anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
ganeshie8
  • ganeshie8
Look at the integrand : \[y = 1+\sqrt{9-x^2}\] does that equation look familiar to you ?
anonymous
  • anonymous
Does it have anything to do with a circle?

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ganeshie8
  • ganeshie8
Yes
anonymous
  • anonymous
I've sort of forgotten it... but it does remind me of some stuff :D what am I supposed to do with it then?
Astrophysics
  • Astrophysics
You know how to graph \[y = \sqrt{9-x^2}\]?
anonymous
  • anonymous
Not exactly :S
Astrophysics
  • Astrophysics
\[y = \sqrt{r^2-x^2}\] this is the top of a circle right c(0,0) so your radius is \[y=\sqrt{3^2-x^2}\] with centre (0,)
Astrophysics
  • Astrophysics
radius = 3
ganeshie8
  • ganeshie8
|dw:1437032388863:dw|
anonymous
  • anonymous
Oh so its a semicircle, going from -3 to 3, ok got it
Astrophysics
  • Astrophysics
Yes, thank god for ganeshies drawing, mine was horrible, but notice the +1 means it shifts 1 unit up vertically.
anonymous
  • anonymous
Lol xD Ohhh true!
ganeshie8
  • ganeshie8
|dw:1437032757654:dw|
anonymous
  • anonymous
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?
ganeshie8
  • ganeshie8
Really ? how about simply using the area of circle formula ?
Astrophysics
  • Astrophysics
Yeah you could just use quarter of a circle area xD
Astrophysics
  • Astrophysics
\[A = \frac{ 1 }{ 4 } \pi r^2\]
anonymous
  • anonymous
Ohhh yeah!! xD i was going the hard way!! True true true!
Astrophysics
  • Astrophysics
and the area of the rectangle
anonymous
  • anonymous
But then, does that still work when it has shifted a unit up?
ganeshie8
  • ganeshie8
|dw:1437033158066:dw|