5.2 Volumes of Revolution: Disk and Washer Methods

5.2 Volumes of Revolution: Disk and Washer Methods We learned how to find the area under a curve. Now, given a curve, we form a 3-dimensional object: we use y-coordinate on the curve (the value of the function that defines the curve) as a radius, and rotate the curve about x-axis. This forms a solid around x-axis, radius of which depends on x. Our question is how to find the volume of this solid between the crossections at x=a and x=b. Approach: split the volume into disks of a small thickness and then sum the volumes of the disks by integration. The volume of the disk (cylinder) is (radius)2(thickness). The radius is the value of the function y or f(x). The thickness is the analog of the base of the rectangle in the computation of the area under a curve. It directed along x-axis, so denote it as dx. 1 Hence, the volume of the disk is 2 dV f ( x) dx To calculate the volume between x=a and x=b, we integrate over this interval: b b b 2 2

V dV f ( x) dx f ( x) dx. a a a The latter two formulas constitute the Disk Method Note: In most of the cases, drawing a sketch of the functions and the typical element is needed to visualize geometry of the object and validate the formula. It is much safer to take all the steps we did than just use the formula! 2 Example: A region is bounded by the curves y x 4 , x 2, x 2, x - axis. Find the volume of the solid of revolution obtained by rotating this region about x-axis. Solution: Draw the region and mark a typical element a thin rectangle stretched perpendicular to the axis of revolution. The radius of the solid is equal to the value of x4. The thickness is dx Hence, the volume of the disk is 4 2 dV x dx and the volume between x=-2 and x=2 is 2 2 2 9

9 9 2 x 2 2 29 210 4 8 2 V x dx x dx . 9 9 9 9 2 2 2 9 3

Example: Find the volume of the solid of revolution obtained by rotating the region defined by the given bounds about the y-axis. y x 2 , y 6 (first quadrant). Solution: We rotate the figure about the y-axis, so chose a thin horizontal rectangle as a typical element so that it forms a disk when rotates. The thickness of the rectangle is dy, and the length is the horizontal distance between the y-axis and the function y x 2. This distance is equal to y. The volume of the disk is dV 2 y dy ydy and the volume is 6 6 2 6 y 62 V ydy ydy 0 18. 2 0 2

0 0 Note: Whenever we have rotation about the y-axis, we need to express the length of the typical element as a function of y. 4 Washer Method If the region is bounded by two curves f(x) and g(x) above the axis about which we rotate it, the result of such rotation is going to be a tube of variable radius and variable wall thickness. In this case, rotation of the typical element forms not a disk, but a washer. Analogous to the previous case, the volume of the washer is 2 2 2 2 dV f ( x) dx g ( x) dx f ( x) g ( x) dx and the volume of the solid can be computed as b

b 2 2 V dV f ( x) g ( x) dx. a a 5 Example: Find the volume of the solid of revolution obtained by rotating the region defined by the given bounds about the x-axis. y x3 / 2 , y x 2 . Solution: Since we rotate the figure about x-axis, a typical element is a rectangle stretched in the vertical direction. The limits of integration are determined by the intersections of the curves: x3 / 2 x 2 x 3 x 4 x 4 x3 0 x3 ( x 1) 0 x 0, x 1. Between these points x3/2>x2, so that in the formula we take f(x)= x3/2 g(x)= x2 dV x 3 x 4 dx 1

and 4 5 1 x x 1 1 0 . V ( x x )dx 4 5 20 4 5 0 0 3 4 6 Homework: Section 5.2: 1,7,13,15,25,27,29. 7

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