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

1. Nov 21, 2013

### Killswitch

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).

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

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 >.>

Last edited: Nov 21, 2013
2. Nov 21, 2013

### SteamKing

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

http://en.wikipedia.org/wiki/Circular_segment

3. Nov 21, 2013

### Killswitch

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.

4. Nov 21, 2013

### iRaid

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.

5. Nov 22, 2013

### SteamKing

Staff Emeritus
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.