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Clarence

  • one year ago

This has to be false right? Suppose we are going to consider the disk of radius r as the region bounded between the graphs of the functions √(r^2-x^2) and -√(r^2-x^2). Then the area of the disk can be expressed as the limit of Riemann sums of rectangles of length Δx and height 2√(r^2-x^2) where xi are a partition of the interval [−r,r].

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  1. anonymous
    • one year ago
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    yea its false

  2. clarence
    • one year ago
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    Think you could help me prove it? I mean, it just sounds false in my head.

  3. anonymous
    • one year ago
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    your right man it definently is false think i spelled that wrong

  4. anonymous
    • one year ago
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    Δx and height 2√(r^2-x^2) where xi are a partition of the interval the height is 3 then the check mark

  5. anonymous
    • one year ago
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    so its deffinetly false

  6. anonymous
    • one year ago
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    so if ur happy with my help click the best response

  7. phi
    • one year ago
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    it sounds true to me

  8. clarence
    • one year ago
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    Think you can prove it phi?

  9. phi
    • one year ago
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    Here is the plot of f(x)= √(r^2-x^2) , but with r=1

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  10. phi
    • one year ago
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    f(x) in that graph is the top part of a circle the other graph g(x) = - sqrt(r^2-x^2) will be the bottom half of the circle

  11. phi
    • one year ago
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    *** Then the area of the disk can be expressed as the limit of Riemann sums of rectangles of length Δx and height 2√(r^2-x^2) where xi are a partition of the interval [−r,r]. *** This means we have lots of rectangles between -1 and 1 (let r=1 in this case) |dw:1444056657537:dw| the height of each thin rectangle is the top y value (i.e. f(x) ) minus the bottom y-value g(x) the width of each rectangle is \( \Delta x \)

  12. phi
    • one year ago
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    Do you know what they mean by Riemann sum ?

  13. clarence
    • one year ago
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    Yeah, a while back some smart people walked me through it all... Actually had fun with maths that time :p

  14. phi
    • one year ago
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    When people introduce Riemann sums, it is often for the "area under the curve" |dw:1444057014048:dw|

  15. phi
    • one year ago
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    in your problem, the bottom is not the x-axis, it is g(x)= - sqrt(r^2-x^2) but it is the same idea: add up the area of lots of thin rectangles (where we let their width become infinitely thin, and we have an infinite number of rectangles) https://en.wikipedia.org/wiki/Riemann_sum

  16. phi
    • one year ago
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    Or see some of Khan's videos https://www.khanacademy.org/math/integral-calculus/indefinite-definite-integrals/riemann-sums/v/simple-riemann-approximation-using-rectangles

  17. phi
    • one year ago
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    in your problem the height of each recangle is f(x) - g(x) (as I defined them up above): \[ \sqrt{r^2 - x^2} - -\sqrt{r^2 - x^2} \\ =\sqrt{r^2 - x^2} +\sqrt{r^2 - x^2} \\ = 2\sqrt{r^2 - x^2} \]

  18. clarence
    • one year ago
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    That makes sense! Thanks!!

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