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Math Help - average height of function

  1. #1
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    average height of function

    find the average height of y=(2/3)*(sqrt (9 - x^2) ) over the interval [-3,3]
    i used the average height formula but i get the answer as 0 while the correct answer is supposed to be Pi/2
    can someone help me out?
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  2. #2
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    \frac{1}{3-(-3)} \int_{-3}^3 \frac{2}{3} \sqrt{9 - x^2} \, dx<br />

    \frac{1}{9} \int_{-3}^3 \sqrt{9 - x^2} \, dx

    \frac{1}{9} \cdot \frac{\pi \cdot 3^2}{2}
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  3. #3
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    how did you integrate to get to the last step?
    i think thats where im erring
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  4. #4
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    you don't need calculus to integrate, only geometry ... the graph of y = \sqrt{9 - x^2} from -3 to 3 is a semicircle of radius 3.
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  5. #5
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    oh i see
    but is it possible to use calculus and find the area because thats what the question wants i think
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  6. #6
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    possible, but a bit more involved.

    use the trig substitution x = 3\sin{\theta}
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