Find the work done in pumping gasoline that weighs 42 pounds per cubic foot. (Hi

Question

Find the work done in pumping gasoline that weighs 42 pounds per cubic foot. (Hint: Evaluate one integral by a geometric formula and the other by observing that the integrand is an odd function.)

26. The top of a cylindrical gasoline storage tank at a service station is 4 feet below the ground level. The axis of the tank is horizontal and its diameter and length are 5 feet and 12 feet, respectivaly. Find the work done in pumping the entire contents of the full tank to a height of 3 feet above the ground level

7 P.S. #4:

A. A torus is formed by revolving the region bounded by the circle

(x-2)^2 + y^2 = 1

about the y-axis. Use the disk method to calculate the volume of the torus.

B. Use the disk method to find the volume of the general torus if the circle has radius r and its center is R units from the axis of rotation.

These two on my homework have stumped me and I do not know where to start. Solutions would be great. Thank you!

Explanation / Answer

1) First, find the volume of a generic slab of gasoline that is h feet from the bottom of the tank and is ∆h feet thick.

We can split the cylindrical storage tank into generic rectangular prism-like slabs. The volume of each slab is:
dV = LWH = 12W∆h.

We need to find the general width of each slab in terms of h. To do this, we need to find how wide the base of the circle is at a distance of h from the bottom of the tank.

The diameter of the tank is 5 feet, so the radius is 5/2 feet. Taking the center of the circle to be at the origin, the equation is:
x^2 + y^2 = (5/2)^2 ==> x^2 + y^2 = 25/4.

Then, the width of the circle in terms of the distance, h, from the bottom of the circle is:
x^2 + (5/2 - h)^2 = 25/4 ==> x = √[25/4 - (5/2 - h)^2] = √(5h - h^2).

Then, W = 2x = 2√(5h - h^2).

Thus:
dV = 12Wh∆h
= 24√(5h - h^2)∆h.
(As a sanity check, h varies from 0 to 5, so the volume of the cylinder is:
V = ∫ 24√(5h - h^2) dh (from h=0 to 5) = 75π,
which agrees with V = πr^2*h = π(5/2)^2(12) = 75π.)

This is the volume of a generic slab of gasoline. The weight can be found by multiplying the volume by the density to yield:
dF = Vd = [24√(5h - h^2)∆h](42) = 1008√(5h - h^2)∆h.

This is also the required force to lift this slab.

Then, the generic slab of gasoline must travel up 5 - h feet to get to the top of the tank, 4 feet to get to ground level, and another 3 feet to move the gasoline 3 feet above ground level. In total, the gasoline travels (5 - h) + 4 + 3 = 12 - h feet.

Thus, the required work to lift this slab up is:
dW = Fd = 1008√(5h - h^2)(12 - h)∆h.

Therefore, since h varies from 0 to 5, the required work is:
W = ∫ 1008√(5h - h^2)(12 - h) dh (from h=0 to 5) = 29925π ft-lbs.

2)Write the equation as x = 2 (+ or -) sqrt(1-y^2) xy.

V = integral(y= -1 to 1) pi[(2 + sqrt(1-y^2))^2 - (2 + sqrt(1-y^2))^2]dy
= integral(y= -1 to 1) 8 pi sqrt(1-y^2) dy
= 8 pi * integral(y= -1 to 1) sqrt(1-y^2) dy
= 8 pi * [Area of half the unit circle]
= 8 pi * (pi * 1^2/2)
= 4 pi.pi

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.