## 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?

1. anonymous

2. ganeshie8

Look at the integrand : $y = 1+\sqrt{9-x^2}$ does that equation look familiar to you ?

3. anonymous

Does it have anything to do with a circle?

4. ganeshie8

Yes

5. 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?

6. Astrophysics

You know how to graph $y = \sqrt{9-x^2}$?

7. anonymous

Not exactly :S

8. 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,)

9. Astrophysics

10. ganeshie8

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11. anonymous

Oh so its a semicircle, going from -3 to 3, ok got it

12. Astrophysics

Yes, thank god for ganeshies drawing, mine was horrible, but notice the +1 means it shifts 1 unit up vertically.

13. anonymous

Lol xD Ohhh true!

14. ganeshie8

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15. 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?

16. ganeshie8

Really ? how about simply using the area of circle formula ?

17. Astrophysics

Yeah you could just use quarter of a circle area xD

18. Astrophysics

$A = \frac{ 1 }{ 4 } \pi r^2$

19. anonymous

Ohhh yeah!! xD i was going the hard way!! True true true!

20. Astrophysics

and the area of the rectangle

21. anonymous

But then, does that still work when it has shifted a unit up?

22. ganeshie8

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