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

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 :)
 

MarkFL

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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?
 

Jomo

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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??
 

MarkFL

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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??
We're trying to find just that portion of the sphere that is below the level of the water. :)
 

Sophdof1

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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?
I am sorry, I don't understand
 

Jomo

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Jomo

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I am sorry, I don't understand
Think about what you get if you rotate a portion of a circle around the correct axis for this problem.
 

Sophdof1

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Think about what you get if you rotate a portion of a circle around the correct axis for this problem.
I don't understand how that will help
 

MarkFL

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Suppose you consider the graph of:

\(\displaystyle f(x)=\sqrt{5^2-x^2}\)

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?
 

Otis

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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.

😎
 

Sophdof1

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Suppose you consider the graph of:

\(\displaystyle f(x)=\sqrt{5^2-x^2}\)

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?
Hi yes that makes a lot of sense
 

HallsofIvy

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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: \(\displaystyle \pi\int_{-10}^8(100- z^2)dz\).
 

MarkFL

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If I were going to answer this question, I would consider the first quadrant part of the curve:

\(\displaystyle f(x)=\sqrt{r^2-x^2}\)

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:

\(\displaystyle V_C=\pi\int_{r-h}^r \left(\sqrt{r^2-x^2}\right)^2\,dx=\pi\int_{r-h}^r r^2-x^2\,dx\)

Applying the FTOC, we obtain:

\(\displaystyle V_C=\frac{\pi}{3}\left[3r^2x-x^3\right]_{r-h}^r=\frac{\pi}{3}\left(2r^3-\left(3r^2(r-h)-(r-h)^3)\right)\right)\)

\(\displaystyle V_C=\frac{\pi}{3}\left(2r^3-(r-h)\left(3r^2-(r-h)^2)\right)\right)\)

\(\displaystyle V_C=\frac{\pi}{3}\left(2r^3-(r-h)\left(2r^2+2rh-h^2)\right)\right)\)

\(\displaystyle V_C=\frac{\pi}{3}\left(2r^3-2r^3-2r^2h+rh^2+2r^2h+2rh^2-h^3\right)\)

\(\displaystyle V_C=\frac{\pi h^2}{3}(3r-h)\)

And so the volume \(V\) of a sphere, less the cap, can be given by:

\(\displaystyle V=\frac{4}{3}\pi r^3-\frac{\pi h^2}{3}(3r-h)=\frac{\pi}{3}\left(4r^3-h^2(3r-h)\right)\)
 
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