anonymous
  • anonymous
Use geometry to evaluate https://i.gyazo.com/82a9844552783f2bbc8a49003e361671.png
Calculus1
katieb
  • katieb
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anonymous
  • anonymous
I know how to solve it but what do they mean "use geometry"
anonymous
  • anonymous
@zzr0ck3r Can you give me a hand?
anonymous
  • anonymous
@dan815 Maybe you could help?

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mathmate
  • mathmate
Hint: see drawing below: |dw:1441330734684:dw|
anonymous
  • anonymous
I don't get it @mathmate :/
anonymous
  • anonymous
anonymous
  • anonymous
You don't have to evaluate an integral. \(y=\sqrt{4-x^2}\) is the equation for the top half of a circle with radius 2, centered at the origin. That's what @mathmate drew. They only want the integral from 0 to 2, so they want the area of ¼ of the circle.|dw:1441403513050:dw|
anonymous
  • anonymous
Oh ok, that makes sense but how would I find that area?
anonymous
  • anonymous
A=πr²
anonymous
  • anonymous
Can you by any chance set it up for me? I'm more of a visual person, so I tend to retrace what someone has done to learn it.
anonymous
  • anonymous
there's nothing to set up here. You just plug in 2 into the formula for area of a circle
anonymous
  • anonymous
Why? I don't get it :/ Even if I plug in 2 for the area I'd get the radius no?
anonymous
  • anonymous
Sorry, I'm really not sure what's confusing you. This is a straight up geometry problem. If the RADIUS of a circle is 2, what's the AREA of a quarter of the circle? \[A = \pi r^2\] Plug in 2 for r to get the area of the whole circle. Then divide the area by 4 to get the area of a quarter of the circle. That's literally all there is to it. Nothing to set up or integrate. No calculus involved here other than knowing that a definite integral gives the area under a curve.
mathmate
  • mathmate
"I know how to solve it but what do they mean "use geometry" If you are not sure, you can also check your answer using geometry with your answer by integration.
anonymous
  • anonymous
You see I'm very bad with geometry, so if I solve using calculus, I'd get the same answer correct?
anonymous
  • anonymous
yes. same result either way

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