integral parsial

Question

use the integral to find the area of a circle

Answer 1

If the equation of the circle is x²+y²=r², the circle has centre at (0,0). Its x intercepts are (-r,0) and (r,0).

Because of symmetry, there is one quadrant between x=0 and x=r, and y=0 and y=r and this quadrant is quarter the area of the circle.

Consider a point (x,√(r²-x²)) on the circle. The height of this point above the x-axis is √(r²-x²​). If we make this into a very thin rectangle of width dx then the sum of the areas (√(r²-x²​)dx) of many of these rectangles will be the area of the quadrant. This gives us the integral we need:

∫√(r²-x²​)dx, the limits of which are [0,r].

Let x=rsinθ then dx=rcosθdθ and √(r²-x²​) becomes √(r²-r²sin²θ)=r√(1-sin²θ)=rcosθ.

When x=0, θ=0, when x=r, sinθ=1, so θ=½π. The integral becomes:

∫rcosθ.rcosθdθ=r²₀∫^½πcos²θdθ. cos(2θ)=2cos²θ-1, so cos²θ=½(cos(2θ)+1).

r²₀∫^½πcos²θdθ=½r²₀∫^½π(cos(2θ)+1)dθ=½r²[½sin(2θ)+θ]₀^½π=½r²(½π) because sin(π)=sin(0)=0.

So the integral is ¼πr². This is the area of the quadrant so the area of the circle is 4 times this=πr², the area of the circle radius r.