The document discusses calculating volumes of solids generated by rotating regions about axes using washers. It provides examples of setting up integrals to find volumes of regions rotated about the x-axis, y-axis, lines other than the axes, and discusses when to use washers versus discs. Example problems demonstrate setting up integrals for various regions rotated about different axes and lines, evaluating the integrals to find the volumes.

1. Volumes Using Washers
By: Melissa Barnhart

2. Volume using Washers
Finding Volume of Rotated Areas using Washers
Volume = ∑ Outer – Inner Region • thickness (∆)
V = ∫ π(R² - r²) • dx or V = ∫ π(R² - r²) • dy
Where outer radius R and the inner radius r are
functions of the variable of integration.
Integration endpoints are the same as before.
dx
dy
Area of a outer circle – inner circle
r r
R
R
f(x)
g(y)

3. Example 1
∆Volume = Area • Thickness
Area = washers (outer - inner)!
= π[(√8x)2
– (x²)²]
Thickness = ∆x
X ranges from 0 out to 2
Volume = ∫ (π(8x – x4
) dx
x = 0
x = 2
= π ∫ (8x – x4
) dx
= π (4x² - (1/5)x5
) |
= π [(16) – (32/5)]
= 48π/5 = 30.159
x = 0
x = 2
Find the volume of the solid generated by
revolving the region bounded by the
parabolas y = x2
and y2
= 8x about the x-axis.

4. Example 2
= π ∫ (4 + 2√4-y² + y²) dy
∆Volume = Area • Thickness
Area = washers (outer - inner)!
= π((1+√4-y²)2
– (1)²)
= π (4 + 2√4-y + y²)
Thickness = ∆y
y ranges from 0 up to 2
Volume = ∫ (π) (4 + 2√4-y² + y²) dy
y = 0
y = 2
y = 0
y = 2
= π (4y + 4sin-1
(y/2) + y√4-y² + ⅓y³)|
= π ((32/3 + 2π) – (0))
= π (32/3 + 2π) = 53.2495
The semicircular region bounded by the y-axis
and x = √4-y² is revolved about the line x = -1.
Setup the integral for its volume.

5. Example 3a
Volume = ∫ π (x) dx
x = 0
x = 4
= π ∫ (x) dx
= π (½x²) |
= π ((½ 16) – (0))
= π (8 – (0) = 8π
= 25.133
x = 0
x = 4
∆Volume = Area • Thickness
Area = discs (not washers)!
= πr² = π(√x)²
= π (x)
Thickness = ∆x
x ranges from 0 out to 4
2
dx
x = y²
√x = y
4
Consider the first quadrant region bounded by
y2
= x, the x-axis and x = 4. Find the volume when
the region is revolved about the x-axis.

6. Example 3b
Volume = ∫ π (16– y4
) dy
y = 0
y = 2
= π ∫ (16 - y4
) dy
= π (16y – (1/5) y5
) |
= π (32 – (32/5) – (0))
= 128π/5 = 80.425
y = 0
y = 2
∆Volume = Area • Thickness
Area = washers (outer - inner)!
= π((4)² – (y²)²)
= π (16 – y4
)
Thickness = ∆y
y ranges from 0 up to 2
2
dy
x = y²
√x = y
4
Consider the first quadrant region bounded by
y2
= x, the x-axis and x = 4. Find the volume when
the region is revolved about the y-axis.

7. Example 3c
Volume = ∫ π (32 – 12y² + y4
) dy
y = 0
y = 2
= π ∫ (32 – 12y² + y4) dy
= π (32y – 4y³ + (1/5) y5
) |
= π (64 – 32 + 32/5) – (0)
= 192π/5 = 120.64
y = 0
y = 2
∆Volume = Area • Thickness
Area = washers (outer - inner)!
= π ((6-y²)² – (2²))
= π (32 – 12y² + y4
)
Thickness = ∆y
y ranges from 0 up to 2
2
dy
x = y²
√x = y
4
6
6-y²
Consider the first quadrant region bounded by
y2
= x, the x-axis and x = 4. Find the volume when
the region is revolved about the line x = 6.

8. Example 3d
Volume = ∫ π (4√x - x) dx
x = 0
x = 4
= π ∫ (4√x - x) dx
= π (8/3)x3/2
- ½x² |
= π ((8/3)(8) – (½)(16)) – (0))
= π ((64/3) - 8) = 40π/3
= 41.888
x = 0
x = 4
∆Volume = Area • Thickness
Area = washers (outer - inner)!
= π ((2²) - (2-√x)²)
= π (4√x - x)
Thickness = ∆x
x ranges from 0 out to 4
2
dx
x = y²
√x = y
4
Consider the first quadrant region bounded by
y2
= x, the x-axis and x = 4. Find the volume when
the region is revolved about the line y = 2.

9. Another Example
Volume = ∫ π(15 - 8x² + x4
) dx
x = 0
x = √3
= π ∫ (15 - 8x² + x4
) dx
= π (15x – (8/3)x3
+ (1/5)x5
) |
= π ((15√3 – (8√3) + (9√3/5)) – (0))
= (44√3/5) π = 47.884
x = 0
x = √3
∆Volume = Area • Thickness
Area = Outer circle – inner circle = washers!
= π(R² - r²) = π[(4 - x²)² - (1)²]
= π(15 - 8x² + x4
)
Thickness = ∆x
x ranges from 0 out to √3 (x = √4-1)
Find the volume of the solid generated by revolving
the first quadrant region bounded by y = 4 – x2
, the
line y = 1, y-axis and the x-axis around the x-axis.
2dx
y = 4 – x2
x = √4-y
4
1

10. Another Example
Volume = ∫ π(15 - 8x² + x4
) dx
x = 0
x = √3
= π ∫ (15 - 8x² + x4
) dx
= π (15x – (8/3)x3
+ (1/5)x5
) |
= π ((15√3 – (8√3) + (9√3/5)) – (0))
= (44√3/5) π = 47.884
x = 0
x = √3
∆Volume = Area • Thickness
Area = Outer circle – inner circle = washers!
= π(R² - r²) = π[(4 - x²)² - (1)²]
= π(15 - 8x² + x4
)
Thickness = ∆x
x ranges from 0 out to √3 (x = √4-1)
Find the volume of the solid generated by revolving
the first quadrant region bounded by y = 4 – x2
, the
line y = 1, y-axis and the x-axis around the x-axis.
2dx
y = 4 – x2
x = √4-y
4
1