Disk Method

3 Key excerpts on "Disk Method"

• 

  eBook - ePub

  Introduction to Calculus

  • Joan Van Glabek(Author)
  • 2011(Publication Date)
  • Collins Reference

    (Publisher)

  .

    Using vertical slices for both integrals, we have: 6.2 VOLUME USING DISKS AND WASHERS In this section, we use the slicing technique to find the volume of various solids. SOLIDS OF REVOLUTION

  A solid of revolution is a solid formed by revolving a plane region about a line. Hold up a piece of paper, revolve it around one edge, and the resulting right circular cylinder is a solid of revolution.

  The Disk Method

  One method for finding the volume of a solid of revolution is called the Disk Method. After a solid has been formed, imagine slicing through it perpendicular to its axis of revolution. If the slice is a solid disk (think of a CD without a hole), the Disk Method is an appropriate choice. The formula is based on the formula for the area of a circle, A = r 2 .

  Procedure 1: If the function is revolved about y = 0, set up the integral

  Procedure 2: If the function is revolved about x = 0, solve the given function for x = g (y ) and set up the integral

  EXAMPLE 6.5

  Find the volume of the solid formed by revolving the region bounded by f (x ) = 4 – x 2 , x = 0, y = 0 about the x -axis.

  SOLUTION 6.5

  Because the equation of the x -axis is y = 0, we use Procedure 1. We find the limits of integration by sketching the plane region f (x ) = 4 – x 2 and determining the x -coordinates where the disks begin and end.

  Because the region is bounded by x = 0 and intersects the x -axis at x = 2, we set up the following integral:

  EXAMPLE 6.6

  Find the volume of the solid formed by revolving the region bounded by f (x ) = 4 – x 2 , x = 0, y = 0 about the y -axis.

  SOLUTION 6.6

  Note that this is the same plane region as in Example 6.5, but now it is revolved about x = 0 (the y -axis). The slices are now dy

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  eBook - PDF

  Calculus of a Single Variable: Early Transcendental Functions, International Metric Edition

  • Ron Larson, Bruce Edwards(Authors)
  • 2018(Publication Date)
  • Cengage Learning EMEA

    (Publisher)

  This portion of the volume could have been determined without using calculus. TECHNOLOGY Some graphing utilities have the capability of generating (or have built-in software capable of generating) a solid of revolution. If you have access to such a utility, use it to graph some of the solids of revolution described in this section. For instance, the solid in Example 4 might appear like that shown in Figure 7.22. Generated by Mathematica Figure 7.22 7 .2 Volume: The Disk Method 459 Manufacturing See LarsonCalculus.com for an interactive version of this type of example. A manufacturer drills a hole through the center of a metal sphere of radius 5 centimeters, as shown in Figure 7.23(a). The hole has a radius of 3 centimeters. What is the volume of the resulting metal ring? Solution You can imagine the ring to be generated by a segment of the circle whose equation is x 2 + y 2 = 25, as shown in Figure 7.23(b). Because the radius of the hole is 3 centimeters, you can let y = 3 and solve the equation x 2 + y 2 = 25 to determine that the limits of integration are x = ±4. So, the inner and outer radii are r(x) = 3 and R(x) = radical.alt225 - x 2 , and the volume is V = π integral.alt1 b a ([R(x)] 2 - [r(x)] 2 ) dx = π integral.alt1 4 -4 [(radical.alt225 - x 2 ) 2 - (3) 2 ] dx = π integral.alt1 4 -4 (16 - x 2 ) dx = π bracketleft.alt2 16x - x 3 3 bracketright.alt2 4 -4 = 256π 3 cubic centimeters. Solids with Known Cross Sections With the Disk Method, you can find the volume of a solid having a circular cross section whose area is A = π R 2 . This method can be generalized to solids of any shape, as long as you know a formula for the area of an arbitrary cross section. Some common cross sections are squares, rectangles, triangles, semicircles, and trapezoids. VOLUMES OF SOLIDS WITH KNOWN CROSS SECTIONS 1. For cross sections of area A(x) taken perpendicular to the x-axis, Volume = integral.alt1 b a A(x) dx.

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  eBook - ePub

  Elementary Calculus

  An Infinitesimal Approach

  • H. Jerome Keisler(Author)
  • 2013(Publication Date)
  • Dover Publications

    (Publisher)

  Figure 6.2.1 ).

  Figure 6.2.1

  We shall work with the region under a curve and the region between two curves. We use one method for rotating about the axis of the independent variable and another for rotating about the axis of the dependent variable.

  For areas our starting point was the formula

  for the area of a rectangle. For volumes of a solid of revolution our starting point is the usual formula for the volume of a right circular cylinder (Figure 6.2.2 ).

  Figure 6.2.2

  DEFINITION

  The volume of a right circular cylinder with height h and base of radius r is

  DISC METHOD: For rotations about the axis of the independent variable.

  Let us first consider the region under a curve. Let R be the region under a curve y = f(x) from x = a to x = b, shown in Figure 6.2.3 (a). x is the independent

  Figure 6.2.3

  variable in this case. To keep R in the first quadrant we assume 0 ≤ a < b and 0 ≤ f(x). Rotate R about the x-axis, generating the solid of revolution S shown in Figure 6.2.3 (b).

  This volume is given by the formula below.

  VOLUME BY DISC METHOD .

  To justify this formula we slice the region R into vertical strips of infinitesimal width ∆x. This slices the solid S into discs of infinitesimal thickness ∆x. Each disc is almost a cylinder of height ∆x whose base is a circle of radius f(x) (Figure 6.2.4 ). Therefore

  Then by the Infinite Sum Theorem we get the desired formula

  Figure 6.2.4

  EXAMPLE 1 Find the volume of a right circular cone with height h and base of radius r.

  It is convenient to center the cone on the x-axis with its vertex at the origin as shown in Figure 6.2.5 . This cone is the solid generated by rotating about the x-axis the triangular region R under the line y = (r/h)x, 0 ≤ x ≤ h.

  Figure 6.2.5

  Since x

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