Volume2 calc II Hobart - m ath 131 application volumes of...

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math 131 application : volumes of revolution , part ii 6 6 . 2 Volumes of Revolution: The Disk Method One of the simplest applications of integration (Theorem 6 . 1 )—and the accumula- tion process—is to determine so-called volumes of revolution. In this section we will concentrate on a method known as the disk method . Solids of Revolution If a region in the plane is revolved about a line in the same plane, the resulting object is a solid of revolution , and the line is called the axis of revolution . The following situation is typical of the problems we will encounter. Solids of Revolution from Areas Under Curves. Suppose that y = f ( x ) is a contin- uous (non-negative) function on the interval [ a , b ] . Rotate the region under the f between x = a and x = b around the the x -axis and determine the volume of the resulting solid of revolution. See Figure 6 . 10 f ( x i ) ax i b . . . . . . . . ........................................................................... . . . . . . . . . . . . . . ................................ y = f ( x ) f ( x i ) i b . . . . . . . . . ............................................................................... . . . . . . . . .............................................................................. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ..... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 6 . 10 : Left: The region under the continuous curve y = f ( x ) on the interval [ a , b ] . Right: The solid generated by rotating the region about the x -axis. Note: The point ( x i , f ( x i )) on the curve traces out a circular cross- section of radius r = f ( x i ) when rotated. Once we know the cross-sectional areas of the solid, we can use Theorem 6 . 1 to determine the volume. But as Figure 6 . 10 shows, when the point ( x i , f ( x i )) on the curve is rotated about the x -axis, it forms a circular cross-section of radius R = f ( x i ) . Therefore, the cross-sectional area at x i is A ( x i )= p R 2 = p [ f ( x i )] 2 . Since f is continuous, so is p [ f ( x 2 and consequently Theorem 6 . 1 applies. Volume of Solid of Revolution = Z b a A ( x ) dx = Z b a p [ f ( x 2 . Of course, we could use this same process if we rotated the region about the y -axis and integrated along the y -axis. We gather these results together and state them as a theorem. THEOREM 6 . 2 ( The Disk Method ) . If V is the volume of the solid of revolution determined by rotating the continuous function f ( x ) on the interval [ a , b ] about the x -axis, then V = p Z b a [ f ( x )] 2 .( 6 . 2 ) If V is the volume of the solid of revolution determined by rotating the continuous function f ( y ) on the interval [ c , d ] about the y -axis, then V = p Z d c [ f ( y )] 2 dy 6 . 3 ) Another Development of the Disk Method Using Riemann Sums Instead of using Theorem 6 . 1 , we could obtain Theorem 6 . 2 directly by using the ‘subdivide and conquer’ strategy once again. Since we will use this strategy in later situations, let’s quickly go through the argument here. 1 1 There are often several ways to prove a result in mathematics. I hope one of these two will resonate with you.
math 131 application : volumes of revolution , part ii 7 As above, we start with a continuous function on [ a , b ] . This time, though, we create a regular partition of [ a , b ] using n intervals and draw the corresponding approximating rectangles of equal width D x . In left half of Figure 6 . 11 we have drawn a single representative approximating rectangle on the i th subinterval.

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