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How can I calculate the area of the external region of $\{ x^2+y^2=4\}$ and internal at $\{ p=2(1+cost)| t\in [0;2\pi] \}$, i know that the result is $8+\pi$, but i don't know the method to get.

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    It appears you want the area bounded by 2 given curves. However, the 1st curve is in the $(x,y)$ plane and the 2nd curve is in the $(t,p)$ plane; there must be some typo in the question... – 2017-01-05
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    Could it be that one curve is in Cartesian coordinates while the other curve is in polar coordinates on the same plane? (So you might need to transform everything into polar and then find the area?) – 2017-01-05

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We will assume that

  • the external region of $x^2+y^2=4 \ $ (this is a circle with radius $2$) is on Cartesian plane;
  • the internal region of $\rho=2(1+\cos\phi)$ is in *polar* coordinates on the same plane.

Hereafter I write $(\rho,\phi)$ for the polar coordinates.

We draw a sketch of both curves, and indeed see that together they bound a crescent-like figure. The entire figure is in the $x\ge0$ half-plane. The figure is symmetric about the horizontal axis. Thus we can compute the area inside the curve $\rho=2(1+\cos\phi)$ in the first quadrant, i.e. for $\phi\in[0,\pi/2]$, and subtract the area ($\pi$) of quarter-circle; this gives us the upper half of the sought-for figure.

If a curve is given by $\rho=f(\phi$), then the area $A_1$ inside the curve within the first quadrant is $$ A_1 = \int_0^{\pi/2}{1\over2}f(\phi)^2 d\phi = \int_0^{\pi/2}{1\over2}(2 + 2 \cos\phi)^2 d\phi = {8+3\pi \over2}. $$ Subtracting the area of quarter-circle, which is $\pi$, we get $\displaystyle{8+\pi \over2}$; this is the upper half of our crescent-like figure located in the first quadrant. Therefore the area of the whole figure is $$8+\pi.$$