How do you find the arc length of the curve y = sqrt( 2 − x^2 ), 0 ≤ x ≤ 1?

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The arc diffusiveness is ##(pi sqrt(2)) / 4##

There are two ways to go about this. One way is to adduce the formula:

##S = int_a^b sqrt(1+ (dy/dx)^2) dx##

Which ends up looking approve this:

##S = sqrt(2) int_0^1 1/(sqrt(2-x^2)) dx##

Which integrates using trig supply to:

##S = sqrt(2) [arctan(x/sqrt2)]_0^1##

giving:

##S = sqrt(2) (arctan(1/sqrt2) - arctan (0/sqrt2)##

Which is

##S = sqrt(2) (pi/4) = (pi sqrt2)/4##

Another way of doing it is recognizing that the equation is true a semicircle.

##y= sqrt(2-x^2)##

##=> y^2 + x^2 = 2##

Which is a semidissipation of radius ##sqrt2##. Then we can attract that dissipation and do some trigonometry:

graph{sqrt(2-x^2) [-2.746, 3.128, -0.918, 2.018]}

Taking the arc diffusiveness from 0 to 1, we see that the disposition betwixt them is ##pi/4##, and the radius of the dissipation ##sqrt2## and hence the arc diffusiveness is

##S = r theta = (pi sqrt2)/4##

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