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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
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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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is this in polar coordinates?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[A=2\int\limits_{0}^{\pi/4}1/2\sin^2\theta d \theta=\int\limits_{0}^{\pi/4}1/2(1\cos2\theta)d \theta\] \[=1/2[\theta1/2\sin2\theta]_{0}^{\pi/4}={1 \over 8} \pi  {1 \over 4}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0@AnwarA: How'd you get that answer, if I may ask?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0From 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 petallike enclosed shape?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Alright, 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
 5 years ago
Best ResponseYou've already chosen the best response.0well you're integrating twice over the same region sinx from 0 to pi/4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that's why I multiplied the integral by 2.. make sense?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yep, I see now. Thanks. :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you haven't done polar curves? you still in high school?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I'm a junior in AP Calc, polar curves is what we're doing immediately after we finish sequences & series. I skipped precalc, so now I'm pretty much going through the rest of the syllabus on my own to cover anything I missed.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I see. wish you the best!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you find the point of intersection?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Plot 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
 5 years ago
Best ResponseYou've already chosen the best response.0alright 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
 5 years ago
Best ResponseYou've already chosen the best response.0Here 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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