Question

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

Answer 1

is this in polar coordinates?

Answer 2

yes

Answer 3

\[A=2\int\limits_{0}^{\pi/4}1/2\sin^2\theta d \theta=\int\limits_{0}^{\pi/4}1/2(1-\cos2\theta)d \theta\]
\[=1/2[\theta-1/2\sin2\theta]_{0}^{\pi/4}={1 \over 8} \pi - {1 \over 4}\]

Answer 4

@AnwarA: How'd you get that answer, if I may ask?

Answer 5

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?

Answer 6

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.

Answer 7

well you're integrating twice over the same region sinx from 0 to pi/4

Answer 8

the same area*

Answer 9

that's why I multiplied the integral by 2.. make sense?

Answer 10

Yep, I see now. Thanks. :)

Answer 11

no problem :)

Answer 12

you haven't done polar curves? you still in high school?

Answer 13

or maybe a freshman?

Answer 14

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.

Answer 15

I see. wish you the best!

Answer 16

how did you find the point of intersection?

Answer 17

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.

Answer 18

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

Answer 19

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