anonymous
  • anonymous
Find the area of the region that lies inside both circles r=sin(theta) and r=cos(theta
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
is this in polar coordinates?
anonymous
  • anonymous
yes
anonymous
  • anonymous
\[A=2\int\limits_{0}^{\pi/4}1/2\sin^2\theta d \theta=\int\limits_{0}^{\pi/4}1/2(1-\cos2\theta)d \theta\] \[=1/2[\theta-1/2\sin2\theta]_{0}^{\pi/4}={1 \over 8} \pi - {1 \over 4}\]

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anonymous
  • anonymous
@AnwarA: How'd you get that answer, if I may ask?
anonymous
  • anonymous
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?
anonymous
  • anonymous
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.
anonymous
  • anonymous
well you're integrating twice over the same region sinx from 0 to pi/4
anonymous
  • anonymous
the same area*
anonymous
  • anonymous
that's why I multiplied the integral by 2.. make sense?
anonymous
  • anonymous
Yep, I see now. Thanks. :)
anonymous
  • anonymous
no problem :)
anonymous
  • anonymous
you haven't done polar curves? you still in high school?
anonymous
  • anonymous
or maybe a freshman?
anonymous
  • anonymous
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.
anonymous
  • anonymous
I see. wish you the best!
anonymous
  • anonymous
how did you find the point of intersection?
anonymous
  • anonymous
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.
anonymous
  • anonymous
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
anonymous
  • anonymous
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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