lec22 - .01 Fall 2006 Lecture 22 Volumes by Disks and...

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Unformatted text preview: Lecture 22 18.01 Fall 2006 Lecture 22: Volumes by Disks and Shells Disks and Shells We will illustrate the 2 methods of finding volume through an example. Example 1. A witch’s cauldron y x 2 Figure 1: y = x rotated around the y-axis. Method 1: Disks y x thickness of dy a Figure 2: Volume by Disks for the Witch’s Cauldron problem. The area of the disk in Figure 2 is πx 2 . The disk has thickness dy and volume dV = πx 2 dy . The volume V of the cauldron is a V = πx 2 dy (substitute y = x 2 ) a 2 a y πa 2 V = πy dy = π = 2 2 1 Lecture 22 18.01 Fall 2006 If a = 1 meter, then V = π a 2 gives 2 V = π m 3 = π (100 cm) 3 = π 10 6 cm 3 ≈ 1600 liters (a huge cauldron) 2 2 2 Warning about units. If a = 100 cm, then V = π (100) 2 = π 10 4 cm 3 = π 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 = x 2 At the top, 100 = x 2 = x = 10 cm. So the second cauldron looks like Figure 3. By contrast, when ⇒...
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lec22 - .01 Fall 2006 Lecture 22 Volumes by Disks and...

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