calculusintegrationpolar coordinates
$ r = 2\cos(\theta)$ has the graph
I want to know why the following integral to find area does not work $$\int_0^{2 \pi } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$
whereas this one does:
$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2} } \frac{1}{2} (2 \cos (\theta ))^2 \, d\theta$$
Why do the limits of integration have to go from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$? Doesn't going from $0$ to $2\pi$ also sweep out the full circle?
Replacing $(r,\theta)\to(r,-\theta)$, we have a same equation so the graph is symmetric with respect to the polar axe.
We can have the following polar point. Since the graph is symmetric so we need to draw a half of the graph:
$$\theta=0\to r=1$$ $$\theta=\pi/6\to r=\sqrt{3}-1$$ $$\theta=\pi/3\to r=0$$ $$\theta=\pi/2\to r=-1$$ $$\theta=2\pi/3\to r=-2$$ $$\theta=5\pi/6\to r=-\sqrt{3}-1$$ $$\theta=\pi\to r=-3$$
If $r=0$, then $\cos(\theta)=1/2$ and so $\theta=\pi/3$ provided $0\le\theta<\pi$. This means that the point $(0,\pi/3)$ is on the graph, and an equation of the tangent line there is $\theta=\pi/3$. It is called limacon
$$r = \cos \theta$$
$$\sqrt{x^2 + y^2} = \frac{x}{\sqrt{x^2 + y^2}}$$
$$x^2 + y^2 = x$$
$$x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4}$$
$$\left(x - \frac{1}{2}\right)^2 + (y - 0)^2 = \left(\frac{1}{2}\right)^2$$
This is the equation for a circle in Cartesian coordinates $(x, y)$ with center $({1 \over 2}, 0)$ and radius $1 \over 2$.
Best Answer
the limit is from $0$ to $\pi$ not $0$ to $2\pi$ if you do the latter, you will go around the circle twice.
at $\theta = 0$ you are $(2,0)$ at $\theta = \pi/3$ you are at $x = 1, y = 1$ at $\theta = \pi/2$ you are at $(0,0)$ and then you move in the lower semicircle for the next $\pi/2$ to $\pi$ if you go farther in $\theta$ you will go back to where you started.
one has to be careful when setting up polar integral about the limits and how the curve is traced. you will lots of one inside other in cardiods and more complicated figures like $r = \cos(5 \theta),$ etc