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April 22nd, 2015, 05:08 AM   #1
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Question HELP!! Maths problem has been bugging me for too long

Hello I am new to this forum and have been driven here by a problem that has been keeping me up for far too long now...

If you were to take a sphere and slice through it at any point, except the center line, what would be the formula for the volume of the section of sphere you have sliced off?

I have spent far too long trying to work out what seems to be a simple problem

Thus far I have established the derivative of the volume of a sphere to get the gradient at any point on the remaining section but am not totally sure this is the right track to be heading down! PLEASE HELP!
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April 22nd, 2015, 07:00 AM   #2
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Have you taken a Calculus class? That is the technique you would want to use.

Setting up a coordinate system so that the origin is at the center of the sphere, you can write its equation as $\displaystyle x^2+ y^2+ z^2= R^2$ where R is the radius of the sphere. We can take the plane slicing the sphere to be $\displaystyle z= z_0$.

That plane cuts the sphere in the circle $\displaystyle x^2+ y^2+ z_0^2= R^2$ or $\displaystyle x^2+ y^2= R^2- z_0^2$, a circle of radius $\displaystyle \sqrt{(R^2- z_0^2}$. That tells us that with x going from $\displaystyle -\sqrt{(R^2- z_0^2}$ to $\displaystyle \sqrt{(R^2- z_0^2}$, y goes from $\displaystyle -\sqrt{R^2- z_0^2- x^2}$ to $\displaystyle \sqrt{R^2- z_0^2- x^2}$. And, for each (x, y), the height will be $\displaystyle z= \sqrt{R^2- x^2- y^2}- y_0$.

We can find the volume of that by $\displaystyle \int_{-\sqrt{(R^2- z_0^2}}^{\sqrt{(R^2- z_0^2}} \int_{-\sqrt{R^2- z_0^2- x^2}}^{\sqrt{R^2- z_0^2- x^2}}\left(\sqrt{R^2- x^2- y^2}- y_0\right) dy dx$
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April 22nd, 2015, 07:29 AM   #3
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Rather than bothering with spherical coordinates, I would look at the volume of revolution about the $x$-axis between $x = -a$ and $x = b$ (where $0 \lt b \lt a$) of the semicircle $f(x) = \sqrt{a^2 - y^2}$.

Also, take a look here.
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