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1 of 3
Methods of Washers:
• Let f and g be continues and nonnegative on
an interval [a, b]. Let R be the region that is
bounded above y = f(x), below y = g(x), and
the sides by x = a and x = b. The Volume of
the object revolved around the x-axis is 𝑉 =
𝑎
𝑏
𝜋 𝑓 𝑥
2
− 𝑔 𝑥
2
𝑑𝑥
• In the washer method you simply need to
identify the top and bottom curve to evaluate.
• Find the volume of the solid of revolution by rotating the region
bounded by the graphs 𝑦 = 𝑥 − 1 and 𝑦 = 𝑥 − 1 2
about the x-
axis
• Notice that the square root function is on top and the parabola on
bottom.
• Notice that the functions intersect at x = 1 and x = 2
• We can know write the following volume formula:
• 𝑉 = 1
2
𝜋 ∙ [ 𝑥 − 1
2
− 𝑥 − 1 2 2
𝑑𝑥
Here is how the problem works out
• 𝑉 = 1
2
𝜋 ∙ [𝑥 − 1 − 𝑥 − 1 4]𝑑𝑥 = 𝜋 0
1
[𝑥 − 1 − 𝑥 − 1 4] 𝑑𝑥
• 𝑉 = 𝜋
𝑥2
2
− 𝑥 −
𝑥−1 5
5
2
1
• 𝑉 = 𝜋
22
2
− 2 −
2−1 5
5
−
12
2
− 1 −
1−1 5
5
• 𝑉 = 𝜋 2 − 2 −
1
5
−
1
2
− 1
• 𝑉 = 𝜋
3
10
=
3𝜋
10

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Method of washer

  • 1. Methods of Washers: • Let f and g be continues and nonnegative on an interval [a, b]. Let R be the region that is bounded above y = f(x), below y = g(x), and the sides by x = a and x = b. The Volume of the object revolved around the x-axis is 𝑉 = 𝑎 𝑏 𝜋 𝑓 𝑥 2 − 𝑔 𝑥 2 𝑑𝑥 • In the washer method you simply need to identify the top and bottom curve to evaluate.
  • 2. • Find the volume of the solid of revolution by rotating the region bounded by the graphs 𝑦 = 𝑥 − 1 and 𝑦 = 𝑥 − 1 2 about the x- axis • Notice that the square root function is on top and the parabola on bottom. • Notice that the functions intersect at x = 1 and x = 2 • We can know write the following volume formula: • 𝑉 = 1 2 𝜋 ∙ [ 𝑥 − 1 2 − 𝑥 − 1 2 2 𝑑𝑥
  • 3. Here is how the problem works out • 𝑉 = 1 2 𝜋 ∙ [𝑥 − 1 − 𝑥 − 1 4]𝑑𝑥 = 𝜋 0 1 [𝑥 − 1 − 𝑥 − 1 4] 𝑑𝑥 • 𝑉 = 𝜋 𝑥2 2 − 𝑥 − 𝑥−1 5 5 2 1 • 𝑉 = 𝜋 22 2 − 2 − 2−1 5 5 − 12 2 − 1 − 1−1 5 5 • 𝑉 = 𝜋 2 − 2 − 1 5 − 1 2 − 1 • 𝑉 = 𝜋 3 10 = 3𝜋 10