   Chapter 5.2, Problem 69E

Chapter
Section
Textbook Problem

# A hole of radius r is bored through the middle of a cylinder of radius R > r at right angles to the cylinder. Set up, but do not evaluate, an integral for the volume cut out.

To determine

To find:

Integral for the volume of cut out

Explanation

1) Concept:

i) Solid of Revolution:

Volume of solid revolution revolving a region around a line is given by

V=abA(x)dx or V=cdA(y)dy

In case of the disk method, the radius is found in terms of x or y and use A=π·radius2

In case of the washer method, find the inner and outer radius and compute the area of washer by subtracting the area of inner disk from outer disk, use

ii) Integrals of Symmetric functions:

Suppose f is continuous on [-a, a]

a) If f is even [f-x=fx], then -aaf(x)dx=20af(x)dx

b) If f is odd [f-x=-fx], then -aaf(x)dx=0

2) Calculation:

Now from the front side, the circle appears to be in a y-z plane,

Therefore, equation of circle is y2+z2=R2

From the top side, the hole lies on a x-y plane,

Therefore, equation of circle is x2+y2=r2

From the front side, the diagram will appear as shown below:

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