areacalculusintegrationpolar coordinates
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 😀
I just added this plot to show what is Raskolnikov trying to tell you. You can find the area. Note the integral he wrote above.
Area of a curve given in polar coordinate is given by $$\rm{A} = \frac 12\int_{\theta_1}^{\theta_2} r^2(\theta) \:\rm{d} \theta$$
We can obtain this formula by dividing the curve into infinitely many thin sectors. (each subtending angle $\rm d \theta$ at the origin)
Then use the formula of area of sector to obtain $$dA=\frac{1}{2} r^2 d \theta$$
Now integrate both the sides,
$$\int \rm dA=A= \frac 12\int_{\theta_1}^{\theta_2} r^2(\theta) \:\rm{d} \theta$$
Best Answer
A picture might help: $r=2\sin t$ is in blue and $r=2\cos t$ is in red for $0\le t\le 2\pi$.
The intersection point is found by setting $2\sin t=2\cos t$ and solving; it occurs at $t=\pi/4$, so you want $$ \underbrace{{1\over 2}\int_0^{\pi/4} (2\sin t)^2\,dt}_\text{area of blue region} + \underbrace{{1\over 2}\int_{\pi/4}^{\pi/2} (2\cos t)^2\,dt}_\text{area of red region}={\pi\over 2}-1. $$