   Chapter 8.4, Problem 70E

Chapter
Section
Textbook Problem

# Fluid Force Evaluate the following two integrals, which yield the fluid forces given in Example 6.(a) F inside = 48 ∫ − 1 0.8 ( 0.8 − y ) ( 2 ) 1 − y 2 d y (b) F outside = 64 ∫ − 1 0.4 ( 0.4 − y ) ( 2 ) 1 − y 2 d y

(a)

To determine

To calculate: The fluid forces with the help of the following integral Finside=4810.8(0.8y)(2)1y2dy.

Explanation

Given:

The force inside the barrel is

Finside=4810.8(0.8y)(2)1y2dy.

Formula used:

a2y2dy=12(ya2y2+(a)2arcsin(ya))+C

Calculation:

Consider the following integral

Finside=4810.8(0.8y)(2)1y2dy;

Finside=4810.8(0.8y)(2)1y2dy=96[0.810.81y2dy10.8y1y2dy]

Now, simplifying the integral 10.8y1y2dy;

10.8y1y2dy=1210.82y1y2dy

Substitute 1y2=t and 2ydy=dt to get;

10.8y1y2dy=1210.8tdt10

(b)

To determine

To calculate: The fluid forces with the help of the following integral Foutside=6410.4(0.4y)(2)1y2dy.

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