Hey so I need help with this question:



So I first tried it using cartesian coordinates but ended up with a nasty integrand after integrating with respect to y.
The next obvious thing to do is use polar coordinates but I am having trouble finding the ranges of r and theta.
The smaller circle (x^2+y^2=2x) is a semi circle (so theta goes from 0 to pi) but the larger circle (x^2+y^2=4) is a quarter-circle (so theta goes from 0 to pi/2).

Can someone help? I think I should be okay with the integrand because I will just use Fubini's theorem to split it up if the bounds of integration allow me to (if r does not depend on theta).

You could probably shortcut this one by completing the square in the second equation, and just using geometry.

Didn't find any easy computation :

I'd write your integral as :

integral from x=0 to x=2 of integral from y=sqrt(2x-x²) to y=sqrt(4-x²) of x (dx, dy)

integral from 0 to 2 of f(x).dx
where :

f(x) = x( sqrt(4-x²) - sqrt(2x-x²) )

notice that :

f(x) = x.sqrt(4-x²) - (x-1).sqrt(2x-x²) - sqrt(2x-x²)

the first two parts are easy to compute, integral from 0 to 2 gives 8/3 (if I do no mistake).

The second part is a lot more tricky :

let x = 2 / (1+t²), t varies from +∞ (as x=0) to 0 (as x=2).
check that sqrt(2x-x²) = 2t / (1+t²)
and that : dx = -4t.dt / (1+t²)²

Hence :

∫ (0;2; f(x) ; dx) = ∫ (+∞ ; 0 ; -8t² / (1+t²)^3 ; dt) = ∫ (0 ; +∞ ; 8t² / (1+t²)^3 ; dt)

You can estimate an anti-derivative of 1 / (1+t²)^n , for any integer n≥1 using integration by parts, starting from ∫ dt / (1+t²) = Arctan(t).

∫ dt / (1+t²) = [ t / (1+t²) ] + ∫ 2t².dt / (1+t²)²
Arctan t = t / (1+t²) +2.Arctan t - 2.∫ dt / (1+t²)²
∫ dt / (1+t²)² = 0.5(Arctan t + t/(1+t²))

and similarly :
∫ dt / (1+t²)^3 = 0.25 ( 3.∫ dt / (1+t²)² + t / (1+t²)²)

Hence, ∫ 8t².dt / (1+t²)^3 = 8.∫ dt / (1+t²)² - 8.∫ dt / (1+t²)^3 = ... = Arctan t + t/(1+t²) - 2t / (1+t²)²
Evaluating this between 0 and +∞ gives π/2.

Final result should be : 8/3 - π/2 ~ 1.096, if I am not mistaken.

Thank you guys!! I managed to use polar coordinates which I found much easier than cartesian (only tricky part was the bounds on r and then integrating cos^4(theta)).

I got 8/3 - pi/2 as well

Congrats !
Indeed, the bounds are not easy with polar coordinates