Integral Calculus: Work Done by a Spring and Pumping Liquids, Slides of Differential and Integral Calculus

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Integral

Calculus

 Work done by a spring

 Work done by pumping a

liquid

Work done by a

spring

1. If the modulus of a spring is 20 lbs./in.,

what is the work required to stretch

the spring a distance of 6 inches?

F(x) = 20x

W = ∫^6 20xdx

= ½ 20 x ] = 10(6)2 - 10(0) = 10 (36)

W = 360 lbs. in.

0

(^0) 6

2. If a force of 50 lbs. stretches a 12 in. spring to 14 in., find the work done in stretching the spring from 15 in. to 17 in. To get the limits of integration 15-12 = 3

17-12 = 5

12 in 15in 17in if x = 2, F = 50

F = kx W = ∫^5 25xdx 50 = 2k = ½ 25x2 ] 35 K = 25 = 25(5)2 /2 - 25(3) 2 / = 625/2 - 225/ W = 200 lbs.-in.

3

x=0 x=3 x=

F(x)= 25x

PART (b)

Thus F(x) = 200x; to calculate the work done

in stretching the spring from 10 inches to 12 inches, use

W  (^)  200 xdx  100 x^2 ] 02  400 in  lbs

2 0

 (^) 

b a

W F ( x ) dx

PART (c)

To determine how far 1600-lb. force will

stretch the spring, we do not need to integrate.

We use F= 200x 1600 = 200x x = 8 inches

To determine the work done to stretch the

spring this far,

W 250 xdx 125 x^2 ]^10.^2180 N m 180 J

1. 2 0

 (^)     

4. A crate is pushed a distance of 15 meters.

If it is pushed with a force equvalent to 4x +

10 newtons, how much work was done to

move the crate?

SOLUTION:

 (^) 

b a

W F ( x ) dx

F ( x )  4 x  10

2 10 ] 600

15 0

2 W  x  x  Joules

Work done by pumping

Work done in Pumping

a Liquid

The total work done in lifting all or

part of the liquid in a container to

any point P above its top is

where w = weight per unit volume of

the liquid

h = distance of the element

from the

point P

dv = volume of the solid



 

 b a

b a W w hdV

W whdV

5

25

15

6 h= 6-y y

EXAMPLE

The inner surface of a tank has the form of a parabola of revolution whose axis is vertical. The depth of the tank and the diameter of the circular top are 12 cm. If the tank is originally full of water, find the work done in pumping all the water: a.To the top b. 3 cm from the top c.Suppose the tank is half-full in (a)

Thus x^^2 ^3 y , substitute in dV^ ^  x^2 dy

dV  ( 3 y ) dy

W w

W w y ydy

W hdV



 864

12 3

12 0

12 0



 







dyne-cm

a.

b. If the water is to be pumped 3 cm above its surface, the only value which will change is h; h = 15-y

W dyne cm

W w y ydy

W w hdV





_______

12 0

12 0



Thus

c. If the tank is half-full, just change the limit of (a) from 0 to 6 since the container is half-full.