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January 12th, 2012, 03:26 PM   #1
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Estimated Volume of a Solid Hemisphere

I wasn't in class today when we were talking about spherical approximation. The question says: To estimate the volume of a solid hemisphere of radius 4, imagine its axis of symmetry to be the interval [0,4] on the axis. Partition [0,4] into 8 sub-intervals of equal length and approximate the solid with cylinders based on the circular cross sections of the hemisphere perpendicular to the x-axis at the subintervals' left endpoints. [attachment=0:37qlb9pm]scan0034.jpg[/attachment:37qlb9pm]

(a)Find the sum of of the volumes of the cylinders. Do you expect to overestimate ? Give reasons for your answer.
(b) Express as a percentage of to the nearest percent.
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January 12th, 2012, 04:24 PM   #2
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Re: Estimated Volume of a Solid Hemisphere

What they want you to do is revolve each of the rectangles about the x-axis 180, forming a stack of cylinders, which will approximate the volume of a hemisphere. The radius of each cylinder will be equal to y(x) for x = 0.5k, where k is in {0,1,3,...,7}. Thus:



We should expect this to be an over-estimate based on the fact that all of the rectangles extend beyond the circle.





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January 12th, 2012, 07:11 PM   #3
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Re: Estimated Volume of a Solid Hemisphere

Thanks for the reply, but I'm totally in the dark. Do you know of a website or something that could give me a rundown on how this works. I can't seem to find one.
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What they want you to do is revolve each of the rectangles about the x-axis 180, forming a stack of cylinders, which will approximate the volume of a hemisphere. The radius of each cylinder will be equal to y(x) for x = 0.5k, where k is in {0,1,3,...,7}. Thus:



We should expect this to be an over-estimate based on the fact that all of the rectangles extend beyond the circle.





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January 12th, 2012, 07:49 PM   #4
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Re: Estimated Volume of a Solid Hemisphere

Let me try to explain it better...

Imagine each of the rectangles is the cross-section of a cylinder whose height is 1/2 unit.

Begin with the leftmost rectangle. It is 8 units in height (2y(0)), so the radius of that cylinder would be 4 units. The volume of a cylinder is With h = 1/2 and r = 4, its volume would be

Continuing, we see that for each cylinder, we have and for the nth cylinder, x = (n - 1)(1/2). Thus, the volume of the nth cylinder is:



Let n be the natural numbers 1-8, then add the volumes all together (you may choose to do it differently, this is how I would do it rather than manually computing 8 separate volumes to save some time):

So now we find:



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January 12th, 2012, 08:12 PM   #5
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Re: Estimated Volume of a Solid Hemisphere

Thank you! Unfortunately, I will have to do all 8 manually because the teacher is strict about things like that, but I appreciate it! To find x, you were solving for the midpoint, correct?
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January 12th, 2012, 08:15 PM   #6
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Re: Estimated Volume of a Solid Hemisphere

No, I was using the left endpoints of each sub-interval. The first sub-interval is from 0 to 1/2, so I used 0, the second sub-interval is 1/2 to 1, so I used 1/2 and so on. The last sub-interval is from 7/2 to 4 so I used 7/2. Using the mid-point would be more accurate, but trust me, you will be introduced to that method as well as several others. Approximating definite integrals is a fascinating and rich subject.
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January 12th, 2012, 08:42 PM   #7
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Re: Estimated Volume of a Solid Hemisphere

Your teacher might just be impressed with you being able to sum a sequence of squares:

(I can show you how to derive this, or prove it using induction, or both, if you're interested)

In my first post above, I wrote:



We may rewrite this as:



Using the above formula, this becomes:



(15)}{6}\)" />

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January 12th, 2012, 08:48 PM   #8
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Re: Estimated Volume of a Solid Hemisphere

I kind of already did all 8, but for the sake of knowledge, I wouldn't mind you explaining how it works.
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January 12th, 2012, 08:52 PM   #9
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Re: Estimated Volume of a Solid Hemisphere

In fact, if I could show my teacher a proof, he would definitely let me use it, so I's love to understand this
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January 12th, 2012, 09:49 PM   #10
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Re: Estimated Volume of a Solid Hemisphere

Great! First, I'll prove it using induction, which requires having the formula:



First, we check it for the base case of n = 0:

this is true, so we assume the following is true:



Now, we add to both sides:













We have derived from thereby completing the proof by induction.

Now, to derive the formula, one may begin with the non-homogeneous recurrences and complete the process of symbolic differencing until a homogeneous relation is obtained:

(1)

(2)

Subtracting (1) from (2) gives:

(3)

(4)

Subtracting (3) from (4) gives:

(5)

(6)

Subtracting (5) from (6) gives:



Now we have a homogeneous relation whose associated auxiliary equation is:





With the repeated root of r = 1 of multiplcity 4, we know then that the closed form will take the form:



We could also have gotten to this point by observing a constant difference in the 3rd degree:

n.....S_n.....?S.....?(?S).....?(?(?S))
0......0....................................
1......1........1...........................
2......5........4........3..................
3......14.......9........5...........2....
4......30.......16.......7...........2....

This tells us will be a cubic polynomial as well.

Next, we use initial conditions to determine the coefficients :

thus we know giving us the following 3X3 system:







Solving the system, we find:

and thus:

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