BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 6.3, Problem 38E

a)

To determine

To find: the more wood content of the ring from two rings.

Expert Solution

Answer to Problem 38E

The both napkin rings are having a same quantity of wood.

Explanation of Solution

The volume only depends up on the height of the napkin ring.

Hence, the both napkin rings are having same quantity of wood.

b)

To determine

To check: The part (a) answer by computing the volume of a napkin ring using cylindrical shells.

Expert Solution

Answer to Problem 38E

The volume of a napkin ring is πh36_

Explanation of Solution

Given:

The height of the napkin ring is h.

The radius of the napkin ring is r.

The radius of the sphere is R.

Calculation:

Consider the equation of napkin ring as follows:

y=R2x2 (1)

The region lies between x=r and x=R.

Sketch the solid region as shown below in Figure 1.

Single Variable Calculus: Concepts and Contexts, Enhanced Edition, Chapter 6.3, Problem 38E

Refer Figure 1

Calculate the volume using the method of cylindrical shell as follows.

V=2×ab2πx[f(x)]dx (2)

Substitute r for a, R for b, and R2x2 for [f(x)] in equation (2).

V=2rR2πxR2x2dx=2πrR2xR2x2dx (3)

Consider u=R2x2 (4)

Differentiate both sides of the equation.

du=2xdxdu=2xdx

Calculate the lower limit value of u using equation (4).

Substitute r for x in equation (4).

u=R2r2

Calculate the upper limit value of u using equation (4).

Substitute R for x in equation (4).

u=R2R2=0

Apply lower and upper limits for u in equation (3).

Substitute u for (R2x2) and (du) for 2xdx dx in equation (3).

V=2πR2+r20u(du)=2πR2+r20u12du (5)

Integrate equation (5).

V=2π[u12+112+1]R2+r20=2π[u3232]R2+r20=2π[2(0)3232(R2+r2)323]=43π(R2+r2)32 (6)

Consider R2+r2=(h2)2

Substitute (h2)2 for R2+r2 in equation (6).

V=43π(h2)2×32=424πh3=16πh3

Hence, the volume of a napkin ring is πh36_.

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