Volume of a Solid of Revolution
See how to find the volume of a solid of revolution
(disc method, washer method).
5 examples and their solutions.
Solid of Revolution
Definition
formed by rotating a 2D figure (line, curve ...)
around an axis.
Disc Method
Formula
V = ∫abπy2 dx
= ∫abπy2 dx
Cross sectional area S(x): πy2
The shape of πy2 is a disc. (= circle)
→ Disc method
Volume from its Cross Sectional Area
Area of a Circle
V = ∫abπx2 dy
Example
A region is bounded by
y = x4, x = 1, and the x-axis.
If the region is rotated around the x-axis,
find the volume of the solid of revolution.
Solution y = x4, x = 1, and the x-axis.
If the region is rotated around the x-axis,
find the volume of the solid of revolution.
V = ∫01π⋅(x4)2 dx - [1]
= π∫01x8 dx - [2]
= π[19x9]01 - [3]
= π9[x9]01
= π9[19 - 09]
= π9[1 - 0]
= π9
[1]
Close
Example
A region is bounded by
y = x4, y = 1, and the y-axis.
If the region is rotated around the y-axis,
find the volume of the solid of revolution.
Solution y = x4, y = 1, and the y-axis.
If the region is rotated around the y-axis,
find the volume of the solid of revolution.
V = ∫01π⋅(y14)2 dy - [1]
= π∫01y12 dy
= π[23x32]01
= 2π3[x32]01
= 2π3(132 - 032)
= 2π3(1 - 0)
= 2π3
[1]
Close
Example
Solution
V = ∫-rrπ(r2 - x2) dx - [2]
= π∫-rr(r2 - x2) dx
= π⋅2∫0r(r2 - x2) dx - [3]
= 2π[r2x - 13x3]0r
= 2π[r2⋅r - 13r3 - [r2⋅0 - 1303]]
= 2π[r3 - 13r3 - [0 - 0]]
= 2π[3r33 - r33]
= 2π⋅2r33
= 43πr3
Close
V = ∫-rrπ(r2 - x2) dx - [2]
= π∫-rr(r2 - x2) dx
= π⋅2∫0r(r2 - x2) dx - [3]
= 2π[r2x - 13x3]0r
= 2π[r2⋅r - 13r3 - [r2⋅0 - 1303]]
= 2π[r3 - 13r3 - [0 - 0]]
= 2π[3r33 - r33]
= 2π⋅2r33
= 43πr3
[1]
[2]
x2 + y2 = r2
y2 = r2 - x2
→ S(x) = πy2
= π(r2 - x2)
y2 = r2 - x2
→ S(x) = πy2
= π(r2 - x2)
[3]
Close
Washer Method
Formula
V = ∫abπ(y1)2 dx - ∫abπ(y2)2 dx
The shape is a washer: ⭗
→ Washer method
Example
A region is bounded by
y = ex, the tangent of y = ex at (1, e), and the y-axis.
If the region is rotated around the x-axis,
find the volume of the solid of revolution.
Solution y = ex, the tangent of y = ex at (1, e), and the y-axis.
If the region is rotated around the x-axis,
find the volume of the solid of revolution.
f(x) = ex
f'(x) = ex - [1]
f'(1) = e1
= e
(1, e)
y = e(x - 1) + e - [2]
= ex - e + e
y = ex
V = ∫01π(ex)2 dx - ∫01π(ex)2 dx - [3]
= π∫01e2x dx - π∫01e2x2 dx - [4]
= π∫01e2x dx - πe2∫01x2 dx
= π[12e2x]01 - πe2[13x3]01 - [5]
= π2[e2x]01 - πe23[x3]01
= π2[e2⋅1 - e2⋅0] - πe23[13 - 03]
= π2[e2 - e0] - πe23[1 - 0]
= π2[e2 - 1] - πe23
= πe22 - π2 - πe23
= 3πe26 - 3π6 - 2πe26
= 3πe2 - 3π - 2πe26
= πe2 - 3π6
= π(e2 - 3)6
f'(x) = ex - [1]
f'(1) = e1
= e
(1, e)
y = e(x - 1) + e - [2]
= ex - e + e
y = ex
V = ∫01π(ex)2 dx - ∫01π(ex)2 dx - [3]
= π∫01e2x dx - π∫01e2x2 dx - [4]
= π∫01e2x dx - πe2∫01x2 dx
= π[12e2x]01 - πe2[13x3]01 - [5]
= π2[e2x]01 - πe23[x3]01
= π2[e2⋅1 - e2⋅0] - πe23[13 - 03]
= π2[e2 - e0] - πe23[1 - 0]
= π2[e2 - 1] - πe23
= πe22 - π2 - πe23
= 3πe26 - 3π6 - 2πe26
= 3πe2 - 3π - 2πe26
= πe2 - 3π6
= π(e2 - 3)6
[2]
Slope of the tangent line: e
Tangent line passes through (1, e).
→ Tangent line: y = e(x - 1) + e
Equation of a Tangent Line (Derivative)
Tangent line passes through (1, e).
→ Tangent line: y = e(x - 1) + e
Equation of a Tangent Line (Derivative)
[3]
Outer function: y = ex
Inner function: y = ex
Inner function: y = ex
Close
Example
A region is bounded by
y = 2√x - 1, the tangent of the function at (2, 2), and the x-axis.
If the region is rotated around the x-axis,
find the volume of the solid of revolution.
Solution y = 2√x - 1, the tangent of the function at (2, 2), and the x-axis.
If the region is rotated around the x-axis,
find the volume of the solid of revolution.
f(x) = 2(x - 1)12
f'(x) = 2⋅12⋅(x - 1)-12⋅1 - [1]
= (x - 1)-12
f'(2) = (2 - 1)-12
= 1-12
= 1
(2, 2)
y = 1(x - 2) + 2 - [2]
y = x
V = ∫02πx2 dx - ∫12π⋅[2(x - 1)12]2 dx - [3]
= π∫02x2 dx - π∫124(x - 1) dx
= π∫02x2 dx - 4π∫12(x - 1) dx
= π[13x3]02 - 4π[12x2 - x]12
= π[13⋅23 - 13⋅03] - 4π[12⋅22 - 2 - [12⋅12 - 1]]
= π[13⋅8 - 0] - 4π[12⋅4 - 2 - [12⋅1 - 1]]
= π⋅83 - 4π[2 - 2 - [12 - 1]]
= 8π3 - 4π[-[12 - 22]]
= 8π3 + 4π[-12]
= 8π3 - 2π
= 8π3 - 6π3
= 2π3
f'(x) = 2⋅12⋅(x - 1)-12⋅1 - [1]
= (x - 1)-12
f'(2) = (2 - 1)-12
= 1-12
= 1
(2, 2)
y = 1(x - 2) + 2 - [2]
y = x
V = ∫02πx2 dx - ∫12π⋅[2(x - 1)12]2 dx - [3]
= π∫02x2 dx - π∫124(x - 1) dx
= π∫02x2 dx - 4π∫12(x - 1) dx
= π[13x3]02 - 4π[12x2 - x]12
= π[13⋅23 - 13⋅03] - 4π[12⋅22 - 2 - [12⋅12 - 1]]
= π[13⋅8 - 0] - 4π[12⋅4 - 2 - [12⋅1 - 1]]
= π⋅83 - 4π[2 - 2 - [12 - 1]]
= 8π3 - 4π[-[12 - 22]]
= 8π3 + 4π[-12]
= 8π3 - 2π
= 8π3 - 6π3
= 2π3
[2]
Slope of the tangent line: 1
Tangent line passes through (2, 2).
→ Tangent line: y = 1(x - 2) + 2
Tangent line passes through (2, 2).
→ Tangent line: y = 1(x - 2) + 2
[3]
Outer function: y = x
Inner function: y = 2(x - 1)1/2
Inner function: y = 2(x - 1)1/2
Close