In previous sections, we learned how to find the volume of functions using the disk and washer method (revolving a function or a set of functions around a vertical or horizontal axis). In this section, we will learn another method of finding the volume – method of cross-sections (or slicing). Before we learn about this method, let us first find the relationship between the area of cross-sections and volume.
For instance, let’s examine the volume of a rectangular prism. We know that the volume of a rectangular cylinder is V = l * w * h.

However, let’s say that we cut a slice from the rectangular prism. What we would obtain is a cross-section of a rectangle, which has an area of A = l * w:

If we stacked enough rectangular cross-sections on top of each other, we will eventually obtain a rectangular prism of height h:

Basically, the volume of a rectangular prism is obtained by taking the sum of the rectangular cross-sections to form a length h, which gives us the volume of:
V = Area of cross section * h = (l*w)*h
Let’s examine another shape – a triangular prism. We know that the volume of a rectangular cylinder is V = (½* b * h) * l

However, let’s say that we cut a slice from the triangular prism. What we would obtain is a cross-section of a triangle:

If we stack enough triangular cross-sections on top of each other, we will eventually obtain a triangular prism of length l:

Basically, the volume of a triangular prism is obtained by taking the sum of the triangular cross-sections to form a length l, which gives us the volume of:
V = Area of cross-section * l = (½ * b * h) * l
Now that we have determined how cross-sections and volumes relate in simple geometric applications, we can connect cross-sectional areas, volume and integration.
| Formula: Volumes of Solids by Cross-Sections
To find the volume of any prism, you need the area of all of the slices of the cross-sections and add them together to find the total volume. A prism can be made up of 2, 10, 1000, or infinitely many cross-sections. Essentially, we are taking a Riemann sum of an infinite number of cross-sections: ![]() In other words, if S is a solid and A(x) is the area of the face formed by a cut at x and perpendicular to the x–axis, then the volume V of the part of S above the interval [a,b] is: ![]() ![]() Of course, if S is a solid and A(y) is the area of the face formed by a cut at y and perpendicular to the y–axis, then the volume of the part of S between cuts at c and d on the y–axis is: ![]() ![]() Guidelines for Finding Volumes of Solids by Cross-Sections
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Example 1: Consider the region enclosed by y =x2, x = 3, and the x axis. Suppose a 3-dimensional cross-section is created that is perpendicular to the x axis. What would be the volume if the cross-section is a square?
Solution:
Sketch the solid and typical cross section.
The curves y =x2, x = 3, and y = 0 (the x axis), as well as the region enclosed by the curves, are drawn below:

Find the formula for A(x) or A(y), the area of the cross-section.
Since the cross-section is perpendicular to the x axis, we need to draw the base of the cross-sections in the region R that is perpendicular to the x axis:

Since the area of the cross-section is a square, the cross-sections will look like this in the x-y-z plane:

The base of the square is from y = 0 to y = x2. Therefore, the base of the rectangle is x2-0=x2. The height will also be x2(since the cross-section is a square).
The area function A(x) = s2=(x2)2= x4
Find the limits of integration.
Since the cross-sections are perpendicular to the x axis, the limits of integration will be along the x axis.
The lower limit of integration is at the intersection of y = 0 and y = x2:
0 = x2
0 = x
The upper limit of integration is at x = 3.
Apply the formula

Example 2: Consider the region enclosed by y =x2, x = 3, and the x axis. Suppose a 3-dimensional cross-section is created that is perpendicular to the y axis. What would be the volume if the cross section is a semicircle?
Solution:
Sketch the solid and typical cross section.
The curves y =x2, x = 3, and y = 0 ( the x axis), as well as the region enclosed by the curves, are drawn below:

Find the formula for A(x) or A(y), the area of the cross-section.
Since the cross-section is perpendicular to the y axis, we need to draw the base of the cross-sections in the region R that is perpendicular to the y axis:

Since the area of the cross-section is a semicircle, the cross-sections will look like this in the x-y-z plane:

Since the cross-sections are perpendicular to the y axis, the functions have to be in terms of y.
Let’s start with the function y = x2. Putting the function in terms of y (which means to isolate x):
- y = x2—>
- x = 3 (since 3 is a constant, the function does not need modification)
- y = 0 (since 0 is also a constant, the function does not need modification)
The length of the diameter of each semicircle is the distance between x = 3 and
d = 3 –
And the radius is:

Therefore, the area of the semicircle is:

Find the limits of integration.
Since the cross-sections are perpendicular to the y axis, the limits of integration will be along the y axis.
From the graph, the lower limit of integration is at y = 0.
The upper limit of integration is at the intersection of y = x2 and x = 3:
y = (3)2
y = 9
Therefore, the upper limit of integration is y = 9.
Apply the formula


Example 3: Consider the region enclosed by y =x2 and y = 4. Suppose a 3-dimensional cross-section is created that is perpendicular to the y axis. What would be the volume if the cross-section is an equilateral triangle?
Solution:
Sketch the solid and typical cross-section.
The curves y =x2, y = 4, and the region enclosed by both curves are drawn below:

Find the formula for A(x) or A(y), the area of the cross-section.
Since the cross-section is perpendicular to the y axis, we need to draw the base of the cross-sections in the region R that is perpendicular to the y axis:

Since the area of the cross-section is an equilateral triangle, the cross-sections will look like this in the x-y-z plane:

Because the cross-section is perpendicular to the y axis, we need to turn each function as a function of y (meaning isolate x):
- y=x2—>
- y = 4 (since 4 is a constant, the function does not need modification).
The base of each triangle is the width of the parabola, which can be partitioned into two segments:

Base of triangle
Since this is an equilateral triangle, each side of the triangle is also
Therefore, the area function is:

Find the limits of integration.
Since the cross-sections are perpendicular to the y axis, the limits of integration will be along the y axis.
From looking at the graph, the lower limit of integration is at the intersection of the x axis and y = x2, which is y = 0.
The upper limit of integration is at y = 4.
Apply the formula


Example 4: Consider the region enclosed by y =x2 and y = x. Suppose a 3-dimensional cross-section is created that is perpendicular to the x axis. What would be the volume if the cross-section is an isosceles right triangle?
Solution:
Sketch the solid and typical cross-section.
The curves y = x2, y = x, and the region enclosed by the curves are drawn below:

Find the formula for A(x) or A(y), the area of the cross-section.
Since the cross-section is perpendicular to the x axis, we need to draw the base of the cross-sections in the region R that is perpendicular to the x axis:

Since the area of the cross-section is an isosceles right triangle, the cross-sections will look like this in the x-y-z plane:

The base of the triangle is the distance between the functions y = x and y = x2. Therefore, the base of the triangle is s = x-x2. Because this is an isosceles right triangle, 2 sides of the triangle will be the same, so two sides will have the length s = x-x2.
The area function is

Find the limits of integration.
The limits of integration will be along the x axis, since the cross-sections are perpendicular to the x axis.
The limits of integration will be at the points of intersection of the graph:
x = x2
0 = x2-x
0 = (x)(x-1)
Setting both factors equal to zero and solve,
x = 0 or x = 1
Therefore, the lower and upper limits of integration are x = 0 and x = 1.
Apply the formula


| Summary of Section
Let S be a solid and suppose that the area of the cross-section in the plane is perpendicular to the x−axis is A(x) on the interval of [a,b]. Then the volume of the solid from x = a to x = b is given the cross-section formula: ![]() Similarly, if the cross-section is perpendicular to the y – axis, and its area is defined by the function A(y), then the volume of the solid from y = c to y = d is given by: ![]() |
References:
https://www.cliffsnotes.com/study-guides/calculus/calculus/applications-of-the-definite-integral/volumes-of-solids-with-known-cross-sections
https://www.math24.net/volume-solid-with-known-cross-section/
https://www.onlinemathlearning.com/volume-rectangular-prism.html
https://slideplayer.com/slide/7420098/
https://www.onlinemathlearning.com/volume-triangular-prism.html
https://www.ck12.org/calculus/cross-section-method/lesson/Volumes-by-Cross-Section-CALC/
https://www.math.upenn.edu/~shilinyu/teaching/Math104Sp15/slides/cross_section.pdf
http://scidiv.bellevuecollege.edu/dh/Calculus_all/CC_5_1_Volumes.pdf









