Calculating the volume of a sphere which is submerged below the water - help please

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Sophdof1

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I am trying to find the volume of a sphere which is submerged below the water. I am given the following information to aid:

- The sphere is solid with diameter 10cm;
- The sphere floats in the water so that the top of the sphere is 2cm above the surface of the water.

I can visualise this, however have no idea how to turn this into a calculus problem - Any ideas?

I'd really appreciate the help :)
 
I would use a solid of revolution. Suppose we observe that the cross-section of a sphere is a circle, and so what curve would we be rotating?
 
Once again I am missing something. Isn't the volume of the sphere just a function of the radius, that is the volume will not change if we put it in water? Isn't the volume V = (4/3)pi*r3, where r = 5cm??
 
Jomo said:
Once again I am missing something. Isn't the volume of the sphere just a function of the radius, that is the volume will not change if we put it in water? Isn't the volume V = (4/3)pi*r3, where r = 5cm??
Click to expand...

We're trying to find just that portion of the sphere that is below the level of the water. :)
 
Suppose you consider the graph of:

[MATH]f(x)=\sqrt{5^2-x^2}[/MATH]
This is the top half of a circle of radius 5 centered at the origin. If we rotate that curve about the \(x\)-axis, we get a sphere. But what if we limit the domain to \([-5,3]\)...then we get a solid that represent the portion of the sphere below the water...does that make sense?
 
If your class has not studied methods for finding solids of revolution, then you could (1) search for online lessons/videos or (2) use the spherical-cap method (that is, look up the formula for the volume of a partial sphere).

However, (1) is a calculus approach, and (2) is not.

?
 
MarkFL said:
Suppose you consider the graph of:

[MATH]f(x)=\sqrt{5^2-x^2}[/MATH]
This is the top half of a circle of radius 5 centered at the origin. If we rotate that curve about the \(x\)-axis, we get a sphere. But what if we limit the domain to \([-5,3]\)...then we get a solid that represent the portion of the sphere below the water...does that make sense?
Click to expand...

Hi yes that makes a lot of sense
 
The equation of a sphere of radius 10 cm is \(\displaystyle x^2+ y^2+ z^2= 100\). So \(\displaystyle x^2+ y^2= 100- z^2\). That is, the cross section at height z is a circle with radius \(\displaystyle r= \sqrt{100- z^2}\). If we imagine that as a thin \piplate of thickness "dz" then its volume is \(\displaystyle \pi r^2 dz= \pi(100- z^2)dz\). As a check, the volume of the entire sphere is \(\displaystyle \pi \int_{-10}^{10} (100- z^2)dz=\)\(\displaystyle 2\pi\int_0^{10}(100- z^2)dz=\)\(\displaystyle 2\pi\left(100z- z^3/3\right)_0^{10}= \)\(\displaystyle 2\pi\left(1000- 1000/3\right)=\)\(\displaystyle \frac{4}{3}\pi (1000)\) which is the same as the standard formula \(\displaystyle V= \frac{4}{3}\pi r^3\).

The point is that, since there are 2 cm above the water, to find the volume below the water, we take the integral up to z= 8 rather than z= 10: \(\displaystyle \pi\int_{-10}^8(100- z^2)dz\).
 
If I were going to answer this question, I would consider the first quadrant part of the curve:

[MATH]f(x)=\sqrt{r^2-x^2}[/MATH]
Now, if we consider some constant \(0<h<r\), where \(h\) represents the height of the cap, then the volume \(V_C\) of the cap can be found from:

[MATH]V_C=\pi\int_{r-h}^r \left(\sqrt{r^2-x^2}\right)^2\,dx=\pi\int_{r-h}^r r^2-x^2\,dx[/MATH]
Applying the FTOC, we obtain:

[MATH]V_C=\frac{\pi}{3}\left[3r^2x-x^3\right]_{r-h}^r=\frac{\pi}{3}\left(2r^3-\left(3r^2(r-h)-(r-h)^3)\right)\right)[/MATH]
[MATH]V_C=\frac{\pi}{3}\left(2r^3-(r-h)\left(3r^2-(r-h)^2)\right)\right)[/MATH]
[MATH]V_C=\frac{\pi}{3}\left(2r^3-(r-h)\left(2r^2+2rh-h^2)\right)\right)[/MATH]
[MATH]V_C=\frac{\pi}{3}\left(2r^3-2r^3-2r^2h+rh^2+2r^2h+2rh^2-h^3\right)[/MATH]
[MATH]V_C=\frac{\pi h^2}{3}(3r-h)[/MATH]
And so the volume \(V\) of a sphere, less the cap, can be given by:

[MATH]V=\frac{4}{3}\pi r^3-\frac{\pi h^2}{3}(3r-h)=\frac{\pi}{3}\left(4r^3-h^2(3r-h)\right)[/MATH]
 
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