Why am I getting negative area?

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Homework Statement



So, I have a cylinder with radius r and length L and it's filled with water that has a variable height H. I'm supposed to find a formula for the volume of the water contained in the cylinder in terms of H r and L.


3. Attempt at a solution
using a circle moved up on the y-axis by r units we get x^2+(y-5)^2=r^2

Then, integrating rectangles with width given by 2\sqrt{x^{2}-(y-r)^{2}}L\Deltay from 0 to H.

After evaluating it using trig substitution I get as my function :

V(H)=(L/2)((r^2-(H-r)^2)^(1/2)(H-r)+2r^2arcsin(H-r)/5)+pi*r^2)
This makes sense because when I let H=2r I get the volume of a cylinder and when H=0 I also get zero.

I derived the exact same formula using trigonometry and geometry as well.

My problem is that when I evaluate it for different height based on the diameters say H=(1/3)2r or H=(1/5)2r I get negative area which does not make sense.

I have a feeling that all of my (H-r) terms should be slightly altered, but I'm not sure, so if someone can point out what I'm missing it would be much appreciated
 
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armolinasf said:

Homework Statement



So, I have a cylinder with radius r and length L and it's filled with water that has a variable height H. I'm supposed to find a formula for the volume of the water contained in the cylinder in terms of H r and L.
How is the cylinder oriented? Is it lying on its side?
armolinasf said:
3. Attempt at a solution
using a circle moved up on the y-axis by r units we get x^2+(y-5)^2=r^2

Then, integrating rectangles with width given by 2\sqrt{x^{2}-(y-r)^{2}}L\Deltay from 0 to H.
Your formula for width should be in terms of one variable only. Since you are integrating with respect to y, the formula should only involve y.

To get y, solve the equation x2 + (y - 5)2 = r2 for x.


armolinasf said:
After evaluating it using trig substitution I get as my function :

V(H)=(L/2)((r^2-(H-r)^2)^(1/2)(H-r)+2r^2arcsin(H-r)/5)+pi*r^2)
This makes sense because when I let H=2r I get the volume of a cylinder and when H=0 I also get zero.

I derived the exact same formula using trigonometry and geometry as well.

My problem is that when I evaluate it for different height based on the diameters say H=(1/3)2r or H=(1/5)2r I get negative area which does not make sense.

I have a feeling that all of my (H-r) terms should be slightly altered, but I'm not sure, so if someone can point out what I'm missing it would be much appreciated
Since you're getting a negative number, I suspect that a distance you are using turns out to be negative, possibly H - r.
 
The cylinder is lying on its side. My mistake with the variables, it should have been (r^-(h-r)^2)^(1/2). Would it matter if H is above or below the middle of the circle? I looked at my model and if H is below the middle of the circle I get the area that's not full.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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