Volumes by Disks and Shells

The area of the disk in Figure 2 is ?x2. The disk has thickness dy and volume dV = ?x2dy.
The volume V of the cauldron is
a
V = ?x2 dy (substitute y = x2)
?0a 2 ?y ?a ?a2
V = ?y dy = ? ? =
2 0 2 0
1
Lecture 22 18.01 Fall 2006
If a = 1 meter, then V = ?
a2 gives
2
V = ?
m3 = ? (100 cm)3 = ? 106 cm3 ? 1600 liters (a huge cauldron)
2 2 2
Warning about units.
If a = 100 cm, then
V = ? (100)2 = ? 104 cm3 = ? 10 ? 16 liters
2 2 2
But 100cm = 1m. Why is this answer different? The resolution of this paradox is hiding in the
equation.
y = x2
At the top, 100 = x2 =? x = 10 cm. So the second cauldron looks like Figure 3. By contrast, when
100 cm
20 cm
Figure 3: The skinny cauldron.
a = 1m, the top is ten times wider: 1 = x2 or x = 1 m. Our equation, y = x2, is not scale-invariant.
The shape described depends on the units used.
Method 2: Shells
This really should be called the cylinder method.
x
y
x
a
a
Figure 4: x = radius of cylinder. Thickness of cylinder = dx. Height of cylinder = a ? y = a ? x2.
2
? ?
Lecture 22 18.01 Fall 2006
The thin shell/cylinder has height a ? x2, circumference 2?x, and thickness dx.
dV = (a ? x2)(2?x)dx
? x=
?
a ? ?
a
V = (a ? x2)(2?x)dx = 2? (ax ? x3)dx
x=?0 0
2 4? ? ? 2 2? ? 2?
x x ?
?
a a a a ?a2
= 2? a 2 ? 4
?
0
= 2? 2 ? 4
= 2? 4
=
2
(same as before)
Example 2. The boiling cauldron
Now, let’s fill this cauldron with water, and light a fire under it to get the water to boil (at 100oC).
Let’s say it’s a cold day: the temperature of the air outside the cauldron is 0oC. How much energy
does it take to boil this water, i.e. to raise the water’s temperature from 0oC to 100oC? Assume the
y
x
70oC
100oC
Figure 5: The boiling cauldron (y = a = 1 meter.)
temperature decreases linearly between the top and the bottom (y = 0) of the cauldron:
T = 100 ? 30y (degrees Celsius)
Use the method of disks, because the water’s temperature is constant over each horizontal disk