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 $$\theta$$|$$\leq$$1
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$$\int$$r^2 d$$theta$$

 

[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 $\theta= 2\pi$.