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Math Help - Polar Coordinate 02

  1. #1
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    Polar Coordinate 02

    Find the area enclose by one loop of the curve
    r= 2 Cos[x] - Sec[x]
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by camherokid View Post
    Find the area enclose by one loop of the curve
    r= 2 Cos[x] - Sec[x]
    We need the limits of integration first. since the loop goes through the origin, we need to find the values of x for which the curve passes through the origin, so set r = 0

    we have, 2 \cos x - \frac {1}{ \cos x} = 0

    \Rightarrow 2 \cos^2 x - 1 = 0

    \Rightarrow \cos x = \pm \frac {1}{ \sqrt { 2 }}

    \Rightarrow x = \frac {\pi}{4}, \frac {3 \pi}{4}, \frac { 5\pi}{4}, \frac {7 \pi}{4}

    looking at the graph, we see we want the region between \frac {\pi}{4} and \frac {7 \pi}{4}

    we want to go anticlockwise from \frac {7 \pi}{4} to \frac {\pi}{4} to get the desired area, but we must go from a smaller to a larger angle. so rewrite \frac {7 \pi}{4} as - \frac { \pi}{4} and we are in business

    So our desired area is given by

    \int_{- \pi / 4}^{ \pi / 4}\frac {1}{2} r^2 ~dx

    = \int_{- \pi / 4}^{ \pi / 4}\frac {1}{2} \left( 2 \cos x - \sec x \right)^2 ~dx
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  3. #3
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    That looks good, Jhevon. You could also just write it as:

    \int_{0}^{\frac{\pi}{4}}(2cos\theta-sec\theta)^{2}d{\theta}

    Same thing.
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by galactus View Post
    That looks good, Jhevon. You could also just write it as:

    \int_{0}^{\frac{\pi}{4}}(2cos\theta-sec\theta)^{2}d{\theta}

    Same thing.
    yes, that is true. thanks
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