Consider the solid object that is obtained when the function: y = 6 (cos(x) - 5) 11T is rotated by 2π radians about the x-axis between the limits x = 9π and x = Find the volume of this object. V = Enter your answer to 3 decimal places.

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Question
Consider the solid object that is obtained when the function:
y = 6 (cos(x) - 5)
is rotated by 2π radians about the x-axis between the limits x =
Find the volume of this object.
V
=
Enter your answer to 3 decimal places.
9π and x = 11π
Transcribed Image Text:Consider the solid object that is obtained when the function: y = 6 (cos(x) - 5) is rotated by 2π radians about the x-axis between the limits x = Find the volume of this object. V = Enter your answer to 3 decimal places. 9π and x = 11π
Expert Solution
Step 1: Introduction:

Given: The function y equals 6 left parenthesis cos left parenthesis x right parenthesis minus 5 right parenthesis.

To find the volume of the solid obtained when the function is rotated about the x-axis between the limits x equals 9 pi and x equals 11 pi.

Concept Used:

The volume of the solid formed by revolving y equals f left parenthesis x right parenthesis about the x-axis between x equals a and x equals b is V equals pi integral subscript a superscript b left square bracket f left parenthesis x right parenthesis right square bracket squared d x.

Algebraic Identityleft parenthesis a plus b right parenthesis squared equals a squared plus b squared plus 2 a b

Trigonometric Identities cos squared left parenthesis x right parenthesis equals fraction numerator 1 plus cos left parenthesis 2 x right parenthesis over denominator 2 end fraction

                                      sin left parenthesis n straight pi right parenthesis equals 0 for any natural number n

Integration formulaintegral cos left parenthesis x right parenthesis d x equals sin left parenthesis x right parenthesis 

                                   integral x to the power of n equals fraction numerator x to the power of n plus 1 end exponent over denominator n plus 1 end fraction.

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