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anonymous

  • 5 years ago

Find the area enclosed by the curve r^2=9cos(5Θ)

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  1. anonymous
    • 5 years ago
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    If you have a function \[r=f(\theta)\]in polar coordinates, the form of the integral you use is \[\int\limits_{}^{}\frac{1}{2}[f(\theta)]^2d \theta=\int\limits_{}{}\frac{1}{2}r^2d \theta\]You have it easy here, since you're given r^2 in terms of theta, so your integral is\[\int\limits_{\theta_1}^{\theta_2}\frac{1}{2}9\cos 5 \theta d \theta=\frac{9}{2}\int\limits_{\theta_1}^{\theta_2}\cos 5 \theta d \theta=\frac{9}{2}\left[ \frac{1}{5} \sin 5 \theta\right]_{\theta_1}^{\theta_2}\]\[=\frac{9}{10}(\sin 5 \theta_2-\sin 5 \theta_2)\]

  2. anonymous
    • 5 years ago
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    You asked for the 'area', which means your integral will be definite. Theta_1 and theta_2 are just arbitrary limits.

  3. anonymous
    • 5 years ago
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    okay that makes sense but how is it that i find the limits?

  4. anonymous
    • 5 years ago
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    They'll be given to you, explicitly or implicitly. For example, thy may say 'Between the angles pi/2 and pi' or they may say 'Find the area of the 'flower' (that's what this thing looks like)' and you'll have to recognize you need to sweep one complete revolution from 0 to 2pi. Is that okay?

  5. anonymous
    • 5 years ago
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    yes thats fine but what if they don't give the limits. The directions really do just say exactly what my questions states. Find the area enclosed by the curve r^2=9cos(5Θ). it gives no limit whatsoever...

  6. anonymous
    • 5 years ago
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    Okay, well that refers to the last part of what I was saying, about *implicit* limits. Here they're expecting you to know something about how the shape is formed - i.e. the angles you have to sweep. Here, you say you have to find the area of the whole thing - so you have to work out from theta = 0 to 'something' that will create the whole figure (or how much of it you want). Then you'll have your limits.

  7. anonymous
    • 5 years ago
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    okay sounds good. Thanks so much!

  8. anonymous
    • 5 years ago
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    No worries. Become a fan! ;)

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