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Mth 174
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Volume of Water in a Cone
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This is a final test problem:
A conical tower has base radius 8 meters and altitude 7.3 meters. It is filled with water to a depth of 3.6 meters. Find the volume of the water.
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wi/16 = (7.3 - hi)/ 7.3
wi = 16 - 2.19 hi
ri = 8 - 1.095 hi
Volume of slice = pi ri^2 delta h = pi(8-1.095hi)^2 delta h
Volume of water in cone = integral ( pi(8-1.095hi)^2 dh, h, 0, 3.6)
= 425.629 m^3
Is this the correct solution?
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The volume of the typical increment will be
pi(8-1.095h_i)^2 delta h
as you say.
Summing all such increments you get
sum(pi(8-1.095h_i)^2 `dh, i from 1 to n),
where h_i goes from 0 for i = 1 to 3.6 for i = n.
In the limit as your `dh intervals approach zero, h_i becomes just h and `dh becomes dh, so the integral you approache is
integra( pi(8-1.095h)^2 dh, h, 0, 3.6).
This is substantially what you got, except for a couple of notational errors.
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You appear to be in good shape with this.
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