So the problem asked me to find the area of the region that lies inside both of the circles
$$r=2\sin\theta, \quad r=\sin\theta +\cos\theta $$
I know that $r=2\sin\theta$ is $x^2+(y-1)^2=1,$but the second one is a little bit harder to me.
$$\begin{align*} x&=r\cos\theta=\cos\theta(\sin\theta + \cos\theta)\\ y&=\sin\theta(\sin\theta+\cos\theta) \end{align*}$$
so $x+y=1+2\cos\theta \sin\theta$....Which gives me no help.
I'm getting used to polar coordinates, but I need some help! Thank you
$\endgroup$ 12 Answers
$\begingroup$Hint: Try multiplying both sides of $r=\sin\theta + \cos\theta$ by $r$, and using the identities $r^2=x^2+y^2$, $x=r\cos\theta$, and $y=r\sin\theta$, to find an equation in terms of $x$ and $y$.
$\endgroup$ $\begingroup$To clarify the common area, you need to plot two graphs. Both are circles.
$\cos{\theta} + \sin{\theta} = \sqrt{2}\sin(\theta + \pi/4)$. So draw a circle $r = \sqrt{2} \sin {\theta}$, and rotate it $\pi/4$ clockwise.
Then, it is the circle whose cernter is (1/2,1/2) and the radius is $\frac{\sqrt{2}}{2}$.
Purely algebraically, $r= \sqrt{x^2 + y^2}$, $\cos\theta = \frac{x}{\sqrt{x^2+y^2}}$, and $\sin\theta = \frac{y}{\sqrt{x^2+y^2}}$.
Then $\sqrt{x^2 + y^2 } = \frac{x}{\sqrt{x^2+y^2}} + \frac{y}{\sqrt{x^2+y^2}}$, i.e. $x^2 + y^2 = x + y$.
So $(x-\frac{1}{2})^2 + (y - \frac12)^2 = \frac12$.
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