You are watching: Find the area of the region that lies inside both curves. r2 = 2 sin(2θ), r = 1

I collection up the area equation together follows:

$$frac12int_0^pi/4((2sinθ)^2-(2cosθ)^2),d heta$$

I could not gain the correct answer with this, i beg your pardon is $fracpi2-1$.

Any help with this difficulty would be appreciated :D

A photo might help: $r=2sin t$ is in blue and also $r=2cos t$ is in red because that $0le tle 2pi$.

The intersection suggest is discovered by setting $2sin t=2cos t$ and also solving; it occurs at $t=pi/4$, so you want$$underbrace1over 2int_0^pi/4 (2sin t)^2,dt_ extarea of blue region+underbrace1over 2int_pi/4^pi/2 (2cos t)^2,dt_ extarea of red region=piover 2-1.$$

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

$r=2sin(x)$

multiply both side $r$

$r^2=2rsin(x)$

$r^2 = 2x$ due to the fact that $x=r sin(x)$

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

therefore$x^2+y^2=2x $ suggests that $y=sqrt (2x-x^2)$ an in similar way for $r=2cos(x)$, $x^2+y^2=2y$ implies that $y= sqrt (2y-x^2)$and combine it by drawing the circle

get the answer.best that luck

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