Difference between revisions of "Determining Volumes by Slicing"

Latest revision as of 10:49, 29 October 2021

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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=h-x} , then Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |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, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A(x)} . Consider a sphere of radius Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R} which is centered at the origin in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \R^3} . If we again integrate vertically then $x$ will vary from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -R} to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R} , Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |x|} , and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r} . Where $r$ is the radius of the circular cross section. Then by the Pythagorean theorem Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=\sqrt{R^2-|x|^2}} and find that Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} 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^2dx-\pi\int\limits_{-R}^R x^2dx\\ &=\pi R^2x\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{align}}

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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=0} around the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [a,b]} , and the lines Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=a} , $x=b$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=0} around the $x$ -axis. We could imagine approximating the volume by approximating Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} with the stepwise function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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.

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V_{\rm cylinder}=\pi r^2 h}

where Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} subdivisions and a length of $b-a$ for the total length of the region, each subdivision has width

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta x=\frac{b-a}{n}}

Since we are using a right-handed approximation, the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} -th sample point will be

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_k=k\Delta x}

So the volume of the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k} -th cylinder is

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

Summing all of the cylinders in the region from Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} to Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} , we have

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

Taking the limit as Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} approaches infinity gives us the exact volume

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=\sqrt{r^2-x^2}} and the line Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=0} . Our limits of integration will be the $x$ -values where the curve intersects the line Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle y=0} , namely, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} -axis. Consider the solid of revolution formed by revolving the region in figure 3 around the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)} around the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_{\rm step}(x)} rather than all the way down to the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x} -axis. Revolving each block around the $x$ -axis forms a washer-shaped solid with outer radius $f_{\rm {step}}(x)$ and inner radius Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g_{\rm step}(x)} . The volume of the Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Delta x=\frac{b-a}{n}} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

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

@@ Line 1: / Line 1: @@
-[https://youtu.be/ODNdwqJ7TAg Ex 2: Volume of a Solid with Known Cross Section Using Integration - Volume by Slices] by James Sousa
+==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.
-[https://youtu.be/LvBZ0jnAMbo Ex 3: Volume of a Solid with Known Cross Section Using Integration - Volume by Slices] by James Sousa
+The volume of a cylinder is calculated using the formula <math>V=\pi r^2h</math> . The base of the cylinder is a circle whose area is given by <math>A=\pi r^2</math> . Notice that the volume of a cylinder is derived by taking the area of its base and multiplying by the height <math>h</math> . For more complicated shapes, we could think of approximating the volume by taking the area of some cross section at some height <math>x</math> and multiplying by some small change in height <math>\Delta x</math> 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 <math>\R^3</math> (3 dimensional space).
-[https://youtu.be/BQ5sg21w-lA Ex 4: Volume of a Solid with Known Cross Section Using Integration - Volume by Slices] by James Sousa
+===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 <math>S</math> is a solid that lies in <math>\R^3</math>
+between <math>x=a</math> and <math>x=b</math> , let <math>A(x)</math> denote the area of a cross section taken in the plane perpendicular to the <math>x</math>-axis, and passing through the point <math>x</math> .
-[https://youtu.be/puSlVA6mwNQ Volumes Using Cross Sectional Slices, Ex 1]  by patrickJMT
+If the function <math>A(x)</math> is  continuous on <math>[a,b]</math> , then the volume <math>V_S</math> of the solid <math>S</math> is given by:
+:{{math|size=1.5em|<math>V_S=\int\limits_a^b A(x)dx</math>}}
-[https://youtu.be/XasIRTb5WH4 Volumes Using Cross Sections Calculus, Square, Semicircles, Rectangles, Equilateral Triangles]  by The Organic Chemistry Tutor
+===Examples===
+====Example 1: A right cylinder====
+[[File:Cylinder with cross section at height x.svg|200px|thumb|right|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 <math>x</math> (see [[:File:Cylinder with cross section at height x.svg|Figure 1]]). Now we find the function, <math>A(x)</math> , which will describe the cross-sectional area of our cylinder at a height of <math>x</math> . The cross-sectional area of a cylinder is simply a circle. Now simply recall that the area of a circle is <math>\pi r^2</math> , and so <math>A(x)=\pi r^2</math> . Before performing the computation, we must choose our bounds of integration. In this case, we simply define <math>x=0</math> to be the base of the cylinder, and so we will integrate from <math>x=0</math> to <math>x=h</math> , where <math>h</math> is the height of the cylinder. Finally, we integrate:
+:<math>\begin{align}
+V_{\mathrm{cylinder}}&=\int\limits_a^b A(x)dx\\
+&=\int\limits_0^h\pi r^2dx\\
+&=\pi r^2\int\limits_0^hdx\\
+&=\pi r^2x\bigg|_{x=0}^h\\
+&=\pi r^2(h-0)\\
+&=\pi r^2h\end{align}</math>
+This is exactly the familiar formula for the volume of a cylinder.
+====Example 2: A right circular cone====
+[[File:Plane intersecting cone 2.png|200px|thumb|right|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 <math>x</math> to be the vertical direction, with the base at <math>x=0</math> and the tip at <math>x=h</math> , and we will let <math>R</math> 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 <math>x</math> .
+[[File:Similar triangles for cone.svg|200px|thumb|right|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 <math>x</math> is similar to the right triangle from the tip to the base at height <math>h</math> .  This tells us that <math>\frac{r}{h-x}=\frac{R}{h}</math> . So that we see that the radius of the circle at height <math>x</math> is <math>r(x)=\frac{R}{h}(h-x)</math> . Now using the familiar formula for the area of a circle we see that <math>A(x)=\pi\frac{R^2}{h^2}(h-x)^2</math> .
+Now we are ready to integrate.
+:<math>\begin{align}
+V_{\mathrm{cone}}&=\int\limits_a^b A(x)dx\\
+&=\int\limits_0^h \pi\frac{R^2}{h^2}(h-x)^2dx\\
+&=\pi\frac{R^2}{h^2}\int\limits_0^h(h-x)^2dx
+\end{align}</math>
+By u-substitution we may let <math>u=h-x</math> , then <math>du=-dx</math> and our integral becomes
+:<math>\begin{align}
+&&=\pi\frac{R^2}{h^2}\left(-\int\limits_h^0 u^2du\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^2h
+\end{align}</math>
+====Example 3: A sphere====
+[[File:Sphere with cross section.svg|200px|thumb|right|Figure 4: Determining the radius of the cross-section of the sphere at a distance <math>|x|</math> 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, <math>A(x)</math> . Consider a sphere of radius <math>R</math> which is centered at the origin in <math>\R^3</math> . If we again integrate vertically then <math>x</math> will vary from <math>-R</math> to <math>R</math> . 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 <math>R</math> , <math>|x|</math> , and <math>r</math> .  Where <math>r</math> is the radius of the circular cross section. Then by the Pythagorean theorem <math>r=\sqrt{R^2-|x|^2}</math> and find that <math>A(x)=\pi(R^2-|x|^2)</math> . It is slightly helpful to notice that <math>|x|^2=x^2</math> so we do not need to keep the absolute value.
+So we have that
+:<math>\begin{align}
+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^2dx-\pi\int\limits_{-R}^R x^2dx\\
+&=\pi R^2x\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{align}</math>
+==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 <math>y=\sqrt{1-x^2}</math> and the line <math>y=0</math> around the <math>x</math>-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===
+[[Image:Generating fx disk method.svg|thumb|397px|right|Figure 1: A solid of revolution is generated by revolving this region around the x-axis.]]
+[[Image:Approx generating fx disk method.svg|thumb|397px|right|Figure 2: Approximation to the generating region in Figure 1.]]
+Consider the solid formed by revolving the region bounded by the curve <math>y=f(x)</math> , which is continuous on <math>[a,b]</math> , and the lines <math>x=a</math> , <math>x=b</math> and <math>y=0</math> around the <math>x</math>-axis. We could imagine approximating the volume by approximating <math>f(x)</math> with the stepwise function <math>g(x)</math> shown in [[:File:Approx generating fx disk method.svg|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.
+:<math>V_{\rm cylinder}=\pi r^2 h</math>
+where <math>r</math> is the radius of the cylinder and <math>h</math> 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 <math>k</math>-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 <math>n</math> subdivisions and a length of <math>b-a</math> for the total length of the region, each subdivision has width
+:<math>\Delta x=\frac{b-a}{n}</math>
+Since we are using a right-handed approximation, the <math>k</math>-th sample point will be
+:<math>x_k=k\Delta x</math>
+So the volume of the <math>k</math>-th cylinder is
+:<math>V_k=\pi f(x_k)^2\Delta x</math>
+Summing all of the cylinders in the region from <math>a</math> to <math>b</math> , we have
+:<math>V_{\rm approx}=\sum_{k=1}^n \pi f(x_k)^2\Delta x</math>
+Taking the limit as <math>n</math> approaches infinity gives us the exact volume
+:<math>V=\lim_{n\to\infty}\sum_{k=1}^n \pi f(x_k)^2\Delta x</math>
+which is equivalent to the integral
+:<math>V=\int\limits_a^b \pi f(x)^2dx</math>
+<blockquote style="background: white; border: 1px solid black; padding: 1em;">
+: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 <math>f(x)=\sqrt{r^2-x^2}</math> and the line <math>y=0</math> . Our limits of integration will be the <math>x</math>-values where the curve intersects the line <math>y=0</math> , namely, <math>x=\pm r</math> . We have
+:<math>\begin{align}V_{\rm sphere}&=\int\limits_{-r}^r \pi(r^2-x^2)dx\\
+&=\pi\left(\int\limits_{-r}^r r^2dx-\int\limits_{-r}^r x^2dx\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{align}</math>
+</blockquote>
+===Washer Method===
+[[Image:Generating fxs washer method.svg|thumb|401px|right|Figure 3: A solid of revolution containing an irregularly shaped hole through its center is generated by revolving this region around the x-axis.]]
+[[Image:Approx generating fxs washer method.svg|thumb|401px|right|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 <math>x</math>-axis. Consider the solid of revolution formed by revolving the region in [[:File:Generating fxs washer method.svg|figure 3]] around the <math>x</math>-axis.  The curve <math>f(x)</math> is the same as that in [[:File:Generating fx disk method.svg|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 <math>g(x)</math> around the <math>x</math>-axis.  Our approximating region has the same upper boundary, <math>f_{\rm step}(x)</math> as in [[:File:Approx generating fx disk method.svg|figure 2]], but now we extend only down to <math>g_{\rm step}(x)</math> rather than all the way down to the <math>x</math>-axis. Revolving each block around the <math>x</math>-axis forms a washer-shaped solid with outer radius <math>f_{\rm step}(x)</math> and inner radius <math>g_{\rm step}(x)</math> . The volume of the <math>k</math>-th hollow cylinder is
+:<math>\begin{align}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{align}</math>
+where <math>\Delta x=\frac{b-a}{n}</math> and <math>x_k=k\Delta x</math> . The volume of the entire approximating solid is
+:<math>V_{\rm approx}=\sum_{k=1}^n \pi\bigl(f(x_k)^2-g(x_k)^2\bigr)\Delta x</math>
+Taking the limit as <math>n</math> approaches infinity gives the volume
+:<math>\begin{align}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{align}</math>
+==Resources==
+===Volume by Slices===
+* [https://youtu.be/Q8c8Gs1OUgk Determine Volume of Solids by Slices] by James Sousa, Math is Power 4U
+* [https://youtu.be/_QhW_pXy0wE Ex 1: Volume of a Solid with Known Cross Section Using Integration - Volume by Slices] by James Sousa, Math is Power 4U
+* [https://youtu.be/ODNdwqJ7TAg Ex 2: Volume of a Solid with Known Cross Section Using Integration - Volume by Slices] by James Sousa, Math is Power 4U
+* [https://youtu.be/LvBZ0jnAMbo Ex 3: Volume of a Solid with Known Cross Section Using Integration - Volume by Slices] by James Sousa, Math is Power 4U
+* [https://youtu.be/BQ5sg21w-lA Ex 4: Volume of a Solid with Known Cross Section Using Integration - Volume by Slices] by James Sousa, Math is Power 4U
+* [https://.youtu.be/2OgAluEqd8s Ex 5: Volume of a Solid with Known Cross Section Using Integration - Volume by Slices] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=4vLy5VoUcQE Volume with Cross Sections: Introduction] by Khan Academy
+* [https://www.youtube.com/watch?v=tf4C8x8e7HQ Volume with Cross Sections: Semicircle] by Khan Academy
+* [https://www.youtube.com/watch?v=S-agS4YaxxU Volume with Cross Sections: Triangle] by Khan Academy
+* [https://youtu.be/XasIRTb5WH4 Volumes Using Cross Sections Calculus, Square, Semicircles, Rectangles, Equilateral Triangles]  by The Organic Chemistry Tutor
+===The Disk Method===
+* [https://www.youtube.com/watch?v=1CbZlM09zF8 Volume of Revolution - The Disk Method] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=oYe4gkKQDE8 Ex 1: Volume of Revolution - The Disk Method] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=e7qNtcWpg2s Ex 2: Volume of Revolution - The Disk Method] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=Ba_q-eu8n34 Ex 3: Volume of Revolution - The Disk Method] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=yTElv1k33rk Ex 1: Volume of Revolution Using the Disk Method] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=xoi-6CyL6cg Ex 2: Volume of Revolution Using the Disk Method] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=2tpyuR-GypA Ex 3: Volume of Revolution Using the Disk Method] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=RPpih8rD-5k Volume of Rotation: Disk Method About the x-axis or y=] by Krista King
+* [https://www.youtube.com/watch?v=DNmHo68wYFY Volume of Rotation: Disk Method About the y-axis or x=] by Krista King
+* [https://www.youtube.com/watch?v=btGaOTXxXs8 Disk Method Around x-axis] by Khan Academy
+* [https://www.youtube.com/watch?v=43AS7bPUORc Disk Method Around y-axis] by Khan Academy
+* [https://www.youtube.com/watch?v=XdzcU5JbVcA Disk Method Around Horizontal Line] by Khan Academy
+* [https://www.youtube.com/watch?v=jxf7XqvZWWg Disk Method Around Vertical Line] by Khan Academy
+===The Washer Method===
+* [https://www.youtube.com/watch?v=rbqWHbxmVUI Volume of Revolution - The Washer Method about the x-axis] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=8gbphumzbSI Volume of Revolution - The Washer Method about the y-axis] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=_FF3CM5MNe8 Volume of Revolution - The Washer Method Not about the x or y axis] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=6zaqRVD90mY Ex 1: Volume of Revolution Using the Washer Method About y=3] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=kivhB247H-c Ex 2: Volume of Revolution Using the Washer Method About y=3] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=IWoEivm5AqM Ex 1: Volume of Revolution Using the Washer Method About y-axis] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=kkVVOStqRmI Ex 2: Volume of Revolution Using the Washer Method About y-axis] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=2yd_Ae-7wls Ex: Volume of Revolution Using the Washer Method About x=5] by James Sousa, Math is Power 4U
+* [https://www.youtube.com/watch?v=zHi1qtf4pGg Volume of Rotation: Washer Method About the x-axis or y=] by Krista King
+* [https://www.youtube.com/watch?v=Sg24zkhG0Vw Volume of Rotation: Washer Method About the y-axis or x=] by Krista King
+* [https://www.youtube.com/watch?v=Thvc2s9aUP4 The Washer Method] by Khan Academy
+* [https://www.youtube.com/watch?v=OFNGpKGg9IQ The Washer Method Rotating Around Horizontal Line (not x-axis) Part 1] by Khan Academy
+* [https://www.youtube.com/watch?v=LKzpw_HUKNQ The Washer Method Rotating Around Horizontal Line (not x-axis) Part 2] by Khan Academy
+* [https://www.youtube.com/watch?v=WAPZihVUmzE The Washer Method Rotating Around Vertical Line (not y-axis) Part 1] by Khan Academy
+* [https://www.youtube.com/watch?v=i-Rb4_n929k The Washer Method Rotating Around Vertical Line (not y-axis) Part 2] by Khan Academy
+===The Disk/Washer Methods===
+* [https://www.youtube.com/watch?v=E5OOMbz5jZk Volumes of Revolution - Disk/Washer Method Example 1] by patrickJMT
+* [https://www.youtube.com/watch?v=ithgZfRKMHI Volumes of Revolution - Disk/Washer Method Example 2] by patrickJMT
+* [https://www.youtube.com/watch?v=lE814ngFt3I Volumes of Revolution - Disk/Washer Method Example 3] by patrickJMT
+==Licensing==
+Content obtained and/or adapted from:
+* [https://en.wikibooks.org/wiki/Calculus/Volume Volume, Wikibooks: Calculus] under a CC BY-SA license
+* [https://en.wikibooks.org/wiki/Calculus/Volume_of_solids_of_revolution Volume of solids of revolution, Wikibooks: Calculus] under a CC BY-SA license

Difference between revisions of "Determining Volumes by Slicing"

Latest revision as of 10:49, 29 October 2021

Contents

Volumes by Slices

Formal Definition

Examples

Example 1: A right cylinder

Example 2: A right circular cone

Example 3: A sphere

Volume of Solids of Revolution

Disk Method

Washer Method

Resources

Volume by Slices

The Disk Method

The Washer Method

The Disk/Washer Methods

Licensing

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