User:Annoyingpi

Find the integral for the arc length of $x^2 + y^2 = 4$ going clockwise from $(-3^{0.5}, 1)$ to $(3^{0.5}, 1)$ .

The function is $y = \sqrt{4-x^2}$ . $x$ is ranging from $-\sqrt{3}$ to $\sqrt3$ .

$\frac{dy}{dx} = -x \frac1{\sqrt{4-x^2}}.$

Therefore the integral is $\boxed{\int_{-\sqrt3}^{\sqrt3} \sqrt{1+ \frac{x^2}{4-x^2}} \, dx}.$

IGNORE EVERYTHING BELOW:

$[asy] draw((0,0)--(0,2)--(2,2)--(2,0)--cycle); draw((0,0)--(0,1)--(1,0)--cycle); draw((0,0)--(0,2)--(1,2)--cycle); draw((0,0)--(2,0)--(2,1)--cycle); fill((0,0)--(0,1)--(1,0)--cycle,gray); fill((0,0)--(0,2)--(1,2)--cycle,gray); fill((0,0)--(2,0)--(2,1)--cycle,gray); label("0",(0,0),SW); label("1",(0,1),W); label("2",(0,2),NW); label("1",(1,0),S); label("2",(2,0),SE); [/asy]$

$[asy] draw((0,0)--(5,12)); draw((5,12)--(8,12)); draw((0,0)--(13,0)); draw((8,12)--(13,0)); draw((5/3,4)--(13/3,0)); draw((13-5/3,4)--(13-13/3,0)); draw((10/3,8)--(6,12)); draw((7,12)--(13-10/3,8)); dot((5,12)); dot((8,12)); dot((13,0)); dot((0,0)); dot((5/3,4)); dot((13/3,0)); dot((13-5/3,4)); dot((13-13/3,0)); dot((10/3,8)); dot((13-10/3,8)); dot((6,12)); dot((7,12)); dot((6+1/3,12)); dot((7-1/3,12)); dot((5/3+5/9,4+4/3)); dot((5/3+10/9,4+8/3)); dot((13-5/3-5/9,4+4/3)); dot((13-5/3-10/9,4+8/3)); dot((13/3 + 26/9,0)); dot((13/3 + 13/9,0)); label("A",(5,12),NW); label("B",(8,12),NE); label("C",(13,0),SE); label("D",(0,0),SW); [/asy]$

$[asy] draw((0,0)--(5,12)); draw((5,12)--(8,12)); draw((0,0)--(13,0)); draw((5,12)--(5,0)); draw((8,12)--(13,0)); draw((8,12)--(8,0)); dot((5,12)); dot((8,12)); dot((13,0)); dot((0,0)); dot((5,0)); dot((8,0)); label("A",(5,12),NW); label("B",(8,12),NE); label("C",(13,0),SE); label("D",(0,0),SW); label("M",(5,0),S); label("N",(8,0),S); label("13",(2.5,6),NW); label("13",(11.5,6),NE); label("13",(6.5,-1),S); label("3",(6.5,12),N); label("12",(5,6),W); label("12",(8,6),E); label("5",(2.5,0),N); label("5",(10.5,0),N); [/asy]$

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User:Annoyingpi