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

Question

Isabella Ackerman · Accepted Answer

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#

Aurora Alvarado · Answer

undefined To find the area of the circle $x^2 + y^2 = 1$, you can use the integral method. Since the circle is symmetric about both the x-axis and the y-axis, you can focus on finding the area of one quarter of the circle (for example, the region where $x \geq 0$ and $y \geq 0$) and then multiply by 4.

The equation of the circle suggests that $y = \sqrt{1 - x^2}$ for $x$ between -1 and 1.

So, the area $A$ of one quarter of the circle can be found by integrating $y = \sqrt{1 - x^2}$ with respect to $x$ from 0 to 1.

$$ A = \int_{0}^{1} \sqrt{1 - x^2} \, dx $$

After integrating this expression, multiply the result by 4 to find the total area of the circle.