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Math Help - Very easy definite integral question

  1. #1
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    Very easy definite integral question

    Evaluate the integral below by interpreting it in terms of areas. In other words, draw a picture of the region the integral represents, and find the area using high school geometry.


    I don't know why, but I'm messing this up. I do x = 3 (so y = 0) and x = -3 (so y = 3 \sqrt 2) to get a graph that looks like a triangle. Using simply geometry to find the answer, I do 6 (the base) * 3 \sqrt 2 (the height) divided by 2 to get 9 \sqrt 2. Which is wrong.

    But why?
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  2. #2
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    Quote Originally Posted by Archduke01 View Post
    Evaluate the integral below by interpreting it in terms of areas. In other words, draw a picture of the region the integral represents, and find the area using high school geometry.


    I don't know why, but I'm messing this up. I do x = 3 (so y = 0) and x = -3 (so y = 3 \sqrt 2) to get a graph that looks like a triangle. Using simply geometry to find the answer, I do 6 (the base) * 3 \sqrt 2 (the height) divided by 2 to get 9 \sqrt 2. Which is wrong.

    But why?
    the graph is not a triangle ...
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    y = \sqrt{9-x^2} \Rightarrow y^2 = 9 - x^2 \Rightarrow x^2 + y^2 = 9, \ y \geq 0

    So what is the area?
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    Would that just be half of the radius of the circle with radius 3?
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    Quote Originally Posted by ninjaku View Post
    Would that just be half of the radius of the circle with radius 3?
    huh?
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    Quote Originally Posted by ninjaku View Post
    Would that just be half of the radius of the circle with radius 3?
    Look at skeeter's graph again..
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  7. #7
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    Quote Originally Posted by ninjaku View Post
    Would that just be half of the radius of the circle with radius 3?
    I presume you mean half the area of the circle with radius 3.
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