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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 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.
Attached Images scan0034.jpg (103.1 KB, 1401 views) 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: We should expect this to be an over-estimate based on the fact that all of the rectangles extend beyond the circle. 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: We should expect this to be an over-estimate based on the fact that all of the rectangles extend beyond the circle. 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 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: 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: (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}\)" /> 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: 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: Tags estimated, hemisphere, solid, volume Search tags for this page
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