Looks like a quarter of a circle with radius 2, so the answer is pi.
If you want to show this using integration, you can use:
ds^2 = dx^2 + dy^2 ------>
s = integral of sqrt[(dx/dt)^2 +
(dy/dt)^2] dt
where we have parametrize x and y using parameter t.
In this case it is simplest to choose
x(t) = 2 sin(t) (because we already now the answer: the curve is a quarter circle). The parameter t ranges from zero to pi/2. If you insert this in y=(sqrt(4-x^2), you find y(t) = 2 cos(t) as expected. Then:
sqrt[(dx/dt)^2 + (dy/dt)^2] = 2
If you integrate this from zero to pi/2 you get pi as the answer.
evaulate an intergral for length of curve with y=(sqrt(4-x^2)) and 0 <_ x <_ 2
im lost, i think i know how to apply the formula but its so difficult to integrate and i need to show all my work!
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