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8.8 Volumes with Cross Sections: Triangles and Semicircles

6 min readfebruary 15, 2024

Recall from the last guide that we can instead break up 3D objects bound by curves into infinitely thin slices that are easier to work with. In this guide, we’ll apply this concept to triangles and semicircles!

🪩 Solids with Cross Sections: Review

To find the volume of a solid with known cross-sections we can use the formula

V=abA(x) dxV = \int_a^b A(x)\ dx

where y=A(x)y=A(x) is a function for the area of a cross-section (some two-dimensional shape) perpendicular to the x-axis on the closed interval [a,b][a,b] and dxdx represents its thickness. In this guide, we’ll explore this same concept using different shapes.

🔺Triangular Cross Sections

The formula for the area of a triangle depends on what kind of triangle it is. This means that the function A(x)A(x) for a solid with triangular cross sections will be different depending on the kind of triangle that makes up the cross sections of the solid. We will develop volume formulas for solids with equilateral and right-angled isosceles triangle cross sections.

🔻 Equilateral Triangles

The formula for the area of an equilateral triangle is given by the equation 34s2\frac{\sqrt{3}}4s^2 where ss is the length of one of the sides of the triangle. This means that the formula for the volume of a solid with equilateral triangle cross sections is given by V=ab34s2dxV = \int_a^b \frac{\sqrt{3}}4s^2 dx.

📐 Right Isosceles Triangles

The formula for the area of a right-angled isosceles triangle is 12s2\frac{1}2s^2 were ss is the length of the two matching sides of the triangle. So for a solid with right isosceles triangle cross sections, its volume will be given by formula V=ab12s2dxV = \int_a^b \frac{1}2s^2 dx.

🚫 Semicircular Cross Sections

To find the area of a circle, we can use the familiar formula πr2\pi r^2, where rr is the radius of the circle. A semicircle only has half the area of a circle though, so we’ll use the equation 12πr2\frac{1}2\pi r^2 for A(x) in our volume formula. That means that for a solid with semicircular cross sections, its volume will be found using the equation V=ab12πr2dxV = \int_a^b \frac{1}2\pi r^2 dx.


🧠 Solving Cross-Section Problems

Now that we have our area functions we need to find ss or rr in order to use them! Let’s work though an example.

Suppose a region bounded by y=x2y = x^2 and y=xy = \sqrt{x} forms the base of a solid with cross sections taken perpendicular to the x-axis. What integral will give the volume of the solid if the cross sections are equilateral triangles? What about semicircles?

Question courtesy of Flipped Math

🧐We’ll begin by drawing a picture of out graphs so we can visualize the problem and identify the bounds we’ll use when evaluating our integral.

Untitled

Image courtesy of Flipped Math

In this image, y=xy = \sqrt{x} forms the upper curve, and y=x2y = x^2 is the bottom curve. Let’s let function h(x)=xh(x)= \sqrt{x} and g(x)=x2g(x) = x^2. The area in grey is the base of our solid. The purple line represents a singular cross section of this solid. You can imagine that the cross section is coming towards you along a z-axis not shown in this drawing. If we rotated our graph and included the z-axis, our graph would look like the one on the top for equilateral triangle cross sections and the one on the bottom for semicircular cross sections.

Untitled

Image courtesy of Flipped Math

Untitled

Image courtesy of Flipped Math

Identifying bounds

👀Graphically, we can see that the two functions intersect for x=0x = 0 and x=1x = 1. This means that our bounds are given by the closed interval [0, 1]. We could also find the bounds algebraically by setting the two functions equal and solving for x as shown below.

h(x)=g(x)x=x2x=x4 x=0 or x=1h(x)=g(x)\\ \sqrt{x}=x^2\\ x=x^4\\ \therefore \ x=0\text{ or }x=1

🔺 Equilateral Triangle

Finding the Side Length

You’ll notice that one side of the equilateral triangle at a point xx is given by h(x)g(x)h(x) - g(x). So, the side length ss of the equilateral triangle cross section is given by xx2\sqrt{x} - x^2.

Setting up Cross Section Area Functions

Now that we have a value for ss, we can identify our A(x)A(x) functions for this solid. For the equilateral triangle, this will be given by 34(xx2)2\frac{\sqrt{3}}4(\sqrt{x} - x^2)^2.

Building the Volume Integral

🪢 Let’s tie it all together now and find our volume integrals. Plugging in our bounds and A(x)A(x) functions into the formula V=abA(x)dxV = \int_a^b A(x) dx, we find that the volume of the solid with equilateral triangle cross sections is V=3401(xx2)2dxV = \frac{\sqrt{3}}4\int_0^1(\sqrt{x} - x^2)^2 dx.

Solving

3401(xx2)2dx=3401x2x5/2+x4 dx=34(x224x7/27+x55)=34(1224(1)7/27+155)34(0224(0)7/27+055)=343540+1470=39280=35/2280\frac{\sqrt{3}}{4}\int_0^1(\sqrt{x} - x^2)^2 dx=\frac{\sqrt{3}}{4}\int_0^1x-2x^{5/2}+x^4\ dx=\frac{\sqrt{3}}{4}\Bigg(\frac{x^2}2-\frac{4x^{7/2}}7+\frac{x^5}{5}\Bigg)=\\ \frac{\sqrt{3}}{4}\Bigg(\frac{1^2}2-\frac{4(1)^{7/2}}7+\frac{1^5}{5}\Bigg)-\frac{\sqrt{3}}{4}\Bigg(\frac{0^2}2-\frac{4(0)^{7/2}}7+\frac{0^5}{5}\Bigg)=\\ \frac{\sqrt{3}}4\cdot \frac{35-40+14}{70}=\frac{\sqrt{3}\cdot 9}{280}=\boxed{\frac{3^{5/2}}{280}}

🚫 Semicircle

Finding the Radius

The diameter of the semicircle at a point xx is defined by h(x)g(x)h(x) - g(x). Recall that the radius of a circle is one half its diameter. Therefore, the radius rr of the semicircle for some xx is xx22\frac{\sqrt{x} - x^2}2.

Setting up Cross Section Area Functions

Now that we have a value for rr, we can identify our A(x)A(x) functions for this solid. For the semicircle, this will be given by 12π(xx22)2=π2(xx2)24=π8(xx2)2\frac{1}2\pi (\frac{\sqrt{x} - x^2}2)^2 = \frac{\pi}2 \frac{(\sqrt{x} - x^2)^2}4 = \frac{\pi}8 (\sqrt{x} - x^2)^2.

Building the Volume Integral

🪢 Let’s tie it all together now and find our volume integrals. Plugging in our bounds and A(x)A(x) functions into the formula V=abA(x)dxV = \int_a^b A(x) dx, we find that the volume of the solid with semicircular cross sections is given by V=π801(xx2)2dxV = \frac{\pi}8\int_0^1 (\sqrt{x} - x^2)^2 dx.

Solving

π801(xx2)2dx=π801x2x5/2+x4 dx=π8(x224x7/27+x55)=π8(1224(1)7/27+155)π8(0224(0)7/27+055)=π83540+1470=9π560\frac{\pi}8\int_0^1 (\sqrt{x} - x^2)^2 dx=\frac{\pi}8\int_0^1x-2x^{5/2}+x^4\ dx=\frac{\pi}8\Bigg(\frac{x^2}2-\frac{4x^{7/2}}7+\frac{x^5}{5}\Bigg)=\\ \frac{\pi}8\Bigg(\frac{1^2}2-\frac{4(1)^{7/2}}7+\frac{1^5}{5}\Bigg)-\frac{\pi}8\Bigg(\frac{0^2}2-\frac{4(0)^{7/2}}7+\frac{0^5}{5}\Bigg)=\\ \frac{\pi}8\cdot \frac{35-40+14}{70}=\boxed{\frac{9\pi}{560}}

📝Summary

When finding the volume of a solid with a known cross section, we can use the equation

V=abA(x) dx.V = \int_a^b A(x)\ dx.

A(x)A(x) is a function for the area of a cross section perpendicular to the x-axis on the interval [a, b].

For equilateral triangle cross sections A(x)=34s2A(x) = \frac{\sqrt{3}}4s^2 where ss is the length of one of the sides of the triangle.

For right isosceles triangle cross sections A(x)=12s2A(x) =\frac{1}2s^2 were ss is the length of the two matching sides of the triangle.

For semicircular cross sections A(x)=12πr2A(x) = \frac{1}2\pi r^2 where rr is the radius of the semicircle at xx.

Using methods for finding the areas between curves we can determine equations to describe ss or rr and find bounds for our integral.

Plugging in our bounds and the function A(x)A(x) results in the volume integral for the solid. Evaluating this definite integral will give a value for the volume of the solid.

Key Terms to Review (11)

Axis of Revolution

: The axis of revolution is an imaginary line around which a two-dimensional shape rotates to create a three-dimensional object. It serves as the central point for rotation.

Cross Sections

: Cross sections refer to the 2D shapes that are obtained when a solid is cut by a plane. They help us understand the shape and properties of the solid.

Definite Integral

: A definite integral is a mathematical tool used to calculate the exact area between a function and the x-axis over a specific interval.

Function

: A relationship between two sets where each input (domain) value corresponds to exactly one output (range) value.

Integration

: Integration is an operation in calculus that finds antiderivatives, or reverse derivatives, allowing us to calculate areas, volumes, and other quantities related to curves and functions.

Limits of Integration

: The limits of integration are the values that determine the range over which an integral is evaluated. They specify the starting and ending points on the x-axis for finding the area or calculating other quantities using integration.

Radius

: The radius is the distance from the center of a circle to any point on its circumference.

Semicircular Cross Sections

: Semicircular cross sections are used to find the volume of a solid by using semicircles as the base for each cross section.

Triangular Cross Sections

: Triangular cross sections refer to the method of finding the volume of a solid by using triangles as the base of each cross section.

y = √(r^2 - x^2)

: This equation represents part of a circle with radius r centered at the origin (0, 0). It gives the y-coordinate of a point on the circle's circumference based on its x-coordinate.

y = f(x)

: This notation represents the relationship between the dependent variable y and the independent variable x in a function. It indicates that y is determined by some rule or equation involving x.

8.8 Volumes with Cross Sections: Triangles and Semicircles

6 min readfebruary 15, 2024

Recall from the last guide that we can instead break up 3D objects bound by curves into infinitely thin slices that are easier to work with. In this guide, we’ll apply this concept to triangles and semicircles!

🪩 Solids with Cross Sections: Review

To find the volume of a solid with known cross-sections we can use the formula

V=abA(x) dxV = \int_a^b A(x)\ dx

where y=A(x)y=A(x) is a function for the area of a cross-section (some two-dimensional shape) perpendicular to the x-axis on the closed interval [a,b][a,b] and dxdx represents its thickness. In this guide, we’ll explore this same concept using different shapes.

🔺Triangular Cross Sections

The formula for the area of a triangle depends on what kind of triangle it is. This means that the function A(x)A(x) for a solid with triangular cross sections will be different depending on the kind of triangle that makes up the cross sections of the solid. We will develop volume formulas for solids with equilateral and right-angled isosceles triangle cross sections.

🔻 Equilateral Triangles

The formula for the area of an equilateral triangle is given by the equation 34s2\frac{\sqrt{3}}4s^2 where ss is the length of one of the sides of the triangle. This means that the formula for the volume of a solid with equilateral triangle cross sections is given by V=ab34s2dxV = \int_a^b \frac{\sqrt{3}}4s^2 dx.

📐 Right Isosceles Triangles

The formula for the area of a right-angled isosceles triangle is 12s2\frac{1}2s^2 were ss is the length of the two matching sides of the triangle. So for a solid with right isosceles triangle cross sections, its volume will be given by formula V=ab12s2dxV = \int_a^b \frac{1}2s^2 dx.

🚫 Semicircular Cross Sections

To find the area of a circle, we can use the familiar formula πr2\pi r^2, where rr is the radius of the circle. A semicircle only has half the area of a circle though, so we’ll use the equation 12πr2\frac{1}2\pi r^2 for A(x) in our volume formula. That means that for a solid with semicircular cross sections, its volume will be found using the equation V=ab12πr2dxV = \int_a^b \frac{1}2\pi r^2 dx.


🧠 Solving Cross-Section Problems

Now that we have our area functions we need to find ss or rr in order to use them! Let’s work though an example.

Suppose a region bounded by y=x2y = x^2 and y=xy = \sqrt{x} forms the base of a solid with cross sections taken perpendicular to the x-axis. What integral will give the volume of the solid if the cross sections are equilateral triangles? What about semicircles?

Question courtesy of Flipped Math

🧐We’ll begin by drawing a picture of out graphs so we can visualize the problem and identify the bounds we’ll use when evaluating our integral.

Untitled

Image courtesy of Flipped Math

In this image, y=xy = \sqrt{x} forms the upper curve, and y=x2y = x^2 is the bottom curve. Let’s let function h(x)=xh(x)= \sqrt{x} and g(x)=x2g(x) = x^2. The area in grey is the base of our solid. The purple line represents a singular cross section of this solid. You can imagine that the cross section is coming towards you along a z-axis not shown in this drawing. If we rotated our graph and included the z-axis, our graph would look like the one on the top for equilateral triangle cross sections and the one on the bottom for semicircular cross sections.

Untitled

Image courtesy of Flipped Math

Untitled

Image courtesy of Flipped Math

Identifying bounds

👀Graphically, we can see that the two functions intersect for x=0x = 0 and x=1x = 1. This means that our bounds are given by the closed interval [0, 1]. We could also find the bounds algebraically by setting the two functions equal and solving for x as shown below.

h(x)=g(x)x=x2x=x4 x=0 or x=1h(x)=g(x)\\ \sqrt{x}=x^2\\ x=x^4\\ \therefore \ x=0\text{ or }x=1

🔺 Equilateral Triangle

Finding the Side Length

You’ll notice that one side of the equilateral triangle at a point xx is given by h(x)g(x)h(x) - g(x). So, the side length ss of the equilateral triangle cross section is given by xx2\sqrt{x} - x^2.

Setting up Cross Section Area Functions

Now that we have a value for ss, we can identify our A(x)A(x) functions for this solid. For the equilateral triangle, this will be given by 34(xx2)2\frac{\sqrt{3}}4(\sqrt{x} - x^2)^2.

Building the Volume Integral

🪢 Let’s tie it all together now and find our volume integrals. Plugging in our bounds and A(x)A(x) functions into the formula V=abA(x)dxV = \int_a^b A(x) dx, we find that the volume of the solid with equilateral triangle cross sections is V=3401(xx2)2dxV = \frac{\sqrt{3}}4\int_0^1(\sqrt{x} - x^2)^2 dx.

Solving

3401(xx2)2dx=3401x2x5/2+x4 dx=34(x224x7/27+x55)=34(1224(1)7/27+155)34(0224(0)7/27+055)=343540+1470=39280=35/2280\frac{\sqrt{3}}{4}\int_0^1(\sqrt{x} - x^2)^2 dx=\frac{\sqrt{3}}{4}\int_0^1x-2x^{5/2}+x^4\ dx=\frac{\sqrt{3}}{4}\Bigg(\frac{x^2}2-\frac{4x^{7/2}}7+\frac{x^5}{5}\Bigg)=\\ \frac{\sqrt{3}}{4}\Bigg(\frac{1^2}2-\frac{4(1)^{7/2}}7+\frac{1^5}{5}\Bigg)-\frac{\sqrt{3}}{4}\Bigg(\frac{0^2}2-\frac{4(0)^{7/2}}7+\frac{0^5}{5}\Bigg)=\\ \frac{\sqrt{3}}4\cdot \frac{35-40+14}{70}=\frac{\sqrt{3}\cdot 9}{280}=\boxed{\frac{3^{5/2}}{280}}

🚫 Semicircle

Finding the Radius

The diameter of the semicircle at a point xx is defined by h(x)g(x)h(x) - g(x). Recall that the radius of a circle is one half its diameter. Therefore, the radius rr of the semicircle for some xx is xx22\frac{\sqrt{x} - x^2}2.

Setting up Cross Section Area Functions

Now that we have a value for rr, we can identify our A(x)A(x) functions for this solid. For the semicircle, this will be given by 12π(xx22)2=π2(xx2)24=π8(xx2)2\frac{1}2\pi (\frac{\sqrt{x} - x^2}2)^2 = \frac{\pi}2 \frac{(\sqrt{x} - x^2)^2}4 = \frac{\pi}8 (\sqrt{x} - x^2)^2.

Building the Volume Integral

🪢 Let’s tie it all together now and find our volume integrals. Plugging in our bounds and A(x)A(x) functions into the formula V=abA(x)dxV = \int_a^b A(x) dx, we find that the volume of the solid with semicircular cross sections is given by V=π801(xx2)2dxV = \frac{\pi}8\int_0^1 (\sqrt{x} - x^2)^2 dx.

Solving

π801(xx2)2dx=π801x2x5/2+x4 dx=π8(x224x7/27+x55)=π8(1224(1)7/27+155)π8(0224(0)7/27+055)=π83540+1470=9π560\frac{\pi}8\int_0^1 (\sqrt{x} - x^2)^2 dx=\frac{\pi}8\int_0^1x-2x^{5/2}+x^4\ dx=\frac{\pi}8\Bigg(\frac{x^2}2-\frac{4x^{7/2}}7+\frac{x^5}{5}\Bigg)=\\ \frac{\pi}8\Bigg(\frac{1^2}2-\frac{4(1)^{7/2}}7+\frac{1^5}{5}\Bigg)-\frac{\pi}8\Bigg(\frac{0^2}2-\frac{4(0)^{7/2}}7+\frac{0^5}{5}\Bigg)=\\ \frac{\pi}8\cdot \frac{35-40+14}{70}=\boxed{\frac{9\pi}{560}}

📝Summary

When finding the volume of a solid with a known cross section, we can use the equation

V=abA(x) dx.V = \int_a^b A(x)\ dx.

A(x)A(x) is a function for the area of a cross section perpendicular to the x-axis on the interval [a, b].

For equilateral triangle cross sections A(x)=34s2A(x) = \frac{\sqrt{3}}4s^2 where ss is the length of one of the sides of the triangle.

For right isosceles triangle cross sections A(x)=12s2A(x) =\frac{1}2s^2 were ss is the length of the two matching sides of the triangle.

For semicircular cross sections A(x)=12πr2A(x) = \frac{1}2\pi r^2 where rr is the radius of the semicircle at xx.

Using methods for finding the areas between curves we can determine equations to describe ss or rr and find bounds for our integral.

Plugging in our bounds and the function A(x)A(x) results in the volume integral for the solid. Evaluating this definite integral will give a value for the volume of the solid.

Key Terms to Review (11)

Axis of Revolution

: The axis of revolution is an imaginary line around which a two-dimensional shape rotates to create a three-dimensional object. It serves as the central point for rotation.

Cross Sections

: Cross sections refer to the 2D shapes that are obtained when a solid is cut by a plane. They help us understand the shape and properties of the solid.

Definite Integral

: A definite integral is a mathematical tool used to calculate the exact area between a function and the x-axis over a specific interval.

Function

: A relationship between two sets where each input (domain) value corresponds to exactly one output (range) value.

Integration

: Integration is an operation in calculus that finds antiderivatives, or reverse derivatives, allowing us to calculate areas, volumes, and other quantities related to curves and functions.

Limits of Integration

: The limits of integration are the values that determine the range over which an integral is evaluated. They specify the starting and ending points on the x-axis for finding the area or calculating other quantities using integration.

Radius

: The radius is the distance from the center of a circle to any point on its circumference.

Semicircular Cross Sections

: Semicircular cross sections are used to find the volume of a solid by using semicircles as the base for each cross section.

Triangular Cross Sections

: Triangular cross sections refer to the method of finding the volume of a solid by using triangles as the base of each cross section.

y = √(r^2 - x^2)

: This equation represents part of a circle with radius r centered at the origin (0, 0). It gives the y-coordinate of a point on the circle's circumference based on its x-coordinate.

y = f(x)

: This notation represents the relationship between the dependent variable y and the independent variable x in a function. It indicates that y is determined by some rule or equation involving x.


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AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.


© 2024 Fiveable Inc. All rights reserved.

AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.