PRACTICE4 Name Section

PRACTICE4

Name

Section

1. Determine the work to empty the water from a filled semi-spherical swimming pool

with radius 10 ft.

x

-10

0

10

0

-5

-10

Dy = -y ft

Fy = 62.4(100 - y2)2y lbs

0

W=

62.4(y3 - 100y) dy

-10

=

62.4

1 4

y4

-

100 2

y2

0 -10

= 62.4(2500) ft-lbs.

2. A cable that weighs 2 lb/ft is used to lift 800 lb of coal up a mine shaft 500 ft deep.

Find the work done.

Wcoal = 800(500) ft-lbs = 400, 000 ft-lb

500

Wrope =

2(500 - y) dy

0

=

2

500y

-

1y2 2

500 0

= 250, 000 ft-lbs

Work=650,000 ft-lb

3. A spring has natural length 20 cm. If a 25-N force is required to keep it stretched to a length of 30 cm (determining k, the spring constant), how much work is required to stretch it from 20 cm to 25 cm? F = kx 25 = k(.3) k = 250/3

W=

1/4 1/5

250 3

x

dx

=

250 3

x2

1/4 1/5

=

250 3

11 16 - 25

J

4. SET UP (do not evaluate) the integral to determine the force on the trapezoidal side of the tank filled with water that has a rectangular bottom with length 12 ft and width 10 ft and rectanular top with length 20 ft and width 10 ft. The tank is 8 ft deep from bottom to top.

-10

11

10

9

8

7

6

5

4

3

2

1

0

-5

0

5

10

15

x

Equation of the line having points (6, 0) and (10, 8) is y = 2x - 12

Ay = (2)

y+2 2

y

Py = 62.4(8 - y)

8

F = 62.4(8 - y)(y + 12) dy

0

5. SET UP (do not evaluate) the integral to determine the work to empty this trapezoidal tank.

Fy = 62.4(10)(2)

y+2 2

y

Dy = (8 - y)

8

W = 624(8 - y)(y + 12) dy

0

6.

Determine

the

arclength

of

the

curve

f (x) =

x2 2

-

ln x 4

for 1 x 4.

f (x)

=

x

-

1 4x

So

f (x)2

=

x2

-

1 2

+

1 16x2

.

A.L. = = =

4 1

1

+

x2

-

1 2

+

1 16x2

dx

=

4 1

x2

+

1 2

+

1 16x2

dx

4 1

x

+

1 4x

2

dx =

4 1

x

+

1 4x

dx

1 2

x2

+

1 4

ln

x

4 1

=

8

+

1 4

ln

4

-

1 2

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download