121616949-math.204 - 190 Chapter 9 Applications of...

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190 Chapter 9 Applications of Integration the horn is Z 1 0 π ( x 2 ) 2 dx = Z 1 0 πx 4 dx = π 1 5 , so the desired volume is π/ 3 - π/ 5 = 2 π/ 15. As with the area between curves, there is an alternate approach that computes the desired volume “all at once” by approximating the volume of the actual solid. We can approximate the volume of a slice of the solid with a washer-shaped volume, as indicated in figure 9.10 . The volume of such a washer is the area of the face times the thickness. The thickness, as usual, is Δ x , while the area of the face is the area of the outer circle minus the area of the inner circle, say πR 2 - πr 2 . In the present example, at a particular x i , the radius R is x i and r is x 2 i . Hence, the whole volume is Z 1 0 πx 2 - πx 4 dx = π x 3 3 - x 5 5 ¶fl fl fl fl 1 0 = π 1 3 - 1 5 = 2 π 15 . Of course, what we have done here is exactly the same calculation as before, except we have in effect recomputed the volume of the outer cone.
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