In the previous section, we discovered the disk method of finding the volume of a function revolved around an axis. The solid of revolution formed a “cylinder” of radius r = F(x) (revolved around a horizontal axis) or F(y) (revolved around a vertical axis) and height b – a on the interval [a,b].
In this section, we will talk about using the washer method to find the volume revolution of a solid bound by 2 functions revolving around the same axis. In order to learn about this method, we first need to derive the volume of a hollow cylinder (similar to how we derived the formula of a solid cylinder for the disk method) and see how the formula applies to functions on the coordinate plane.
As we learned, the equation of a circle is A = r^{2}. If we cut a small hole from the circle, the resulting shape is called a washer:
The area of the washer is simple: find the area of the outer circle and subtract the area of the inner circle:
You can imagine that a hollow cylinder is a collection of washers which, if stacked on top of each other, will form a hollow cylinder with height h. That means we have to find the area of each washer (A_{washer} =[R^{2}r^{2}]) and add them up.
We can find the sum of all the washers by taking the integral of the area function on the interval x = 0 to x = h, integrating:
Since [R^{2}r^{2}] is a constant, it will be factored from the integrand:
Integrate the constant of 1:
In this way, we have derived the volume of a hollow cylinder: h[R^{2}r^{2}]
This same pattern can be applied to two functions being revolved around an axis.
Formula: Volume of Revolution – Washer Method
X Axis Rotation Suppose we have two functions F(x) and G(x), and F(x) > G(x) and both functions are nonnegative and continuous on the interval [a,b]. G(x) is the inner radius “r” and F(x) is the outer radius “R”. When both functions are rotated about the x axis, the volume of the solid will look like a hollow ‘cylinder’. Notice how the crosssection of the volume of revolution looks like a washer: The formula for the volume of solid that we have obtained by rotating the both functions around the xaxis will be: Y Axis Rotation Of course, this applies to two functions revolving around the y axis. Suppose we have two functions F(y) and G(y), and F(y) > G(y) and both functions are continuous on the interval [c,d]. The solid of revolution will look like the following: The formula for the volume of solid that we have obtained by rotating the both functions around the yaxis will be: Guidelines for Applying the Washer Method

Example 1:
Suppose you have a solid obtained by rotating the regions of parabola x = y^{2} and the other function is x = around the yaxis. Find the Volume of the enclosed region.
Solution:
Draw the curves on the coordinate plane. Determine axis of rotation (with respect to x or y).
The functions x = y^{2} and x = , as well as the region bounded by both curves, are graphed on the coordinate plane below:
As indicated in the problem, we will revolve around the y axis.
Determine the limits of integration
Since we are revolving both functions around the y axis, the limits of integration will be:
Set both functions equal to each other to find the points of intersection:
y^{2}=
(y^{2})^{2}=()^{2}
y^{4}=y
y^{4}y=0
(y)(y^{3}1)= 0
Set both factors equal to zero and solve:
y = 0 or y = 1
Therefore, c = 0 and d = 1, as is shown on the graph:
Apply the formula
The larger radius is F(y), x = y^{2}, and G(y) is the smaller radius, x = .
Since we are revolving both functions around the y axis,
The diagram of the solid of revolution is shown below. In the diagram, the y axis is the horizontal axis, and the x axis is the vertical axis:
Example 2:
Suppose you have the curves x = y^{2}+6, x = y+4 find the volume of the enclosed region rotated about the y axis.
Solution:
Draw the curves on the coordinate plane. Determine axis of rotation (with respect to x or y).
The curves x = y^{2}+6, x = y+4, as well as the region bounded by both curves, are drawn below:
As indicated in the problem, we are revolving the region about the y axis.
Determine the limits of integration
Because we are revolving around the y axis, the limits of integration will be vertical. Set the two curves equal to each other to determine the limits of integration:
y^{2}+6= y+4
y^{2}+y+2=0
Using the quadratic formula,
a = 1, b = 1, and c = 2. Substituting these numbers into the formula,
x = 1 or 2
Therefore, c = 1 and d = 2.
Apply the formula
The inner radius is G(y) = y +4, and the outer radius F(y) = y^{2}+6. This can be determined by drawing an arrow from the axis of rotation to the curves:
The diagram of the solid of revolution is below. The y axis is the horizontal axis and the x axis the vertical axis.
Example 3:
Suppose you have the curves y = x^{2}+6, y = x+2, x = 0, and x = 1. Find the volume of the enclosed region rotated about the x axis.
Solution:
Draw the curves on the coordinate plane. Determine axis of rotation (with respect to x or y).
The curves y = x^{2}+6, y = x+2, x = 0, and x = 1, as well as the region enclosed between the curves, are drawn below:
As indicated in the problem, the solid of revolution is rotated about the x axis.
Determine the limits of integration
As indicated in the problem and shown on the graph, the limits of integration will be a = 0 and b = 1.
Apply the formula
The inner radius is G(x) = x^{2}+2 and the outer radius is F(x) = x^{2}+6
The volume of revolution is shown below:
Example 4:
Suppose you have the curves x = y^{2}+2, x =2, and y = 2. Find the volume of the enclosed region rotated about the y axis.
Solution:
Draw the curves on the coordinate plane. Determine axis of rotation (with respect to x or y).
The curves x = y^{2}+2, x =2, and y = 2, as well as the region enclosed by the curves, are drawn below:
As indicated in the problem, the y axis is the axis of rotation.
Determine the limits of integration
Since we are revolving the solid around the y axis, the limits of integration will be vertical.
As shown in the diagram, the lower limit is at the intersection of x = 2 and x = y^{2}+2:
2 = y^{2}+2
0 = y^{2}
y = 0
The upper limit of integration is at y = 2.
Therefore, c = 0 and d = 2.
Apply the formula
The inner radius G(y) = 2 and the outer radius F(y) = y^{2}+2.
The volume of revolution is shown below. The y axis is the horizontal axis and the x axis is the vertical axis.
Example 5:
Suppose you have the curves y = x^{2}3, y = 3 Find the volume of the enclosed region rotated about y = 2.
Solution:
Draw the curves on the coordinate plane. Determine the axis of rotation (with respect to x or y).
The curves y = x^{2}3, y = 3, and the axis y = 2 is drawn in the coordinate plane below. The region enclosed between the curves is also shaded.
As indicated, the solid will be rotated about the horizontal axis y = 2, therefore, we are revolving with respect to x.
Determine the limits of integration
Because the axis of rotation is horizontal, the limits of integration are horizontal.
The limits of integration are at the points of intersection:
x^{2}3= 3
x^{2}=
x^{4}= x
x^{4}x=0
x(x^{3}1)= 0
Set the first factor equal to zero:
x = 0
Set the second factor equal to zero:
x^{3}1= 0
x^{3}=1
(x^{3})^{1/3}=(1)^{1/3}
x=1
Therefore, the limits of integration are a = 0 and b = 1
Apply the formula
The inner radius G(x) = 2 – (3) and the outer radius is F(x) = 2 – (x^{2}3)
The solid of revolution is shown below:
Example 6:
Suppose you have the curves y = x^{2}+4, x = 1, x = 0, y = 2. Find the volume of the enclosed region rotated about y = 1.
Solution:
Draw the curves on the coordinate plane. Determine the axis of rotation (with respect to x or y).
The curves y = x^{2}+4, y = 2, x = 1, and x = 0, as well as the region enclosed by the curves are shaded below.
As indicated, the axis of rotation is y = 1, which is a horizontal axis. Therefore, we are revolving the solid with respect to y.
Determine the limits of integration
Because the axis of rotation is horizontal, the limits of integration will be horizontal.
As shown in the graph and indicated in the problem, the limits of integration is x = 0 and x = 1.
Apply the formula
The outer radius is F(x) = x^{2}+4(1) = x^{2}+5 and the inner radius G(x) = 2 – (1) = 2 + 1 = 3.
The solid of revolution is shown below:
Summary of Section
The washer method is used to calculate the volume of a hollow solid of revolution when integrating along an axis parallel to the axis of revolution. The volume of the solid formed by rotating the area between the curves of F(x) and G(x) and the lines x = a and x = b about the xaxis is given by: Of course, the volume of the solid formed by rotating the area between the curves of F(y) and G(y) and the lines x = c and x = d about the yaxis is given by: 
References:
https://www.math24.net/volumesolidofrevolutiondiskswashers/#example5
http://scidiv.bellevuecollege.edu/dh/Calculus_all/CC_5_1_Volumes.pdf
https://xaktly.com/VolumesWasher.html
https://www.mathsisfun.com/calculus/solidsrevolutiondiskwasher.html
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