Why am I getting negative area?

In summary, the conversation is about finding a formula for the volume of water contained in a cylinder with variable height, radius, and length. The attempted solution involves integrating rectangles and using trigonometry and geometry. However, there is a discrepancy when evaluating for different heights and it is suspected that a negative distance is being used. The orientation of the cylinder and the placement of H above or below the middle of the circle may also affect the results.
  • #1
armolinasf
196
0

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[tex]\sqrt{x^{2}-(y-r)^{2}}[/tex]L[tex]\Delta[/tex]y 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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  • #2
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[tex]\sqrt{x^{2}-(y-r)^{2}}[/tex]L[tex]\Delta[/tex]y 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.
 
  • #3
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.
 

FAQ: Why am I getting negative area?

Why am I getting negative area when calculating the volume of a shape?

Negative area can occur when the shape being calculated is not closed or when the coordinates of the shape are not entered correctly. This results in the formula producing a negative value for the area, which is not possible in real life.

Why does my graph show negative area?

In a graph, negative area can occur when the function being graphed produces values below the x-axis. This creates a portion of the graph that is below the x-axis and therefore, has a negative area.

Why am I getting a negative area for a real-life object?

In real life, negative area can occur due to measurement errors or inaccuracies. It could also be due to the limitations of the measuring instrument or technique used to determine the area of the object.

Can a negative area have a physical meaning?

No, a negative area cannot have a physical meaning as area is a measure of the space occupied by an object and it cannot have a negative value in physical terms.

How can I avoid getting negative area in my calculations?

To avoid negative area, ensure that the shape being calculated is closed and all the coordinates are entered correctly. In graphs, make sure the function being graphed does not produce values below the x-axis. Also, double-check your measurements and use precise measuring instruments and techniques.

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