INTEGRAL CALCULUS 3

VOLUME OF SOLID REVOLUTION

CIRCULAR DISK METHOD
CIRCULAR DISK METHOD
CYLINDRICAL SHELL METHOD
CYLINDRICAL SHELL METHOD
SECOND THEOREM OF PAPPUS
The volume V of a solid of revolution generated by the revolution about an
external axis is equal to the product of the Area and the distance traveled by the
geometric centroid

𝑉 = 𝐴(2𝜋𝑟)

Solid Plane Area Radius Volume

Cone Right Triangle 1 1 1
ℎ𝑟 𝑟 𝜋𝑟 ℎ
2 3 3
Cylinder Rectangle ℎ𝑟 1 𝜋𝑟 ℎ
2𝑟
Sphere Semicircle 1 4𝑟 4
𝜋𝑟 𝜋𝑟
2 3𝜋 3
QUESTION 1
 The area bounded by the curve y = sin x from x = 0 to x =  is revolved about
the x-axis. What is the volume generated?

𝐵𝑦 𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑟 𝑑𝑖𝑠𝑘 𝑚𝑒𝑡ℎ𝑜𝑑 (𝑟𝑒𝑐𝑜𝑚𝑚𝑒𝑛𝑑𝑒𝑑):

𝑉= 𝜋 𝑦 − 𝑦 𝑑𝑥

𝑦
𝑉= 𝜋 sin 𝑥 − 0 𝑑𝑥
𝑑𝑥 𝜋 1 − cos 2𝑥
𝑉= 𝜋 𝑑𝑥
2

𝑥 sin 2𝑥
𝑉= 𝜋 − = 4.935
2 4
QUESTION 1
 The area bounded by the curve y = sin x from x = 0 to x =  is revolved about
the x-axis. What is the volume generated?

𝐵𝑦 𝑐𝑦𝑙𝑖𝑛𝑑𝑟𝑖𝑐𝑎𝑙 𝑠ℎ𝑒𝑙𝑙 𝑚𝑒𝑡ℎ𝑜𝑑 (𝑛𝑜𝑡 𝑟𝑒𝑐𝑜𝑚𝑚𝑒𝑛𝑑𝑒𝑑):

𝑉 = 2𝜋 𝑦 𝑥 − 𝑥 𝑑𝑦
𝑑𝑦 𝑥
𝜋
𝑦 𝑉 = 2(2𝜋 𝑦 − arcsin(𝑦) 𝑑𝑦)
2
𝜋 𝜋
2 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑖𝑜𝑛 𝑏𝑦 𝑝𝑎𝑟𝑡𝑠? ? 𝑇_𝑇
𝜋𝑦 1
𝑉 = 4𝜋 − 𝑦 1 − 𝑦 + 2𝑦 − 1 arcsin 𝑦
2 4

𝑉 = 4.935
QUESTION 1
 The area bounded by the curve y = sin x from x = 0 to x =  is revolved about
the x-axis. What is the volume generated?

𝐵𝑦 𝑠𝑒𝑐𝑜𝑛𝑑 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑜𝑓 𝑝𝑎𝑝𝑝𝑢𝑠 (ℎ𝑖𝑔ℎ𝑙𝑦 𝑟𝑒𝑐𝑜𝑚𝑚𝑒𝑛𝑑𝑒𝑑):

𝑉 = 𝐴(2𝜋𝑟)

𝑦
𝑦 𝑉= 𝑦𝑑𝑥 (2𝜋)( ) =𝜋 𝑦 𝑑𝑥
𝑦 2
2
𝑑𝑥 𝑉= 𝜋 sin 𝑥 𝑑𝑥
𝜋

1 − cos 2𝑥
𝑉= 𝜋 𝑑𝑥
2

𝑥 sin 2𝑥
𝑉= 𝜋 − = 4.935
2 4
Question 2
 The area bounded by y2 = x3 and x = 4 is revolved about the line x = 4. Find the volume generated.

X 0 1 2 3 4
Y 0 ±1 ±2√2 ±3√3 ±8
Question 2
 The area bounded by y2 = x3 and x = 4 is revolved about the line x = 4. Find the volume generated.

𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 ℎ𝑎𝑙𝑓 𝑎𝑟𝑒𝑎:

𝐵𝑦 𝑠𝑒𝑐𝑜𝑛𝑑 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑜𝑓 𝑝𝑎𝑝𝑝𝑢𝑠 (𝑟𝑒𝑐𝑜𝑚𝑚𝑒𝑛𝑑𝑒𝑑):

𝑉 = 𝐴(2𝜋𝑟)
(4,8)
𝑉
= 𝑦𝑑𝑥 (2𝜋)(4 − 𝑥) = 2𝜋 𝑦(4 − 𝑥)𝑑𝑥
4−𝑥 2
𝑥
𝑉
= 2𝜋 𝑥 4 − 𝑥 𝑑𝑥 = 91.91
𝑦 2
𝑉 8 2
= 2𝜋 𝑥 − 𝑥 = 91.91
𝑑𝑥 2 5 7

𝑆𝑡𝑟𝑖𝑝 𝑃𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝐴𝑥𝑖𝑠 𝑜𝑓 𝑅𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑉 = 183.82

Question 2
 The area bounded by y2 = x3 and x = 4 is revolved about the line x = 4. Find the volume generated.

𝐶𝑜𝑛𝑠𝑖𝑑𝑒𝑟 ℎ𝑎𝑙𝑓 𝑎𝑟𝑒𝑎: 𝐵𝑦 𝑠𝑒𝑐𝑜𝑛𝑑 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑜𝑓 𝑝𝑎𝑝𝑝𝑢𝑠 (𝑟𝑒𝑐𝑜𝑚𝑚𝑒𝑛𝑑𝑒𝑑):

𝑉 = 𝐴(2𝜋𝑟)

(4,8) 𝑉 2𝜋 4 − 𝑥
= (4 − 𝑥)𝑑𝑦 =𝜋 4 − 𝑥 𝑑𝑦
2 2
4−𝑥 𝑉
=𝜋 4−𝑦 𝑑𝑦 = 91.91
2 2
𝑑𝑦 𝑉
4−𝑥 =𝜋 16 − 8𝑦 + 𝑦 𝑑𝑦
2
𝑉 3 3
= 𝜋 16𝑦 − 8 ∗ 𝑦 + 𝑦 = 91.91
𝑆𝑡𝑟𝑖𝑝 𝑃𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝐴𝑥𝑖𝑠 𝑜𝑓 𝑅𝑒𝑣𝑜𝑙𝑢𝑡𝑖𝑜𝑛 2 5 7
𝑉 = 183.82
Question 3
 The area bounded by x2 = 8y, the lines x = 8 and y = 0 is revolved about the line x = 16.
Compute the volume generated.
𝐵𝑦 𝑠𝑒𝑐𝑜𝑛𝑑 𝑡ℎ𝑒𝑜𝑟𝑒𝑚 𝑜𝑓 𝑝𝑎𝑝𝑝𝑢𝑠 (𝑟𝑒𝑐𝑜𝑚𝑚𝑒𝑛𝑑𝑒𝑑):

(8,8) 𝑉 = 𝐴(2𝜋𝑟)

𝑉= 𝑦𝑑𝑥 (2𝜋)(16 − 𝑥) = 2𝜋 𝑦(16 − 𝑥)𝑑𝑥

𝑑𝑥 1
𝑥 𝑉 = 2𝜋 𝑥 16 − 𝑥 𝑑𝑥 = 1340.41
8
16 − 𝑥

𝑦 𝜋 16𝑥 𝑥
𝑉= − = 1340.41
4 3 4
SURFACE AREA OF REVOLUTION
𝑑𝑦
𝑑𝑥
𝑑𝑠 = 𝑑𝑥 + 𝑑𝑦

𝑑𝑠 𝑑𝑥 𝑑𝑦
= +
𝑑𝑦 𝑑𝑦 𝑑𝑦

𝑑𝑠 𝑑𝑥
= +1
𝑑𝑦 𝑑𝑦

𝑑𝑥
𝑑𝑠 = + 1 𝑑𝑦
𝑑𝑦
Question 4
 Find the surface area generated by rotating the first quadrant portion of the
curve x2 = 16  8y about the y-axis.

𝑆𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝐹𝑜𝑟𝑚: 𝑑𝑦
𝐴 = 2𝜋 𝑟𝑑𝑠
𝑑𝑠 = 1+ 𝑑𝑥
1 𝑑𝑥
𝑦−2=− 𝑥
8
1
𝐴 = 2𝜋 𝑥 1+ − 𝑥 𝑑𝑥 = 61.27
4

𝑟 =𝑥
𝑥 𝑥 𝑥
𝐴 = 2𝜋 𝑥 1 + 𝑑𝑥 𝑢 =1+ 𝑑𝑢 = 𝑑𝑥
16 16 8

2
𝐴 = 2𝜋 8 𝑢𝑑𝑢 𝐴 = 16𝜋 𝑢
(4,0) 3

𝐴 = 61.27