I am trying to find the integral to this equation in order to obtain the cross section of a circle that has a radius = 2 and is filled with water up to 1.5.(adsbygoogle = window.adsbygoogle || []).push({});

For some reason, no matter what I do I cannot find the correct solution to this integral. Solving it myself I obtain:

∫√(4-(x-2)^2) dx = (x-2)/2 * √(4-(x-2)^2) + 2arcsin((x-2)/2).

[itex]\int[/itex][itex]\sqrt{4-(x-2)^{2}} dx[/itex] = [itex]((x-2)/2)\sqrt{(4-(x-2)^2)}[/itex] [itex]+ 2arcsin((x-2)/2)[/itex]

Then solving for the definite integral of a=0 and b=1.5 I get: -29.439.

Using Wolfram Alpha I get:

http://www.wolframalpha.com/input/?i=integral+of+(4-(x-2)^2)^0.5

But I cannot solve the definite integral from a=0 and b=1.5 because the equation square roots a negative value.

Using Symbolab I get:

http://symbolab.com/math/solver/ste...-calculator/\int\sqrt{4-\left(x-2\right)^{2}}

But unfortunately when I solve for the definite integral from a=0 and b=1.5, I get the exact same result I did when I formulated the equation myself.

Can anyone tell me what I am screwing up? I am so confused >.>

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# Integral of (4-(x-2)^2)^0.5

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