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Math Help - Double integration

  1. #1
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    Double integration

    double Integral (sin(theta)*x)dx dtheta

    0 to 1/2pi and 1-cos(theta) to 0
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  2. #2
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    Hi,

    The question is:

    \int^{\frac{\pi}{2}} _ 0 \int ^{0} _{1-cos(\theta)}sin(\theta)\times {x}.dxd\theta
    <br />
= \int ^{\frac{\pi}{2}} _ 0 \left ( \frac {x^2}{2} \times sin(\theta )\right ) |^{0} _{1-cos(\theta)}.d\theta<br />
    <br />
= \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 \left ( x^2\times sin(\theta )\right ) |^{0} _{1-cos(\theta)}.d\theta<br />
    <br />
= \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 \left (0\times sin(\theta ) - (1-cos(\theta))^2\times sin(\theta )\right ) .d\theta<br />
    <br />
= \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 \left ( - (1-2cos (\theta)+cos^2(\theta))\times sin(\theta )\right ) .d\theta<br />
    <br />
= \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0  - sin(\theta )+2cos (\theta)sin(\theta )-cos^2(\theta)sin(\theta) .d\theta<br />
    <br />
= \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0  - sin(\theta )+sin(2\theta )-\frac{1}{2}(cos(2\theta) +1)sin(\theta) .d\theta<br />
    <br />
= \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0  - sin(\theta )+sin(2\theta )-\frac{1}{2}sin(\theta)cos(2\theta) -\frac{1}{2}sin(\theta) .d\theta<br />
    <br />
= \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 - sin(\theta )+sin(2\theta )-\frac{1}{4}(sin(\theta+2\theta)+sin(\theta-2\theta)) -\frac{1}{2}sin(\theta) .d\theta<br />
    <br />
= \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 - sin(\theta )+sin(2\theta )-\frac{1}{4}sin(3\theta)-\frac{1}{4}sin(-\theta) -\frac{1}{2}sin(\theta) .d\theta<br />
    <br />
= \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 - sin(\theta )+sin(2\theta )-\frac{1}{4}sin(3\theta)+\frac{1}{4}sin(\theta) -\frac{1}{2}sin(\theta) .d\theta<br />
    <br />
= \frac {1}{2}\int ^{\frac{\pi}{2}} _ 0 - \frac {5}{4} sin(\theta )+sin(2\theta )-\frac{1}{4}sin(3\theta) .d\theta<br />
    <br />
= \frac {1}{2}\left( - \frac{5}{4}cos(\theta )+\frac{1}{2}cos(2\theta )-\frac{1}{4}\times\frac{1}{3}cos(3\theta) \right )|^\frac{\pi}{2} _0<br />
    <br />
= \frac {1}{2}\left( -\frac{5}{4} cos(\theta )+\frac{1}{2}cos(2\theta )-\frac{1}{12}cos(3\theta) \right )|^\frac{\pi}{2} _0<br />

    <br />
=\frac {1}{2}(-\frac{1}{2} +\frac{5}{4} -\frac{1}{2} +\frac{1}{12})<br />
    <br />
=\frac{1}{6}<br />

    the calculation is long forgive me if i made some mistakes.
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  3. #3
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    It takes less time by reversing integration order.
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  4. #4
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    Hello, cchyfly!

    Another approach . . .


    \int^{\frac{\pi}{2}}_0 \underbrace{\int^{1-\cos\theta}_0\!\! x\cdot\sin\theta\,dx}\,d\theta


    \int^{1-\cos\theta}_0\!\!x\cdot\sin\theta\,dx \;=\;\frac{1}{2}x^2\cdot\sin\theta\,\bigg]^{1-\cos\theta}_0 \;=\;\frac{1}{2}(1-\cos\theta)^2\sin\theta - \frac{1}{2}(0^2)\sin\theta

    . . =\;\frac{1}{2}\left(1 - 2\cos\theta + \cos^2\!\theta\right)\sin\theta \;=\;\frac{1}{2}\left(\sin\theta - 2\sin\theta\cos\theta + \cos^2\!\theta\sin\theta\right)


    Then we have: . \frac{1}{2}\int^{\frac{\pi}{2}}_0\left(\sin\theta - 2\sin\theta\cos\theta + \cos^2\theta\sin\theta\right)d\theta . = \;\frac{1}{2}\bigg[-\cos\theta - \sin^2\!\theta -\frac{1}{3}\cos^3\!\theta\,\bigg]^{\frac{\pi}{2}}_0

    . . = \;\frac{1}{2}\left(-\cos\frac{\pi}{2} - \sin^2\frac{\pi}{2} - \frac{1}{3}\cos^3\frac{\pi}{2}\right) - \frac{1}{2}\left(-\cos0 - \sin^20 - \frac{1}{3}\cos^30\right)

    . . = \;\frac{1}{2}\left(\text{- }0 - 1 - 0\right) - \frac{1}{2}\left(\text{-}1 - 0 - \frac{1}{3}\right) \;=\;\frac{1}{2}\left(-1\right) - \frac{1}{2}\left(-\frac{4}{3}\right) \;=\;-\frac{1}{3} + \frac{2}{3}

    . . = \;\frac{1}{6}

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  5. #5
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    Quote Originally Posted by cchyfly View Post
    double Integral (sin(theta)*x)dx dtheta

    0 to 1/2pi and 1-cos(theta) to 0
    <br />
\begin{aligned}<br />
   \int_{0}^{\pi /2}\!{\int_{1-\cos \theta }^{0}{x\sin \theta \,dx}\,d\theta }&=-\int_{0}^{1}\!{\int_{\arccos (1-x)}^{\pi /2}{x\sin \theta \,d\theta }\,dx} \\ <br />
 & =\int_{0}^{1}{\left( x^{2}-x \right)\,dx} \\ <br />
 & =-\frac{1}{6}. \\ <br />
\end{aligned}
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