- #1
Gotcha
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Can anyone please direct me in the right way on working out the approximate area of a semi-circle with equation y = (r^2 - x^2)^0.5, by using a Riemann Sum
How to Approximate the Area of a Semi-Circle Using Riemann Sums
- Context: Undergrad
- Thread starter Gotcha
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- Riemann Riemann sum Sum
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Discussion Overview
The discussion revolves around approximating the area of a semi-circle using Riemann sums, focusing on the mathematical formulation and partitioning of the domain. Participants explore the application of Riemann sums to this specific geometric problem.
Discussion Character
- Mathematical reasoning, Technical explanation, Homework-related
Main Points Raised
- One participant requests guidance on using Riemann sums to approximate the area of a semi-circle defined by the equation y = (r^2 - x^2)^0.5.
- Another participant discusses the general formula for Riemann sums and suggests modifying the coordinate system to center at the left end of the semi-circle.
- A participant proposes finding the domain of the function as [-3, 3], creating an arbitrary partition, and evaluating the Riemann sum using selected points in that domain.
- One participant argues that all values of the partition should be positive to avoid potential problems, while another questions the significance of this approach.
- A participant mentions using a graphing calculator program to assist with the calculations, offering it to others.
- Another participant clarifies that they do not have access to a graphing calculator, making the offered software irrelevant to them.
Areas of Agreement / Disagreement
There is no consensus on the best approach to partitioning the domain for the Riemann sum, as participants express differing opinions on whether to use only positive values. The discussion remains unresolved regarding the implications of these choices.
Contextual Notes
Participants have not fully explored the implications of their partitioning choices or the potential effects on the accuracy of the Riemann sum approximation.
- #2
dextercioby
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What do you know about Riemann sums (meaning their general formula for the case of simple definite integrals)...?
Chose a system of coordinates with the center at the left end of the semicircle,so that the Ox axis in its posotive part to comprise entire diameter.Therefore your equation for the curve will be slightly modified.
Daniel.
Chose a system of coordinates with the center at the left end of the semicircle,so that the Ox axis in its posotive part to comprise entire diameter.Therefore your equation for the curve will be slightly modified.
Daniel.
- #3
quasar987
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Here's what I suggest..
First find the domain of that function ([-3,3]). Then create an arbitrary partition of that domain. That is to say, select an arbitrary number of points in that domain and label them [itex]\{x_0, x_1,...x_n\}[/itex]. (x0 has to be -3 and xn has to be 3). In principle, the more points you chose, the better the approximation.
Then construct and evaluate the Riemann sum
[tex]\Sigma_{i=1}^{n} y(t_i)(x_{i}-x_{i-1})[/tex]
Where the ti are arbitrarily chosen points in the interval [itex][x_{i-1},x_{i}][/tex][/itex]
First find the domain of that function ([-3,3]). Then create an arbitrary partition of that domain. That is to say, select an arbitrary number of points in that domain and label them [itex]\{x_0, x_1,...x_n\}[/itex]. (x0 has to be -3 and xn has to be 3). In principle, the more points you chose, the better the approximation.
Then construct and evaluate the Riemann sum
[tex]\Sigma_{i=1}^{n} y(t_i)(x_{i}-x_{i-1})[/tex]
Where the ti are arbitrarily chosen points in the interval [itex][x_{i-1},x_{i}][/tex][/itex]
- #4
dextercioby
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Yes,but i suggested him that all values of the partition be positive,the way you took'em half are and half are not...I think that should create some avoidable problems...
Daniel.
Daniel.
- #5
quasar987
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Really? Like what? I don't see what the difference will be, since [itex]x_{i}-x_{i-1}[/itex] will be positive anyway.
- #6
Aki
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I got a cool program on my graphing calculator that does that for me. It's handy when it comes to test. If you want it, just pm me.
- #7
dextercioby
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I don't know who you were talking to,but if,by chance,u were talking to me,learn that i don't have a graphing computer and that's why the software would be totally useless...
Thanks for the offer,though...
Daniel.
Thanks for the offer,though...
Daniel.
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