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anonymous

  • one year ago

For this question, does it mean that I can use the Fundamental Theory of Calculus? Or is this question looking for a different kind of solution?

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  1. anonymous
    • one year ago
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  2. ganeshie8
    • one year ago
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    Look at the integrand : \[y = 1+\sqrt{9-x^2}\] does that equation look familiar to you ?

  3. anonymous
    • one year ago
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    Does it have anything to do with a circle?

  4. ganeshie8
    • one year ago
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    Yes

  5. anonymous
    • one year ago
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    I've sort of forgotten it... but it does remind me of some stuff :D what am I supposed to do with it then?

  6. Astrophysics
    • one year ago
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    You know how to graph \[y = \sqrt{9-x^2}\]?

  7. anonymous
    • one year ago
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    Not exactly :S

  8. Astrophysics
    • one year ago
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    \[y = \sqrt{r^2-x^2}\] this is the top of a circle right c(0,0) so your radius is \[y=\sqrt{3^2-x^2}\] with centre (0,)

  9. Astrophysics
    • one year ago
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    radius = 3

  10. ganeshie8
    • one year ago
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    |dw:1437032388863:dw|

  11. anonymous
    • one year ago
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    Oh so its a semicircle, going from -3 to 3, ok got it

  12. Astrophysics
    • one year ago
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    Yes, thank god for ganeshies drawing, mine was horrible, but notice the +1 means it shifts 1 unit up vertically.

  13. anonymous
    • one year ago
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    Lol xD Ohhh true!

  14. ganeshie8
    • one year ago
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    |dw:1437032757654:dw|

  15. anonymous
    • one year ago
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    Thats a great drawing! Thanks so much @ganeshie8! So now I can use the idea of summing the area of rectangles to find the area under the curve from -3 to 0, right?

  16. ganeshie8
    • one year ago
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    Really ? how about simply using the area of circle formula ?

  17. Astrophysics
    • one year ago
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    Yeah you could just use quarter of a circle area xD

  18. Astrophysics
    • one year ago
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    \[A = \frac{ 1 }{ 4 } \pi r^2\]

  19. anonymous
    • one year ago
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    Ohhh yeah!! xD i was going the hard way!! True true true!

  20. Astrophysics
    • one year ago
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    and the area of the rectangle

  21. anonymous
    • one year ago
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    But then, does that still work when it has shifted a unit up?

  22. ganeshie8
    • one year ago
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    |dw:1437033158066:dw|