volumes calculas 2 shell method, Summaries of Calculus

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• MATH 154 Week

Goals

• (^) Volume of a cylinder.
• (^) Volume as a definite integral.
• (^) Volumes by slicing and rotation about an axis.
• (^) Disk by disk method.

Volumes We start with a simple type of solid called a cylinder (or, more precisely, a right cylinder ). As illustrated in the Figure, a cylinder is bounded by a plane region B 1 , called the base , and a congruent region B 2 in a parallel plane. The cylinder consists of all points on line segments that are perpendicular to the base and join B 1 to B 2

. If the area of the base is A and the height of the cylinder (the distance from B 1 to B 2 ) is h , then the volume V of the cylinder is defined as V = Ah.

Calculation of volumes

Volumes Let A ( x ) be the area of the cross-section of S in a plane P x perpendicular to the x -axis and passing through the point x , where a  x  b. (See figure. Think of slicing S with a knife through x and computing the area of this slice.) The cross-sectional area A ( x ) will vary as x increases from a to b.

Volumes Let’s divide S into n “slabs” of equal width  x by using the planes P x 1

, P

x 2 ,... to slice the solid. (Think of slicing a loaf of bread.) If we choose sample points x i  in [ x i – 1 , x i ], we can approximate the i th slab S i (the part of S that lies between the planes P xi – 1 and P xi ) by a cylinder with base area A ( x i ) and “height”  x.

Volumes This approximation appears to become better and better as n . (Think of the slices as becoming thinner and thinner.) Therefore, we define the volume as the limit of these sums as n . But we recognize the limit of Riemann sums as a definite integral and so we have the following definition.

Volumes When we use the volume formula it is important to remember that A ( x ) is the area of a moving cross-section obtained by slicing through x perpendicular to the x -axis. Notice that, for a cylinder, the cross-sectional area is constant: A ( x ) = A for all x. So, our definition of volume gives ; this agrees with the formula V = Ah.

Volumes If the cross-section is a washer, we find the inner radius r in and outer radius r out from a sketch (as in Figure 10) and compute the area of the washer by subtracting the area of the inner disk from the area of the outer disk: A =  (outer radius) 2

•  (inner radius) 2 Figure 10

Example 1 Show that the volume of a sphere of radius r is. Solution: If we place the sphere so that its center is at the origin, then the plane P x intersects the sphere in a circle whose radius (from the Pythagorean Theorem) is So, the cross-sectional area is A ( x ) =  y 2 =  ( r 2

• x 2 )

Volumes Figure 5 illustrates the definition of volume when the solid is a sphere with radius r = 1. From the result of Example 1, we know that the volume of the sphere is , which is approximately 4.18879. Approximating the volume of a sphere with radius 1

Volumes Here the slabs are circular cylinders, or disks , and the three parts of Figure show the geometric interpretations of the Riemann sums when n = 5, 10, and 20 if we choose the sample points x i  to be the midpoints Notice that as we increase the number of approximating cylinders, the corresponding Riemann sums become closer to the true volume.

Example

Example