# Volume of a rectangle through a sphere

1. Nov 13, 2011

### Locoism

1. The problem statement, all variables and given/known data

1. The problem statement, all variables and given/known data
Suppose that a square hole with sides of length 2 is cut symmetrically through
the center of a sphere of radius 2. Show that the volume removed is given by

where

I'm not sure how to approach this, but I figure you can express the sphere in terms of f(x,y,z) = $x^2+y^2+z^2=4$
then I express z in terms of x and y
$\sqrt{4-y^2-x^2}$

But now I'm guessing I need to set up a double integral. Looking at bounds, -2<x<2 and -1<y<1,

$\int_{-2}^{2}\int_{-1}^{1}y\sqrt{4-x^2-y^2}dydx$

Is this the right way to go???
How then would I show it is given by the above integral?

2. Nov 13, 2011

### Dick

You are thinking about the hole going parallel to the z axis, right? Did you mean the integrand to be 2*sqrt(4-x^2-y^2)? And why should the x limits be different from the y limits if you want to cut out a square hole? The original question is a little confusing. You could just put F(x)=V. But I suspect they just want you do the dy integration and then use symmetry.

3. Nov 14, 2011

### HallsofIvy

Staff Emeritus

4. Nov 14, 2011

### jackmell

Show, yeah, that's right. Now you livin' in the big house:

$$\text{myblackarea(x)}=\int_{-1}^{1} 2\sqrt{4-x^2-y^2}dy$$

$$\text{myvolume}=\int_{0}^{1} \text{2 myblackarea(x)}dx$$

And if you take the time to learn how to do this, you'll never have a problem with these integrals ever again. My point is, the act of creating the plot, cultivates an intutitive understanding of the underlying mathematics. So tell your teacher I said ask everyone in class to draw this picture even if it takes them 6 hours to figure out how.

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Last edited: Nov 14, 2011
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