ns am make the efforts to discover the area in between the polar curve \$r = 2 sin θ\$ and \$r = 2 cos θ\$.

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  Thanks because that contributing response to medtox.orgematics Stack Exchange!

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