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

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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.

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/step-by-step/calculus/definite-integral-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 >.>
 
Last edited:

SteamKing

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I'm not sure you have your integral set up correctly (always include the variable of integration dx in your integral expression.)

However, maybe this article can give you some ideas about the correct approach:

http://en.wikipedia.org/wiki/Circular_segment
 
It seems I did just forget the notation for dx in my equation. However, I solved using with dx in my original equation. That link you posted is useful for calculating the area of the circlular segment as well, but it does not take a calculus approach in order to solve it. It simple takes the area of the circle segment, subtracted by triangle within that segment of the circle.
 
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I think I'm getting somewhere, probably can use more trig identities to solve it completely:
u=x-2, du=dx
$$\int \sqrt{4-u^{2}}du$$
Then trig sub. u=2siny, du=2cosydy and end up with:
$$\int 4cos^{2}(y)dy$$

Hmm can keep going I think that's right so far.
 

SteamKing

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It seems I did just forget the notation for dx in my equation. However, I solved using with dx in my original equation. That link you posted is useful for calculating the area of the circlular segment as well, but it does not take a calculus approach in order to solve it. It simple takes the area of the circle segment, subtracted by triangle within that segment of the circle.
That's why when setting up an integral, it's sometimes helpful to draw a sketch of what it is you are trying to determine. I think if you do that for your problem, you will see that your integral formulation requires some adjustment.
 

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