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anonymous

  • 5 years ago

Find the area of the region that lies inside both circles r=sin(theta) and r=cos(theta

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  1. anonymous
    • 5 years ago
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    is this in polar coordinates?

  2. anonymous
    • 5 years ago
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    yes

  3. anonymous
    • 5 years ago
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    \[A=2\int\limits_{0}^{\pi/4}1/2\sin^2\theta d \theta=\int\limits_{0}^{\pi/4}1/2(1-\cos2\theta)d \theta\] \[=1/2[\theta-1/2\sin2\theta]_{0}^{\pi/4}={1 \over 8} \pi - {1 \over 4}\]

  4. anonymous
    • 5 years ago
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    @AnwarA: How'd you get that answer, if I may ask?

  5. anonymous
    • 5 years ago
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    From what I've gathered, we can see that the point of intersection is at (theta)=π/4. Then we can get one integral of 1/2 * sin^2(theta) from 0 to π/4, added to an integral of 1/2 * cos^2(theta) from π/4 to π/2. Wouldn't that be enough to account for both halves of the petal-like enclosed shape?

  6. anonymous
    • 5 years ago
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    Alright, so we've gotten exactly the same answers...xD But can you show me how they're the same? I haven't done polar curves yet and I'm curious.

  7. anonymous
    • 5 years ago
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    well you're integrating twice over the same region sinx from 0 to pi/4

  8. anonymous
    • 5 years ago
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    the same area*

  9. anonymous
    • 5 years ago
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    that's why I multiplied the integral by 2.. make sense?

  10. anonymous
    • 5 years ago
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    Yep, I see now. Thanks. :)

  11. anonymous
    • 5 years ago
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    no problem :)

  12. anonymous
    • 5 years ago
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    you haven't done polar curves? you still in high school?

  13. anonymous
    • 5 years ago
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    or maybe a freshman?

  14. anonymous
    • 5 years ago
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    I'm a junior in AP Calc, polar curves is what we're doing immediately after we finish sequences & series. I skipped pre-calc, so now I'm pretty much going through the rest of the syllabus on my own to cover anything I missed.

  15. anonymous
    • 5 years ago
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    I see. wish you the best!

  16. anonymous
    • 5 years ago
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    how did you find the point of intersection?

  17. anonymous
    • 5 years ago
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    Plot the two polar curves (or set the two equations equal to each other) and based on simple trigonometry, the angle at which they both equal each other is π/4. I think you'll see exactly what I mean when you graph it though.

  18. anonymous
    • 5 years ago
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    alright thank you, it makes sense. btw im review sequences and series right now for my calculus exam can you reccemend a video or website which covers the section well? thank you

  19. anonymous
    • 5 years ago
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    Here you have links of tons of different types of convergence tests, basic concepts; basically anything you could want to see about these. :) No problem, http://tutorial.math.lamar.edu/Classes/CalcII/SeriesIntro.aspx

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