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Math Help - integral need help !

  1. #1
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    Question integral need help !

    the integral (from 0 to a) of sqrt(a^2 -r^2) 2r dr.

    Thank you very much.
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  2. #2
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    <br />
\int_0^a \ {\sqrt{a^2-r^2} 2r dr} = \frac{2}{3}a^3<br />

    I might be wrong though
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  3. #3
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    Good work. That's correct.
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  4. #4
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    hi nertil1 ,

    Thanks. Could you please show how you come up with this answer? Thank you very much.
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  5. #5
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    Quote Originally Posted by kittycat View Post
    hi nertil1 ,

    Thanks. Could you please show how you come up with this answer? Thank you very much.
    Use the substitution t=a^2-r^2
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  6. #6
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    Let  r = a\sin \theta .
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  7. #7
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    2\int_{0}^{a}r\sqrt{a^{2}-r^{2}}dr

    Let r=asin{\theta}, \;\ dr=acos{\theta}d{\theta}

    2\int_{0}^{a}{asin{\theta}\sqrt{a^{2}(1-sin^{2}{\theta})}acos{\theta}}d{\theta}

    Remember, 1-sin^{2}{\theta}=cos^{2}{\theta}

    This leads to:

    2a^{3}\int_{0}^{a}{sin{\theta}cos^{2}{\theta}}d{\t  heta}

    -2a^{3}\frac{cos^{3}{\theta}}{3}

    But, \theta=sin^{-1}(\frac{r}{a})

    Sub back in and get:

    \frac{2(r^{2}-a^{2})\sqrt{a^{2}-r^{2}}}{3}

    Sub in the limits of integration and get:

    0-\frac{2a^{3}}{3}=\frac{2a^{3}}{3}
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  8. #8
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    Why use such a scary substitution? The derivative already appears on the outside.
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