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
