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anonymous

  • 5 years ago

How do I show that the area of a circle is pi r^2 with integration?

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  1. amistre64
    • 5 years ago
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    you intefrate the circumference :)

  2. amistre64
    • 5 years ago
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    integrate even lol

  3. amistre64
    • 5 years ago
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    2pi r from 0 to radius of circle

  4. anonymous
    • 5 years ago
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    integrate \[\sqrt{r ^{2}-x ^{2}}\] from 0 to r . Multiply by 4

  5. amistre64
    • 5 years ago
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    what that amounts to is adding up all the circumferences from 0 to the radius whih = area of the circle

  6. amistre64
    • 5 years ago
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    \[\int_{0}^{r}2\pi.r dr\] \[2\pi \int_{0}^{r}r.dr\] \[2\pi \frac{r^2}{2}-2\pi \frac{0}{2}=\pi r^2\]

  7. anonymous
    • 5 years ago
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    Thank you soo much!!!!

  8. amistre64
    • 5 years ago
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    youre welcome :) i accidently discovered that when I tried finding the volume of a solid the wrong way lol

  9. watchmath
    • 5 years ago
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    Hi, Amistre I think what you want to say is \[\int_0^{2\pi}r\,dr\]

  10. anonymous
    • 5 years ago
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    watchmath I integrated your formula, but I got 2 pi^2

  11. watchmath
    • 5 years ago
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    Ah sorry, it should be double integral \(\int_0^{2\pi}\int_0^r r\,drd\theta\) But I guess you haven't learn double integral

  12. anonymous
    • 5 years ago
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    not yet

  13. watchmath
    • 5 years ago
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    I think the more traditional one is the following \[2\int_{-r}^r\sqrt{r^2-x^2}\,dx \] But to compute this integral you need to use the trigonometric substitution.

  14. anonymous
    • 5 years ago
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    r is a constant right?

  15. amistre64
    • 5 years ago
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    i typed it as i see it :) \[2\pi \int_{0}^{r}r.dr\] its the shell method for area instead of volume

  16. amistre64
    • 5 years ago
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    area is the sum of all the circumferences of circles from 0 radius to full radius

  17. amistre64
    • 5 years ago
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    top r might be better expreesed as x tho

  18. watchmath
    • 5 years ago
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    I see that know amistre :D

  19. amistre64
    • 5 years ago
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    \[\frac{2\pi r^2}{2}|_{0}^{x}\]

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