Daniel Yim & Drew Salisbury & Greg Wenrich
MTH 234.002
Calculus II
Porject

In[1]:=

<< Graphics`

Our function for this tower

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f[x_] := (x - 35.78)^2

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Cubic Feet:

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Pi * Integrate[f[x], {x, 0, 35.78}]

Out[28]=

47967.8

Gallons we can hold

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V[x_] = (Pi * Integrate[f[x], {x, 0, 35.78}]) * 7.48  //N

Out[29]=

358799.

Surface area

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SurfaceArea := 2 * Pi * Integrate[f[x] * (1 + f '[x]^2)^(1/2), {x, 0, 35.78}] //N

SurfaceArea = SurfaceArea + Pi * 35.78^2

Out[31]=

5.15389*10^6

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Inverse function

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InverseFunct[x_] := x^(1/2) + 35.78

Work required to fill up the tower

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7.48 * 8

Out[32]=

59.84

59.84 pounds in a cubic foot of water

Integrate[59.84 * f[x], {x , 0, 2665.03}]

Out[33]=

3.62549*10^11

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Integrate[59.84 * InverseFunct[x] * 7.48 * (2665.03 - x), {x , 0, 2665.03}]

Out[39]=

1.00637*10^11

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We define our 50,000 ft interval markers for the tower

We multiply the inverse of the function by two because we are only using the positive leg of the function

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Markers[x_]    := Integrate[2 * InverseFunct[x], {x, 0 , x}]

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Solve[Markers[x] == 50000, x]

Out[13]=

{{x→494.084}}

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Solve[Markers[x] == 100000, x]

Out[14]=

{{x→896.928}}

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Solve[Markers[x] == 150000, x]

Out[15]=

{{x→1261.41}}

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Solve[Markers[x] == 200000, x]

Out[16]=

{{x→1601.13}}

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Solve[Markers[x] == 250000, x]

Out[17]=

{{x→1922.71}}

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Solve[Markers[x] == 300000, x]

Out[18]=

{{x→2230.07}}

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Solve[Markers[x] == 350000, x]

Out[19]=

{{x→2525.8}}

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The height of our tower, in feet

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Solve[Markers[x] == 374149., x]

Out[21]=

{{x→2665.03}}

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In[36]:=

SurfaceOfRevolution[f[x], {x, 0, 35.78}]

[Graphics:HTMLFiles/index_40.gif]

Out[36]=

-Graphics3D -

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