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 crosssections (or slicing). Before we learn about this method, let us first find the relationship between the area of crosssections 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 crosssection of a rectangle, which has an area of A = l * w:
If we stacked enough rectangular crosssections 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 crosssections 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 crosssection of a triangle:
If we stack enough triangular crosssections 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 crosssections to form a length l, which gives us the volume of:
V = Area of crosssection * l = (½ * b * h) * l
Now that we have determined how crosssections and volumes relate in simple geometric applications, we can connect crosssectional areas, volume and integration.
Formula: Volumes of Solids by CrossSections
To find the volume of any prism, you need the area of all of the slices of the crosssections and add them together to find the total volume. A prism can be made up of 2, 10, 1000, or infinitely many crosssections. Essentially, we are taking a Riemann sum of an infinite number of crosssections: 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 CrossSections

Example 1: Consider the region enclosed by y =x^{2}, x = 3, and the x axis. Suppose a 3dimensional crosssection is created that is perpendicular to the x axis. What would be the volume if the crosssection is a square?
Solution:
Sketch the solid and typical cross section.
The curves y =x^{2}, 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 crosssection.
Since the crosssection is perpendicular to the x axis, we need to draw the base of the crosssections in the region R that is perpendicular to the x axis:
Since the area of the crosssection is a square, the crosssections will look like this in the xyz plane:
The base of the square is from y = 0 to y = x^{2}. Therefore, the base of the rectangle is x^{2}0=x^{2}. The height will also be x^{2}(since the crosssection is a square).
The area function A(x) = s^{2}=(x^{2})^{2}= x^{4}
Find the limits of integration.
Since the crosssections 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 = x^{2}:
0 = x^{2}
0 = x
The upper limit of integration is at x = 3.
Apply the formula
Example 2: Consider the region enclosed by y =x^{2}, x = 3, and the x axis. Suppose a 3dimensional crosssection 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 =x^{2}, 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 crosssection.
Since the crosssection is perpendicular to the y axis, we need to draw the base of the crosssections in the region R that is perpendicular to the y axis:
Since the area of the crosssection is a semicircle, the crosssections will look like this in the xyz plane:
Since the crosssections are perpendicular to the y axis, the functions have to be in terms of y.
Let’s start with the function y = x^{2}. Putting the function in terms of y (which means to isolate x):
 y = x^{2}—>
 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 crosssections 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 = x^{2} 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 =x^{2} and y = 4. Suppose a 3dimensional crosssection is created that is perpendicular to the y axis. What would be the volume if the crosssection is an equilateral triangle?
Solution:
Sketch the solid and typical crosssection.
The curves y =x^{2}, 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 crosssection.
Since the crosssection is perpendicular to the y axis, we need to draw the base of the crosssections in the region R that is perpendicular to the y axis:
Since the area of the crosssection is an equilateral triangle, the crosssections will look like this in the xyz plane:
Because the crosssection is perpendicular to the y axis, we need to turn each function as a function of y (meaning isolate x):
 y=x^{2}—>
 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 crosssections 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 = x^{2}, which is y = 0.
The upper limit of integration is at y = 4.
Apply the formula
Example 4: Consider the region enclosed by y =x^{2} and y = x. Suppose a 3dimensional crosssection is created that is perpendicular to the x axis. What would be the volume if the crosssection is an isosceles right triangle?
Solution:
Sketch the solid and typical crosssection.
The curves y = x^{2}, 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 crosssection.
Since the crosssection is perpendicular to the x axis, we need to draw the base of the crosssections in the region R that is perpendicular to the x axis:
Since the area of the crosssection is an isosceles right triangle, the crosssections will look like this in the xyz plane:
The base of the triangle is the distance between the functions y = x and y = x^{2}. Therefore, the base of the triangle is s = xx^{2}. Because this is an isosceles right triangle, 2 sides of the triangle will be the same, so two sides will have the length s = xx^{2}.
The area function is
Find the limits of integration.
The limits of integration will be along the x axis, since the crosssections are perpendicular to the x axis.
The limits of integration will be at the points of intersection of the graph:
x = x^{2}
0 = x^{2}x
0 = (x)(x1)
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 crosssection 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 crosssection formula: Similarly, if the crosssection 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/studyguides/calculus/calculus/applicationsofthedefiniteintegral/volumesofsolidswithknowncrosssections
https://www.math24.net/volumesolidwithknowncrosssection/
https://www.onlinemathlearning.com/volumerectangularprism.html
https://slideplayer.com/slide/7420098/
https://www.onlinemathlearning.com/volumetriangularprism.html
https://www.ck12.org/calculus/crosssectionmethod/lesson/VolumesbyCrossSectionCALC/
https://www.math.upenn.edu/~shilinyu/teaching/Math104Sp15/slides/cross_section.pdf
http://scidiv.bellevuecollege.edu/dh/Calculus_all/CC_5_1_Volumes.pdf
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