Find the area enclosed by the polar curve r= 2-cos (theta)

Question

mrnood · Answer

$$Area = \int\limits_{0}^{2\pi} (2-\cos \theta ) d \theta $$

anonymous · Answer

@MrNood it's $$Area=\int\limits_{0}^{2\pi} \frac{ 1 }{ 2 } f(\theta)^2 $$ isn't it?

anonymous · Answer

forgot to tag $$d \theta $$ at the end

mrnood · Answer

Not quite sure where that's coming from.

I was working on integral r dtheta

mrnood · Answer

I'm a bit rusty on this stuff - so happy to be overridden!

anonymous · Answer

I got an answer of 4pi, so I think that's right

mrnood · Answer

seems unlikely value if you think of the function. In fact the answer MUST include r since the area obviously depends on r

If the radius was constant then the area would be pi r^2

In this function the radius decreases at points around the polar sweep so I would expect the area to be < pi r^2

As I say - a little rusty but...

mrnood · Answer

the plot looks something like:

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