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Math Help - Calculus III: Area and Arc Length in Polar Coordinates I

  1. #1
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    Calculus III: Area and Arc Length in Polar Coordinates I

    Hi! I attempted to solve the following problems and I would really appreciate it if someone checks them and show me the exacts steps so I know where I went wrong. ( I forgot how to find the limits, please show me how) Thanks!

    1) Write the Integral that represents the area of the shaded region shown in the figure (please see attachment) of the following equation.

    r= 1- cos 2θ

    My Answer: 1/2 ∫(1-cos2θ)^2 dθ with limits pi/2, 2pi (pure guess)

    2) Find the area of the region by finding the limits and evaluating the integral
    of the following equations:

    a) One petal of r= cos 5θ
    b) Interior of r= 1- sin θ

    My answers:

    a) limits: 0, pi/5
    integral equation: 1/2 ∫ (cos 5θ)^2 dθ
    area: pi/20 or 0.157

    b) limits (these were a pure guess) : pi/6 , 5pi/6
    integral equation: 1/2 ∫ (1- sin θ)^2 dθ
    area: pi/6 or 0.523
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  2. #2
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    One petal of rose, r=cos(5{\theta}):

    Multiply by two because one petal is symmetric about the x-axis.

    \int_{0}^{\frac{\pi}{10}}cos^{2}(5{\theta})d{\thet  a}
    Last edited by galactus; November 24th 2008 at 06:39 AM.
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  3. #3
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    how do you find the limits by hand? Does that mean my answers are incorrect? If it is then then how would you solve it?
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  4. #4
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    For the rose, I just used:

    cos(5{\theta})=0

    5{\theta}=\frac{\pi}{2}

    {\theta}=\frac{\pi}{10}

    Your answer to that one is correct. I just showed another way.

    For the other, it would appear you should integrate over the whole region. 0 to 2Pi.

    \frac{1}{2}\int_{0}^{2\pi}(1-sin({\theta}))^{2}d{\theta}=\frac{3\pi}{2}
    Last edited by galactus; November 24th 2008 at 06:39 AM.
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