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itzmashy

  • 2 years ago

a. find the area enclosed in the inner loop of the limacon, r=1+2costheta. b. find hte area inside the limacon but outside the inner loop

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  1. eliassaab
    • 2 years ago
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    b) \[2 \int_0^{\frac{2 \pi }{3}} \frac{1}{2} (2 \cos (\theta )+1)^2 \, d\theta=\frac{3 \sqrt{3}}{2}+2 \pi \]

  2. eliassaab
    • 2 years ago
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    a)\[2 \int_{\frac{2 \pi }{3}}^{\pi } \frac{1}{2} (2 \cos (\theta )+1)^2 \, d\theta=\pi -\frac{3 \sqrt{3}}{2} \]

  3. itzmashy
    • 2 years ago
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    i'm not supposed to use a double integral?

  4. colorful
    • 2 years ago
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    the way I see it the first step is to find the bounds on theta I think having the picture ion front of you helps

  5. eliassaab
    • 2 years ago
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    Both of them are single integrals

  6. colorful
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=plot%20r%3D1%2B2costheta.%20%20&t=crmtb01

  7. colorful
    • 2 years ago
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    sorry those bounds are wrong \[r=0=1+2\cos\theta\implies \theta=\frac{2\pi}3,\frac{4\pi}3\]which means the bounds on theta are\[\frac{2\pi}3\le\theta\le\frac{4\pi}3\]

  8. colorful
    • 2 years ago
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    The bounds on r are just the polar function itself\[0\le r\le1+2\cos\theta\]and the area differential in polar coordinates is\[dA=rdrd\theta\]hence the intergal for the area of the inner loop should be\[\large \int\int dA=\int_{\frac{2\pi}3}^{\frac{4\pi}3}\int_{0}^{1+2r\cos\theta}rdr\theta\]

  9. colorful
    • 2 years ago
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    ...I assumed we are working in the interval \([0,2\pi]\)

  10. itzmashy
    • 2 years ago
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    and @colorful i'm at work so i can't be checking this all the time :( but it was a homework problem last night and it blew my mind cuz i've never even seen a limacon before last night.. -_-"

  11. eliassaab
    • 2 years ago
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    The bounds for my solutions above are fine

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