The area of the slab depends upon y; the area at height y from the base is given by the product of the square of the radius of the slice and pi, i.e., A(y) = pi*r^2. However, we need to determine r as a function of y so that the area can be expressed as a function of y. To do this, we can use similar triangles.
To determine r(y), consider a cross-sectional view of the hemisphere. This cross-section is a semicircle of radius 5. Vertical lines at heights y and 0 are similar. The radius of the slice at height y can be called r. Therefore, we have the proportion:
(5 - y) / r = 5 / 5
which simplifies to r = 5 - y.
Now we can write the area A(y) as:
A(y) = pi*(5-y)^2.
Next, we find the volume of the differential slab of height dy as:
dV = A(y)*dy.
Now, to find the work required to pump the slab of water out of the tank, we need to multiply its weight by the distance that it needs to be lifted. The weight of the slab is given by:
dW = dV * 62.5
and the distance it needs to be lifted is (5 - y).
Thus, the total work required to pump out that slab is:
dU = dW * (5 - y) = 62.5 * A(y) * dy * (5 - y).
To find the total work required to pump out all the water, we integrate dU over the interval [0, 5]:
U = integral(62.5 * pi * (5 - y)^2 * dy * (5 - y)) over [0, 5].
U = 62.5 * pi * integral((5 - y)^3 * dy) over [0, 5].
Now, we integrate and find the value of the integral:
U = 62.5 * pi * [(-1/4)*(5 - y)^4] evaluated from 0 to 5.
U = 62.5 * pi * ([-(1/4)*(0)^4] - [(-1/4)*(5)^4])
U = 62.5 * pi * (-(0) + 625/4)
U = 62.5 * 625 * (pi / 2)
U = 19531.25 * pi ft*lb.
Therefore, the work required to pump the water out of the tank is 19531.25 * pi ft*lb.
The hemispherical tank shown is full of water. Given that water weighs 62.5 lb/ft3, find the work required to pump the water out of the tank.
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What is shown is just the tank (a hemisphere) with a radius of 5 ft.
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First I calculated the Volume of the hemisphere, V = (2/3)*pi*r^3
V = (2/3)*pi*125 = (250/3)*pi
Then I took the integral of: Volume*5y*dy from 0 to 5.
Which equals: ((250/3)*pi)*(5/2)y^2 evaluated at 5 and 0.
I came up with 16362.5 ft*lb.
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Am I using the wrong method?
The work required depends upon where the water is extracted. I assume you are pumping out the top.
I don't see why you claim that the energy is
Volume*5y*dy from 0 to 5
The work required is the weight of each differential slab of height dy, multiplied by the distance it must be lifted, 5 - y, integrated from 0 to 5. The area of each slab is different. It depends upon y.
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