Volumes with Cross Sections: Triangles and Semicircles

#8.8 Volumes with Cross Sections: Triangles and Semicircles

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

$$V = \int_a^b A(x)\ dx$$

where $y=A(x)$ is a function for the area of a cross-section (some two-dimensional shape) perpendicular to the x-axis on the closed interval $[a,b]$ and $dx$ 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 $A(x)$ 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 $\frac{\sqrt{3}}4s^2$ where $s$ 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 $V = \int_a^b \frac{\sqrt{3}}4s^2 dx$.

#📐 Right Isosceles Triangles

The formula for the area of a right-angled isosceles triangle is $\frac{1}2s^2$ were $s$ 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 $V = \int_a^b \frac{1}2s^2 dx$.

#🚫 Semicircular Cross Sections

To find the area of a circle, we can use the familiar formula $\pi r^2$, where $r$ is the radius of the circle. A semicircle only has half the area of a circle though, so we’ll use the equation $\frac{1}2\pi r^2$ 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 $V = \int_a^b \frac{1}2\pi r^2 dx$.

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#🧠 Solving Cross-Section Problems

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

Suppose a region bounded by $y = x^2$ and $y = \sqrt{x}$ 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.

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Visualization of provided problem.

Image courtesy of Flipped Math

In this image, $y = \sqrt{x}$ forms the upper curve, and $y = x^2$ is the bottom curve. Let’s let function $h(x)= \sqrt{x}$ and $g(x) = x^2$. 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

Visualization of 3D cross-section

Image courtesy of Flipped Math

!Untitled

Visualization of 3D cross-section

Image courtesy of Flipped Math

#Identifying bounds

👀Graphically, we can see that the two functions intersect for $x = 0$ and $x = 1$. 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.

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

#🔺 Equilateral Triangle

#Finding the Side Length

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

#Setting up Cross Section Area Functions

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

#Building the Volume Integral

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

#Solving

$$\frac{\sqrt{3}}{4}\int_0^1(\sqrt{x} - x^2)^2 dx=\frac{\sqrt{3}}{4}\int_0^1x-2x^{5/2}+x^4\ dx=\frac{\sqrt{3}}{4}\Bigg(\frac{x^2}2-\frac{4x^{7/2}}7+\frac{x^5}{5}\Bigg)=\\ \frac{\sqrt{3}}{4}\Bigg(\frac{1^2}2-\frac{4(1)^{7/2}}7+\frac{1^5}{5}\Bigg)-\frac{\sqrt{3}}{4}\Bigg(\frac{0^2}2-\frac{4(0)^{7/2}}7+\frac{0^5}{5}\Bigg)=\\ \frac{\sqrt{3}}4\cdot \frac{35-40+14}{70}=\frac{\sqrt{3}\cdot 9}{280}=\boxed{\frac{3^{5/2}}{280}}$$

#🚫 Semicircle

#Finding the Radius

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

#Setting up Cross Section Area Functions

Now that we have a value for $r$, we can identify our $A(x)$ functions for this solid. For the semicircle, this will be given by $\frac{1}2\pi (\frac{\sqrt{x} - x^2}2)^2 = \frac{\pi}2 \frac{(\sqrt{x} - x^2)^2}4 = \frac{\pi}8 (\sqrt{x} - x^2)^2$.

#Building the Volume Integral

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

#Solving

$$\frac{\pi}8\int_0^1 (\sqrt{x} - x^2)^2 dx=\frac{\pi}8\int_0^1x-2x^{5/2}+x^4\ dx=\frac{\pi}8\Bigg(\frac{x^2}2-\frac{4x^{7/2}}7+\frac{x^5}{5}\Bigg)=\\ \frac{\pi}8\Bigg(\frac{1^2}2-\frac{4(1)^{7/2}}7+\frac{1^5}{5}\Bigg)-\frac{\pi}8\Bigg(\frac{0^2}2-\frac{4(0)^{7/2}}7+\frac{0^5}{5}\Bigg)=\\ \frac{\pi}8\cdot \frac{35-40+14}{70}=\boxed{\frac{9\pi}{560}}$$

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#📝Summary

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

$$V = \int_a^b A(x)\ dx.$$

$A(x)$ 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 $A(x) = \frac{\sqrt{3}}4s^2$ where $s$ is the length of one of the sides of the triangle.

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

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

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

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