• Volumes Using Cross Sections Calculus, Square, Semicircles, Rectangles, Equilateral Triangles by The Organic Chemistry Tutor

The Disk Method

• Volume of Revolution - The Disk Method by James Sousa, Math is Power 4U
• Ex 1: Volume of Revolution - The Disk Method by James Sousa, Math is Power 4U
• Ex 2: Volume of Revolution - The Disk Method by James Sousa, Math is Power 4U
• Ex 3: Volume of Revolution - The Disk Method by James Sousa, Math is Power 4U
• Ex 1: Volume of Revolution Using the Disk Method by James Sousa, Math is Power 4U
• Ex 2: Volume of Revolution Using the Disk Method by James Sousa, Math is Power 4U
• Ex 3: Volume of Revolution Using the Disk Method by James Sousa, Math is Power 4U

• Volume of Rotation: Disk Method About the x-axis or y= by Krista King
• Volume of Rotation: Disk Method About the y-axis or x= by Krista King

• Disk Method Around x-axis by Khan Academy
• Disk Method Around y-axis by Khan Academy
• Disk Method Around Horizontal Line by Khan Academy
• Disk Method Around Vertical Line by Khan Academy

The Washer Method

• Volume of Revolution - The Washer Method about the x-axis by James Sousa, Math is Power 4U
• Volume of Revolution - The Washer Method about the y-axis by James Sousa, Math is Power 4U
• Volume of Revolution - The Washer Method Not about the x or y axis by James Sousa, Math is Power 4U
• Ex 1: Volume of Revolution Using the Washer Method About y=3 by James Sousa, Math is Power 4U
• Ex 2: Volume of Revolution Using the Washer Method About y=3 by James Sousa, Math is Power 4U
• Ex 1: Volume of Revolution Using the Washer Method About y-axis by James Sousa, Math is Power 4U
• Ex 2: Volume of Revolution Using the Washer Method About y-axis by James Sousa, Math is Power 4U
• Ex: Volume of Revolution Using the Washer Method About x=5 by James Sousa, Math is Power 4U

• Volume of Rotation: Washer Method About the x-axis or y= by Krista King
• Volume of Rotation: Washer Method About the y-axis or x= by Krista King

• The Washer Method by Khan Academy
• The Washer Method Rotating Around Horizontal Line (not x-axis) Part 1 by Khan Academy
• The Washer Method Rotating Around Horizontal Line (not x-axis) Part 2 by Khan Academy
• The Washer Method Rotating Around Vertical Line (not y-axis) Part 1 by Khan Academy
• The Washer Method Rotating Around Vertical Line (not y-axis) Part 2 by Khan Academy

The Disk/Washer Methods

• Volumes of Revolution - Disk/Washer Method Example 1 by patrickJMT
• Volumes of Revolution - Disk/Washer Method Example 2 by patrickJMT
• Volumes of Revolution - Disk/Washer Method Example 3 by patrickJMT

Licensing

Content obtained and/or adapted from:

• Volume, Wikibooks: Calculus under a CC BY-SA license
• Volume of solids of revolution, Wikibooks: Calculus under a CC BY-SA license