4
$\begingroup$

I am trying to find the area between the polar curves $r = 2 \sin θ$ and $r = 2 \cos θ$.

I set up the area equation as follows:

$$\frac12\int_0^{\pi/4}((2\sinθ)^2-(2\cosθ)^2)\,d\theta$$

I could not get the correct answer with this, which is $\frac\pi2-1$.

Any help with this problem would be appreciated :D

$\endgroup$
  • $\begingroup$ i accidentally put the wrong correct answer, it is (pi/2)-1 $\endgroup$ – nshah Apr 7 '14 at 23:33
5
$\begingroup$

A picture might help: $r=2\sin t$ is in blue and $r=2\cos t$ is in red for $0\le t\le 2\pi$.

enter image description here

The intersection point is found by setting $2\sin t=2\cos t$ and solving; it occurs at $t=\pi/4$, so you want $$ \underbrace{{1\over 2}\int_0^{\pi/4} (2\sin t)^2\,dt}_\text{area of blue region} + \underbrace{{1\over 2}\int_{\pi/4}^{\pi/2} (2\cos t)^2\,dt}_\text{area of red region}={\pi\over 2}-1. $$

$\endgroup$
-1
$\begingroup$

this is just a hint and you have to do it by yourself so see below..... by using 12th method so we know by polar co-ordinates $x= r \sin(x)$ and $y=r \cos(x)$ here in question

$r=2\sin(x)$

multiply both side $r$

$r^2=2r\sin(x)$

$r^2 = 2x$ since $x=r \sin(x)$

since a point circle equation is $x^2+y^2=r^2$

therefore $x^2+y^2=2x $ implies that $y=\sqrt (2x-x^2)$ similarly for $r=2\cos(x)$, $x^2+y^2=2y$ implies that $y= \sqrt (2y-x^2)$ and integrate it by drawing the circle

get the answer. best of luck

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.