121616949-math.205 - π Z 1(1 √ y 2(1-√ y 2 dy π Z 4...

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9.3 Volume 191 Not only does this accomplish the task with only one integral, the integral is somewhat easier than those in the previous calculation. Things are not always so neat, but it is often the case that one of the two methods will be simpler than the other, so it is worth considering both before starting to do calculations. 0 1 2 3 0 1 2 3 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 0 1 2 3 0 1 2 3 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ........................ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 9.11 Computing volumes with “shells”. ( JA ) EXAMPLE 9.11 Suppose the area under y = - x 2 + 1 between x = 0 and x = 1 is rotated around the x -axis. Find the volume by both methods. Disk method: Z 1 0 π (1 - x 2 ) 2 dx = 8 15 π . Shell method: Z 1 0 2 πy p 1 - y dy = 8 15 π . Exercises 1. Verify that
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Unformatted text preview: π Z 1 (1 + √ y ) 2-(1-√ y ) 2 dy + π Z 4 1 (1 + √ y ) 2-( y-1) 2 = 8 3 π + 65 6 π = 27 2 π . 2. Verify that Z 3 2 πx ( x + 1-( x-1) 2 ) dx = 27 2 π . 3. Verify that Z 1 π (1-x 2 ) 2 dx = 8 15 π . 4. Verify that Z 1 2 πy p 1-y dy = 8 15 π . 5. Use integration to find the volume of the solid obtained by revolving the region bounded by x + y = 2 and the x and y axes around the x-axis. ⇒ 6. Find the volume of the solid obtained by revolving the region bounded by y = x-x 2 and the x-axis around the x-axis. ⇒...
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