6.10: Determining Volumes by Slicing

Section Preview

Note to the Instructor (click to expand)

The purpose of this section is for the student to get used to Riemann Sums applied to three-dimensional objects. All volumes in this section are non-rotational.

Core Prerequisites (i.e., "Understanding the Lesson"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are required to understand the core concepts in this lesson.

(click to expand)

• Geometry
  • Similar Triangles: The cross-sectional dimension is related to its position along the axis by recognizing similar triangles and equating ratios; in Example 1.2.1 the proportion \(\frac{s}{a} = \frac{x}{h}\) gives \(s = \frac{ax}{h}\), and Checkpoint 1.2.1 uses the same idea for the cone.
  • Volumes and Surface Areas of Common Shapes: The slicing method models each slice as a thin cylinder of volume \(A(x^*_i)\,\Delta x\), so understanding that a cylinder's volume is base-area times height (\(V = A \cdot h\)) is the conceptual anchor for the entire method; the section also recalls the sphere, cone, and pyramid volume formulas as known from prior courses.
• Quadratic Functions
  • Graphing Quadratic Functions Using Properties: The base region in Examples 1.2.2 and 1.2.3 is bounded by the parabola \(f(x) = 4 - x^2\); visualizing this region (downward parabola, vertex \((0,4)\), \(x\)-intercepts at \(\pm 2\)) and its axis of symmetry is essential for setting up the integral and for exploiting the symmetry used in Example 1.2.2.

Operational Prerequisites (i.e., "Completing the Homework"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are needed to complete the homework.

(click to expand)

• Geometry
  • Circles — Properties, Area, and Circumference: The semicircular cross-sections in Example 1.2.3 require the area \(\frac{1}{2}\pi r^2\), and deriving the cone's volume in Checkpoint 1.2.1 requires the circular cross-section area \(\pi r^2\).
  • Area and Perimeter of Triangles and Quadrilaterals: Building the cross-sectional area function \(A(x)\) uses the area of a square in Example 1.2.1 (\(A = s^2\)) and the area of a triangle in Example 1.2.2 (\(A = \frac{1}{2}\,\text{base}\cdot\text{height}\)); the homework also draws on rectangle and other triangle areas.
• Solving Equations
  • Solving Literal Equations: When the cross-sections are perpendicular to the \(y\)-axis (Example 1.2.3), the boundary \(y = 4 - x^2\) must be solved for \(x^2 = 4 - y\) so the area can be written in terms of \(y\).
• Conic Sections
  • Ellipses: Homework problem #16 involves a solid with an elliptical base, which requires the standard equation of an ellipse to set up the cross-sectional area.

The following is a list of learning objectives for this section.

Learning Objectives (click to expand)

• Determine the volume of a solid by integrating a cross-section (the slicing method).

In the preceding section, we used definite integrals to find the area between two curves. In this section, we use definite integrals to find volumes of three-dimensional solids. We start by developing a basic approach based on slicing.

Volume and the Slicing Method

Just as area is the numerical measure of a two-dimensional region, volume is the numerical measure of a three-dimensional solid. Most of us have computed volumes of solids by using basic geometric formulas. The volume of a rectangular solid, for example, can be computed by multiplying length, width, and height: \(V = lwh\). The following formulas for volumes have been introduced in previous courses.

• a sphere\[V_{\text{sphere}}=\dfrac{4}{3} \pi r^3,\nonumber\]
• a cone\[V_{\text{cone}}=\dfrac{1}{3} \pi r^2h\nonumber\]
• and a pyramid\[V_{\text{pyramid}}=\dfrac{1}{3}Ah\nonumber\]

Although some of these formulas were derived using Geometry alone, all these formulas can be obtained using integration.

We can also calculate the volume of a cylinder. Although most of us think of a cylinder as having a circular base, such as a soup can or a metal rod, in mathematics, the word cylinder has a more general meaning. We first need to define some vocabulary to discuss cylinders in this more general context.

We define a solid's cross-section as the intersection of a plane with the solid. A cylinder is defined as any solid that can be generated by translating a plane region along a line perpendicular to the region called the axis of the cylinder. Thus, all cross-sections perpendicular to the axis of a cylinder are identical. The solid shown in Figure \(\PageIndex{1}\) is an example of a cylinder with a noncircular base. To calculate the volume of a cylinder, we multiply the area of the cross-section by the height of the cylinder: \(V=A \cdot h\). In the case of a right circular cylinder (soup can), this becomes \(V= \pi r^2h\).

Figure \(\PageIndex{1}\): Each cross-section of a particular cylinder is identical to the others.

If a solid does not have a constant cross-section (and it is not one of the other basic solids), we may not have a formula for its volume. In this case, we can use a definite integral to calculate the volume of the solid. We do this by slicing the solid into pieces, estimating the volume of each slice, and then adding those estimated volumes together. The slices should all be parallel to one another, and when we put all the slices together, we should get the whole solid. Consider, for example, the solid \(S\) shown in Figure \(\PageIndex{2}\), extending along the \(x\)-axis.

Figure \(\PageIndex{2}\): A solid with a varying cross-section.

We want to divide \(S\) into slices perpendicular to the \(x\)-axis. As we see later in the chapter, there may be times when we want to slice the solid in some other direction—say, with slices perpendicular to the \(y\)-axis. The decision of which way to slice the solid is vital. If we make the wrong choice, the computations can get quite messy. Later in the chapter, we examine some of these situations in detail and decide which way to slice the solid. For the purposes of this section, however, we use slices perpendicular to the \(x\)-axis.

Because the cross-sectional area is not constant, we let \(A(x)\) represent the cross-sectional area of the cross-section at point \(x\). Now let \(P=\{x_0,x_1,\ldots,x_n\}\) be a regular partition of \([a,b]\), and for \(i=1,2,\ldots,n\), let \(S_i\) represent the slice of \(S\) stretching from \(x_{i-1}\) to \(x_i\). The following figure shows the sliced solid with \(n=3\).

Figure \(\PageIndex{3}\): The solid \(S\) has been divided into three slices perpendicular to the \(x\)-axis.

Finally, for \(i=1,2,\ldots,n\), let \(x^*_i\) be an arbitrary point in \([x_{i-1},x_i]\). Then the volume of slice \(S_i\) can be estimated by \(V(S_i) \approx A(x^*_i)\, \Delta x\). Adding these approximations together, we see the volume of the entire solid \(S\) can be approximated by\[V(S) \approx \sum_{i=1}^nA(x^*_i)\, \Delta x.\nonumber\]By now, we can recognize this as a Riemann sum, and our next step is to take the limit as \(n \to \infty \). Then we have\[V(S)=\lim_{n \to \infty}\sum_{i=1}^nA(x^*_i)\, \Delta x= \int _a^b A(x)\,dx.\nonumber\]The technique we have just described is called the slicing method. To apply it, we use the following strategy.

Problem-Solving Strategy: Finding Volumes by the Slicing Method

1. Examine the solid and determine the shape of a cross-section of the solid. It is often helpful to draw a picture if one is not provided.
2. Determine a formula for the area of the cross-section.
3. Integrate the area formula over the appropriate interval to get the volume.

This section assumes the slices are perpendicular to the \(x\)-axis. Therefore, the area formula is in terms of \(x\), and the limits of integration lie on the \(x\)-axis. However, the problem-solving strategy shown here is valid regardless of how we choose to slice the solid.

Example \(\PageIndex{1}\)

We know from Geometry that the formula for the volume of a pyramid is \(V=\frac{1}{3}Ah\). If the pyramid has a square base, this becomes \(V=\frac{1}{3}a^2h\), where \(a\) denotes the length of one side of the base. We are going to use the slicing method to derive this formula.

Solution

We want to apply the slicing method to a pyramid with a square base. To set up the integral, consider the pyramid shown in Figure \(\PageIndex{4}\), oriented along the \(x\)-axis.

Figure \(\PageIndex{4}\): (a) A pyramid with a square base is oriented along the \(x\)-axis. (b) A two-dimensional view of the pyramid is seen from the side.

We first want to determine the shape of a cross-section of the pyramid. We know the base is a square, so the cross-sections are squares as well. Now, we want to determine a formula for the area of one of these cross-sectional squares. Looking at Figure \(\PageIndex{4}\)(b), and using a proportion, since these are similar triangles, we have\[\dfrac{s}{a}=\dfrac{x}{h}\nonumber\]or\[s=\dfrac{ax}{h}.\nonumber\]Therefore, the area of one of the cross-sectional squares is\[A(x)=s^2=\left(\dfrac{ax}{h}\right)^2.\nonumber\]Then we find the volume of the pyramid by integrating from \(0\) to \(h\):\[V= \int _0^hA(x)\,dx= \int _0^h\left(\dfrac{ax}{h}\right)^2\,dx=\dfrac{a^2}{h^2} \int _0^hx^2\,dx=\Big[\dfrac{a^2}{h^2}\left(\dfrac{1}{3}x^3\right)\Big]\bigg|^h_0=\dfrac{1}{3}a^2h.\nonumber\]This is the formula we were looking for.

Checkpoint \(\PageIndex{1}\)

Use the slicing method to derive the formula \[V=\frac{1}{3} \pi r^2h\nonumber\] for the volume of a circular cone.

Hint

Use similar triangles, as in Example \(\PageIndex{1}\).

While Example \(\PageIndex{1}\) is a shape we are familiar with, it behooves us to experiment with finding the volume of a solid having a more challenging design. For this discussion, we consider solids having a base region defined by a given boundary. This boundary is often an area between two functions on a specified interval, however, it can be a region bounded by a non-functional curve such as a conic section. Onto this base region, we will slide in cross-sectional slices perpendicular to the base (see the Interactive Element below).

Interactive Element: Volumes of Known Cross-sections

 

Description: This Geogebra applet allows you to visualize how a solid of known cross-sections is generated by placing cross-sectional slices perpendicular to the \(x\)-axis on a base defined by the boundary between two functions (red and blue) from \(x = 0\) to \(X\), where you can adjust \(X\) using the bottom slider.

Interact: Adjust the rightmost boundary, \(X\), of the region to whatever value you want, select a single cross-sectional shape (square, equilateral triangle, or semi-circle), increase \(n\) to a decent number of slices to help you visualize what is going on, and (finally) use your mouse to adjust the graph of the right side of the applet to see the resulting solid.

Let's take a look at two examples.

Example \(\PageIndex{2}\)

Find the volume of the solid with base region bounded by \(f(x) = 4 - x^2\) and the \(x\)-axis, where the cross-sections perpendicular to the \(x\)-axis are isosceles right triangles with hypotenuse in the base region.

Solution

We begin by drawing a top-down view of the base region, including the \(i^{\text{th}}\) slice (see Figure \(\PageIndex{5}\) below).

Figure \(\PageIndex{5}\): A top-down view of the base region along with the \(i^{\text{th}}\) slice.

It will also be beneficial to sketch a side view of the \(i^{\text{th}}\) slice (see Figure \(\PageIndex{6}\) below).

Figure \(\PageIndex{6}\): A side view of the \(i^{\text{th}}\) slice.

Interactive Element: The Solid with Triangular Cross-sections

The following Interactive Element might help you visualize the solid whose cross-sections perpendicular to the \(x\)-axis are isosceles right triangles.

Interact: Click and drag empty space to rotate; use the middle mouse scroll wheel to zoom in and out; click the blue "curve" button in the function list to reveal the triangular slice; click the magenta marker to step the triangular slice across the solid—it sweeps and repeats.

Observation: Every cross-section perpendicular to the \(x\)-axis is an isosceles right triangle whose hypotenuse \(4-x^2\) is widest at the center and vanishes at \(x=\pm 2\); the symmetry about \(x=0\) is why the volume runs from \(0\) to \(2\) and is doubled.

The volume of this \(i^{\text{th}}\) slice can be found using the fact that the surface area of a triangle is one-half the product of the base and the height. Hence,\[V_i = \dfrac{1}{2} \left(\text{base}\right)\left(\text{height}\right) \Delta x = \dfrac{1}{2} \left( f(x_i^*) - 0 \right) \left(\text{height}\right) \Delta x = \dfrac{1}{2} \left( 4 - (x_i^*)^2 \right) \left(\text{height}\right) \Delta x.\nonumber\]To determine the height of this \(i^{\text{th}}\) slice, we use the fact that the base has length \(4 - (x_i^*)^2\), which means half of the base has length \(\frac{1}{2}\left(4 - (x_i^*)^2\right)\). Since the triangle is an isosceles right triangle, the height equals this half-base length. Hence,\[V_i = \dfrac{1}{4} \left( 4 - (x_i^*)^2 \right)^2 \Delta x.\nonumber\]From Figure \(\PageIndex{5}\), we can see that symmetry can be used so that we only need to add up the volumes of the slices starting at \(x = 0\) and ending at \(x = 2\). We double this result to approximate the total area. That is,\[V \approx 2 \sum_{i = 0}^n V_i = \dfrac{1}{2} \sum_{i = 0}^n \left( 4 - (x_i^*)^2 \right)^2 \Delta x.\nonumber\]Taking the limit as \(n \to \infty\), we get\[\begin{array}{rcl} V & = & \dfrac{1}{2} \displaystyle \int_0^2 (4 - x^2)^2 \, dx \\ \\ & = & \dfrac{1}{2} \displaystyle \int_0^2 16 - 8 x^2 + x^4 \, dx \\ \\ & = & \dfrac{1}{2} \left( 16x - \dfrac{8}{3} x^3 + \dfrac{1}{5} x^5 \right) \bigg|_0^2 \\ \\ & = & \dfrac{1}{2} \left( 32 - \dfrac{64}{3} + \dfrac{32}{5} \right) \\ \\ & = & \dfrac{128}{15} \, \text{cubic units} \\ \end{array}\nonumber\]

Example \(\PageIndex{3}\)

Find the volume of the solid with base region bounded by \(f(x) = 4 - x^2\) and the \(x\)-axis, where the cross-sections perpendicular to the \(y\)-axis are semi-circles.

Solution

Just as we did in Example \(\PageIndex{2}\), we begin by sketching a top-down view of the base region with the \(i^{\text{th}}\) slice, and a side view of the \(i^{\text{th}}\) slice (see Figure \(\PageIndex{7}\)).

Figure \(\PageIndex{7}\)A: A top-down view of the base region along with the \(i^{\text{th}}\) slice.
Figure \(\PageIndex{7}\)B: A side view of the \(i^{\text{th}}\) slice.

Interactive Element: The Solid with Semicircular Cross-sections

The following Interactive Element might help you visualize the solid whose cross-sections perpendicular to the \(y\)-axis are semi-circles.

Interact: Click and drag empty space to rotate; click the purple marker to step the semicircular slice through the solid — it sweeps bottom-to-top and repeats.

Observation: Every cross-section perpendicular to the \(y\)-axis is a semicircle whose radius \(\sqrt{4-y}\) shrinks from \(2\) at the base \((y=0)\) to \(0\) at the top \((y=4)\), so the volume integrates \(\tfrac{1}{2}\pi(4-y)\) with respect to \(y\).

The volume of the \(i^{\text{th}}\) slice is\[V_i = \dfrac{1}{2} \pi \left(r_i^*\right)^2 \Delta y = \dfrac{1}{2} \pi \left( x_i^* \right)^2 \Delta y.\nonumber\]Using the relation \(y_i^* = 4 - \left(x_i^*\right)^2\), we get\[V_i = \dfrac{1}{2} \pi \left( 4 - y_i^* \right) \Delta y.\nonumber\]Therefore\[\begin{array}{rcl} V & = & \displaystyle \lim_{n \to \infty} \sum_{i = 0}^n V_i \\ \\ & = & \displaystyle \int_{y = 0}^{y = 4} \dfrac{1}{2} \pi \left( 4 - y \right) \, dy \\ \\ & = & \dfrac{\pi}{2} \left( 4y - \dfrac{y^2}{2} \right) \bigg|_{y = 0}^{y = 4} \\ \\ & = & 4 \pi \, \text{cubic units} \\ \end{array}\nonumber\]

---

Time-Sensitive Advice: HawkConnect

What is HawkConnect?

In the past, navigating through all the student resources (EOPS, TRIO, MI CASA, etc.) required you to go to individual sites and apply individually for each program. HawkConnect now unifies the entire process into a single, guided application process.

To learn more about HawkConnect and to apply, go to the CRC Support Services website.

---

Success in Mathematics: Show Your Work

Many mathematics exams include the statement, "Show Your Work" (if it's not explicitly written, then it is implied). The exact meaning of these words can change from one professor to another; however, they usually mean that you need to show more than just the answer to a problem. You need to explain or show how you found the answer. Your responsibility is to learn what "Show Your Work" means to your instructor.

Why do you need to show your work?

Your authentic work (e.g., without use of A.I., tutors, or any other external help) demonstrates what you know. It shows what you understand and what you do not understand. By looking at your work and the work of others, an instructor can help you improve your understanding.

Everyone makes mistakes. To learn from these mistakes, you need to be able to find them. If you show your work, you can go back and find mistakes. This helps you identify misunderstandings and bad habits. Without your work, all you can do is start over.

Finally, although you may be able to do a problem "in your head," this will not always be possible. The problems get more challenging and involve more detail as we move forward. Showing your work now develops the habits and procedures you will need to do math in the future.

Here are some questions about this topic that may help you succeed in this class (and possibly many other classes as well):

1. A student did a homework exercise. She checked her answer with the back of the book. It was wrong. So, she erased her work and started over. Why should she not erase her work?
2. A student did a homework exercise. He checked his answer with the back of the book. It was wrong. He marked the problem on his homework with an asterisk. He went on with the assignment. He never went back to the problem. Why is it essential for this student to finish the problem?

If you are unsure about the answers to either of these questions, feel free to ask your instructor - that's what they are there for!

---

Miscellaneous Topics: What Are Your Top Degree Choices?

I have seen all the statistics regarding the correlation between income and education. In general, the higher your education, the higher your income.

1. Name three benefits, besides money, that a higher degree can give you.
2. What are your top three choices for degrees (e.g., English, Mathematics, Psychology)?