Using integrals, find the area of the circle $x^2 + y^2 = 1$ ?

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Answer 1 · 2018-05-17T22:59:34

Using polar coordinates:
$x= rho cos theta$
$y= rho sin theta$

we have to integrate for $0<=rho <=1$ and $0 <= theta <= 2pi$ , so:

$S = int_0^1int_0^(2pi) rho d rho d theta$

$S = int_0^1 rho d rho int_0^(2pi) d theta$

$S = 2pi int_0^1 rho d rho$

$S = 2pi [rho^2/2]_0^1$

$S = pi$

Alternatively we have that for $x,y >=0$ :

$x^2+y^2 = 1$

$y=sqrt(1-x^2)$

For symmetry reasons:

$S = 4int_0^1 sqrt(1-x^2)dx$

substitute:

$x = sint$ with $t in [0,pi/2]$

$dx = cost dt$

As in the interval the cosine is positive:

$sqrt(1-x^2) = sqrt(1-sin^2t) = sqrt(cos^2t) = cost$

So:

$S = 4int_0^(pi/2) cos^2tdt$

$S = 4int_0^(pi/2) (1+cos(2t))/2dt$

$S = 2int_0^(pi/2) dt + 2int_0^(pi/2)cos(2t)dt$

$S = 2[t]_0^(pi/2) + [sin2t]_0^(pi/2)$

$S = 2(pi/2-0) + sinpi-sin0 = pi$

Using integrals, find the area of the circle x^2 + y^2 = 1x^2 + y^2 = 1 ?