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Area Interpretation to Evaluate The Integral

  1. Jun 30, 2006 #1
    From 0 to 4. ( I don't know how to make the s sign.)

    [tex] squareroot of (16-x^2)[/tex]

    I set it equal to y and squared it to get.

    [tex] y^2 = 16 - x^2[/tex]

    Then I changed it to:

    [tex] Y^2 + X^2 = 16[/tex]

    I know it's a semi circle on the x axis at 0 to 4.

    What should my final answer be?
  2. jcsd
  3. Jun 30, 2006 #2

    Well do you know how to find the area of a semi-circle with a radius of 4?
  4. Jun 30, 2006 #3
    No. But how do you know the radius is 4?

    Are ou taking the difference from the x points? Wouldn't the radius be y axis.
  5. Jun 30, 2006 #4
    No? As in you don't know how to find the area of a circle? As to how I know it has radius 4, what is the general equation of a circle?

    (x - x0)2 + (y - y0)2 = r2

    look at the equation you ahve in your original post and look at teh above, now do you see why it has a radius of 4?
  6. Jun 30, 2006 #5
    Ok Thanks.
    2 more questions

    In the equation of a circle. What do the x subzero's stand for.

    And Now I figured out the area to be 25.13 but why do they give me from 0 to 4.
  7. Jun 30, 2006 #6
    The point (x0, y0) would be the center of the circle. They give you from 0 to 4 because they want you to find the integral from 0 to 4 or the area under teh curve from 0 to 4, which in terms of this circle would actually be the area of a quarter circle.
  8. Jun 30, 2006 #7
    Ah ok, So do I assume that the center of the circle is the origin here?
  9. Jun 30, 2006 #8
    There's no need to ASSUME that it is, because it is and that's all there is to it, compare your equation to the general equation of a circle and the center is obviously the origin.
  10. Jun 30, 2006 #9
    Yeah, well my equation is not in your form.

    I don't have those x and y subzeroes.
  11. Jun 30, 2006 #10
    It more or less is, depending how picky you want to be. The general equation of a circle is as I posted above

    (x - x0)2 + (y - y0)2 = r2

    So if the center is the origin then x0 = y0 = 0
    so the equation becomes

    (x - 0)2 + (y - 0)2 = r2

    Now note that x - 0 = x and y - 0 = y so now we have

    (x)2 + (y)2 = r2

    and now removing the parentheses we have

    x2 + y2 = r2

    Which is the same form as your equation, and note that this will be the general equation for a circle of radius r centered at the origin.
  12. Jul 1, 2006 #11


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    You do understand, I hope, that you should already have learned what you are talking about- the equation of a circle, and what the numbers in it mean, how to find the area of a circle, or a half or quarter circle, before you start studying calculus!
  13. Jul 1, 2006 #12
    It's interesting that you say that because with my school's math program, you do not learn certain important things like this before taking calculus. It's because we have a ridiculously bad integrated program (Math I, Math II, Math III, and Math IV) where all areas of math are covered lightly and gradually throughout high school. Zero rigour. Zero self-teaching (our books only have questions...no answers, examples, etc.).

    I guess many kids are going to be very lost next year in my AP Calculus class...hopefully I'll be fine.
  14. Jul 1, 2006 #13


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    It's hard to believe one would study calculus before basic geometry or even analytic geometry. It looks to me like the whole point of this question is to demonstrate that you can use basic geometry formulas to do integrals. If you don't know the geometry, the whole point is lost!
  15. Jul 2, 2006 #14
    Thanks for the help d_leet.
  16. Jul 2, 2006 #15
    Your welcome, glad I could help.
  17. Jul 2, 2006 #16


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    25.13 is the area of the whole semicircle, which is the integral from -4 to 4. They only want half the semicircle, the half that lies between 0 and 4.
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