A hemispherical tank, placed so that the top is a circular region of radius 6ft, is filled with water to a depth of 4 ft. Find the work done in pumping the water to the top of the tank.

Ans. 50,185 lb-ft

Printable View

- May 2nd 2007, 04:17 PM^_^Engineer_Adam^_^help with work: Integral applications
A hemispherical tank, placed so that the top is a circular region of radius 6ft, is filled with water to a depth of 4 ft. Find the work done in pumping the water to the top of the tank.

Ans. 50,185 lb-ft - Oct 4th 2007, 05:06 AM^_^Engineer_Adam^_^
does anyon know how to solve this problem?

- Oct 4th 2007, 11:36 AMticbol
Work = weight * (delta height)

weight = volume * density

density of water is 62.4 lbs/cu.ft.

Hemisphere.

Diameter = 2(6)

If the origin (0,0) is the center of the circle of the hemisphere, the equation of the circle is

x^2 +y^2 = 6^2 --------------(i)

Our infinitesimal volume, dV--your rectangular element-- is a horizontal disc whose radius is x, and whose thickness is dy.

So, dV = pi(x^2) *dy

Its weight, dW, is

dW = (pi(x^2)*dy)(62.4) = (62.4pi)(x^2)dy

To raise this dW to the top of the bowl, the work done is, say, d(work).

d(work) = dW * (-y) <-----negative because the the y's below (0,0) are negative.

So,

d(work) = [(62.4pi)(x^2)dy]*(-y)

From (i), x^2 = 36 -y^2, so,

d(work) = [(62.4pi)(36 -y^2)dy]*(-y)

d(work) = (62.4pi)[y^3 -36y]dy

Because the water is 4ft deep, then the integration with dy is from y = -6 to y = -2.

Hence,

Work = (62.4pi)INT.(-6 --> -2)[y^3 -36y]dy

Work = (62.4pi)[(1/4)(y^4) -18y^2]|(-6 --> -2)

Work = (62.4pi)[{(1/4)(-2)^4 -18(-2)^2} - {(1/4)(-6)^4 -18(-6)^2}]

Work = (62.4pi)[{4 -72} - {324 -648}]

Work = (62.4pi)[4 -72 -324 +648]

Work = (62.4pi)[256]

Work = 50,185 ft-lbs. ---------------answer.