What is the area enclosed by the curve r=(1+Cos\theta),0\leq \theta \leq 2\pi?

[fan_103]

Find the area enclosed by the curve r=(1+Cos\theta),0\leq \theta \leq 2\pi
|cos \theta|\leq1
Maximum r =2(1+1)=4

When r=0,
2+2Cos\theta=0
Cos\theta=-1
Key angle=0
\theta=\pi,3/2\pi

Area of curve =1/2\intr^2 dtheta

 

[CompuChip]

So what did you try already?

 

[fan_103]

I don't know how to sketch the graph?!?

 

[lanedance]

try setting up the area integral and we can discuss

here's a hint & some example tex for you try clicking below (another hint)
hint: as a start what is an infintesimal area element in polar coordinates?
\int_0^{2 \pi} d \theta \int_0^{r(\theta)} dr f(r,\theta)

 

[HallsofIvy]

When \theta= 0, r= 1+1= 2 so one point on the graph is at (2,0) When \theta= \pi/2, r= 1+ 0= 1 so another point is at (0, 1). When \theta= \pi, r= (1+ (-1))= 0 so a third point is (0,0). When \theta= 3\pi/2, r= 1+ 0= 1 so a fourth point is (0, -1). When \theta= 2\pi r= 1+1= 2[/itex] so we are back at (2, 0). That's all you need to know: we go around one complete circuit of the figure as \theta goes from 0 to 2\pi. You don't need to actually draw the graph. (It is a figure known as a "cardioid".)
Integrate rd\theta= (1+ cos(\theta)d\theta from \theta= 0 to \theta= 2\pi.