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Math Help - Integral

  1. #1
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    Integral

    Hello everyone, I have this procedure correct?

    I have used trigonometric substitution:

     \displaystyle\int_{0}^{2\pi}  \displaystyle\int_{0}^{a}  r^2 cos ( \phi  ) \sqrt[2 ]{a^2 -r^2} dr d\phi

    r = asen(\theta), dr = acos\theta.d\theta

    And replacing the integral I is:

    cos(\phi). a^2.sen^2(\theta) \sqrt[ ]{a^2 - (asen.\theta)^2} .acos\theta.d\theta

    Factoring a^2 and taking square root:

    cos(\phi). a^2.sen^2(\theta).a. \sqrt[ ]{1 - sen^2\theta} .acos\theta.d\theta

    = cos(\phi).a^4.sen^2(\theta).cos^2(\theta).d\theta


    A greeting and thank you very much!
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by Dogod11 View Post
    Hello everyone, I have this procedure correct?

    I have used trigonometric substitution:

     \displaystyle\int_{0}^{2\pi} \displaystyle\int_{0}^{a} r^2 cos ( \phi ) \sqrt[2 ]{a^2 -r^2} dr d\phi

    r = asen(\theta), dr = acos\theta.d\theta

    And replacing the integral I is:

    cos(\phi). a^2.sen^2(\theta) \sqrt[ ]{a^2 - (asen.\theta)^2} .acos\theta.d\theta

    Factoring a^2 and taking square root:

    cos(\phi). a^2.sen^2(\theta).a. \sqrt[ ]{1 - sen^2\theta} .acos\theta.d\theta

    = cos(\phi).a^4.sen^2(\theta).cos^2(\theta).d\theta


    A greeting and thank you very much!
    If I remember correctly sin(x)=sen(x) if so

    Use the identity

    \sin^2(\theta)=1-\cos^2(\theta)

    Then use \cos^2(\theta)=\frac{1}{2}(1+\cos(2\theta))

    Three times to expand out

    (cos^2(\theta))^2

    I hope this helps
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  3. #3
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    Smile

    If I remember correctly sin(x)=sen(x) if so
    Hello, you are right, sorry, my natural language is not English.

    But I've been very helpful, also confirmed that it was rather the beginning,

    Thanks again,

    Greetings
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