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January 12th, 2012, 02:26 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 $S\small{8}$ of the volumes of the cylinders. Do you expect $S\small{8}$ to overestimate $V$? Give reasons for your answer.
(b) Express $|V\,-\,S\small{8}|$ as a percentage of $V$ to the nearest percent.
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 January 12th, 2012, 03:24 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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: $S_8=\frac{\pi}{8}\sum_{k=0}^7$$64-k^2$$=\frac{93\pi}{2}$ We should expect this to be an over-estimate based on the fact that all of the rectangles extend beyond the circle. $V=\frac{2}{3}\pi(4)^3=\frac{128\pi}{3}$ $S_8-V=\frac{23\pi}{6}$ $\frac{S_8-V}{V}\cdot100\approx9%$
January 12th, 2012, 06:11 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.
Quote:
 Originally Posted by MarkFL 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: $S_8=\frac{\pi}{8}\sum_{k=0}^7$$64-k^2$$=\frac{93\pi}{2}$ We should expect this to be an over-estimate based on the fact that all of the rectangles extend beyond the circle. $V=\frac{2}{3}\pi(4)^3=\frac{128\pi}{3}$ $S_8-V=\frac{23\pi}{6}$ $\frac{S_8-V}{V}\cdot100\approx9%$

 January 12th, 2012, 06:49 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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 $\pi r^2h$ With h = 1/2 and r = 4, its volume would be $\pi(4)^2$$\frac{1}{2}$$=8\pi$ Continuing, we see that for each cylinder, we have $r^2=16-x^2$ and for the nth cylinder, x = (n - 1)(1/2). Thus, the volume of the nth cylinder is: $C_n=\frac{\pi}{2}$$16-\(\frac{n-1}{2}$$^2\)=\frac{\pi}{8}$$64-(n-1)^2$$$ 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: $S_8=\sum_{k=1}^8C_k=\sum_{k=1}^8$$\frac{\pi}{8}\(6 4-(n-1)^2$$\)=\frac{\pi}{8}\sum_{k=1}^8$$64-(n-1)^2$$=$ $\frac{\pi}{8}$$8^3-\sum_{k=1}^8(n-1)^2$$=\frac{\pi}{8}$$8^3-140$$=\frac{372\pi}{8}=\frac{93\pi}{2}$
 January 12th, 2012, 07:12 PM #5 Member   Joined: Aug 2011 From: Alabama Posts: 93 Thanks: 0 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?
 January 12th, 2012, 07:15 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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.
 January 12th, 2012, 07:42 PM #7 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Estimated Volume of a Solid Hemisphere Your teacher might just be impressed with you being able to sum a sequence of squares: $S_n=\sum_{k=0}^nk^2=\frac{n(n+1)(2n+1)}{6}$ (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: $S_8=\frac{\pi}{8}\sum_{k=0}^7$$64-k^2$$$ We may rewrite this as: $S_8=\frac{\pi}{8}$$64\cdot8-\sum_{k=0}^7k^2$$$ Using the above formula, this becomes: $S_8=\frac{\pi}{8}$$8^3-\frac{7(7+1)(2\cdot7+1)}{6}$$$ $S_8=\frac{\pi}{8}$$8^3-\frac{7((15)}{6}$$" /> $S_8=\frac{\pi}{8}$$8^3-7(4)(5)$$=\frac{\pi}{8}$$512-140$$=\frac{\pi}{8}$$372$$=\frac{93\pi}{2}$
 January 12th, 2012, 07:48 PM #8 Member   Joined: Aug 2011 From: Alabama Posts: 93 Thanks: 0 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.
 January 12th, 2012, 07:52 PM #9 Member   Joined: Aug 2011 From: Alabama Posts: 93 Thanks: 0 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
 January 12th, 2012, 08:49 PM #10 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Estimated Volume of a Solid Hemisphere Great! First, I'll prove it using induction, which requires having the formula: $S_n=\sum_{k=0}^nk^2=\frac{n(n+1)(2n+1)}{6}$ First, we check it for the base case of n = 0: $S_0=\sum_{k=0}^0k^2=\frac{0(0+1)(2\cdot0+1)}{6}=0$ this is true, so we assume the following is true: $S_n=\sum_{k=0}^nk^2=\frac{n(n+1)(2n+1)}{6}$ Now, we add $(n+1)^2$ to both sides: $\sum_{k=0}^nk^2+(n+1)^2=\frac{n(n+1)(2n+1)}{6}+(n+ 1)^2$ $\sum_{k=0}^{n+1}k^2=\frac{n(n+1)(2n+1)+6(n+1)^2}{6 }$ $\sum_{k=0}^{n+1}k^2=\frac{(n+1)$$n(2n+1)+6(n+1)$$} {6}$ $\sum_{k=0}^{n+1}k^2=\frac{(n+1)$$2n^2+7n+6$$}{6}$ $\sum_{k=0}^{n+1}k^2=\frac{(n+1)(n+2)(2n+3)}{6}$ $S_{n+1}=\frac{(n+1)$$(n+1)+1)(2(n+1)+1)}{6}$ We have derived $S_{n+1}$ from $S_n$ 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) $S_{n+1}=S_n+(n+1)^2$ (2) $S_{n+2}=S_{n+1}+(n+2)^2$ Subtracting (1) from (2) gives: (3) $S_{n+2}=2S_{n+1}-S_{n}+2n+3$ (4) $S_{n+3}=2S_{n+2}-S_{n+1}+2(n+1)+3$ Subtracting (3) from (4) gives: (5) $S_{n+3}=3S_{n+2}-3S_{n+1}+S_{n}+2$ (6) $S_{n+4}=3S_{n+3}-3S_{n+2}+S_{n+1}+2$ Subtracting (5) from (6) gives: $S_{n+4}=4S_{n+3}-6S_{n+2}+4S_{n+1}-S_{n}$ Now we have a homogeneous relation whose associated auxiliary equation is: $r^4-4r^3+6r^2-4r+1=0$ $(r-1)^4=0$ With the repeated root of r = 1 of multiplcity 4, we know then that the closed form will take the form: $S_n=k_0+k_1n+k_2n^2+k_3n^3$ 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 $S_n$ will be a cubic polynomial as well. Next, we use initial conditions to determine the coefficients $k_i$: $S_0=k_0=0$ thus we know $k_0=0$ giving us the following 3X3 system: $S_1=k_1+k_2+k_3=1$ $S_2=2k_1+4k_2+8k_3=5$ $S_3=3k_1+9k_2+27k_3=14$ Solving the system, we find: $k_1=\frac{1}{6},\,k_2=\frac{1}{2},\,k_3=\frac{1}{3 }$ and thus: $S_n=\frac{1}{6}n+\frac{1}{2}n^2+\frac{1}{3}n^3=\fr ac{n+3n^2+2n^3}{6}=\frac{n\(1+3n+2n^2$$}{6}=\frac{ n(n+1)(2n+1)}{6}$

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