Please find the area enclosed by this parametric function:
$$\begin{gathered}
x = 2a\cos t - a\cos 2t \hfill \
y = 2a\sin t - a\sin 2t \hfill \
\end{gathered} $$

Question

Please find the area enclosed by this parametric function:
$$\begin{gathered}
  x = 2a\cos t - a\cos 2t \hfill \
  y = 2a\sin t - a\sin 2t \hfill \ 
\end{gathered} $$

amistre64 · Answer

the area of a section of a circle is 1/2 pi t^2  right?

amistre64 · Answer

2pi
---  r^2 .. yeah
  2

anonymous · Answer

right

amistre64 · Answer

and r here is the function provided

anonymous · Answer

Could you show me the details?

amistre64 · Answer

if i could remember the details i would :)

anonymous · Answer

But I don't think your answer is right...

amistre64 · Answer

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amistre64 · Answer

my answer is neither right nor wrong, its simply incomplete

amistre64 · Answer

im thinking this is a double integral

amistre64 · Answer

dr dt

amistre64 · Answer

we measure the value or r thru each degree, and add up all the degrees and their little rs

amistre64 · Answer

have you done doubles yet?

amistre64 · Answer

r^2 = x^2 + y^2 so that part is "simple" enough ...

anonymous · Answer

Yes I have learn

amistre64 · Answer

$$r^2 = (2a\ cos(t)−acos(2t))^2\ +(2a\ sin(t)−a\ sin(2t))^2$$
unless im reading the material wrong

amistre64 · Answer

t = 0 to 2pi

amistre64 · Answer

i assume "a" is considered to be a constant?

anonymous · Answer

$$dA = \frac{1}
{2}r*rdt$$
Is it you are thinking?

amistre64 · Answer

yes, or something vaguely similar to it..

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