Find Volume Inside x^2+y^2+z^2=4 and (x+2)^2+(y-1)^2+(z+2)^2=4

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Discussion Overview

The discussion focuses on finding the volume of the intersection between two spheres defined by the equations \(x^2+y^2+z^2=4\) and \((x+2)^2+(y-1)^2+(z+2)^2=4\). The scope includes mathematical reasoning and integration techniques to determine the volume common to both spheres.

Discussion Character

  • Mathematical reasoning, Technical explanation

Main Points Raised

  • One participant suggests drawing and integrating to find the volume, proposing a result of \(\frac{11 \pi}{12}\).
  • Another participant clarifies that they are interested in the volume of the intersection specifically, asking for details on the bounds for the integral.
  • A different participant proposes a method to find the region of intersection by equating both equations and suggests that the "height" of each point in the intersection can be expressed as the difference between the two equations, leading to the formulation of the volume integral \(V=\iint\limits_{D}\left(z_1-z_2\right)dA\).

Areas of Agreement / Disagreement

Participants do not reach a consensus on the volume calculation, and multiple approaches to the problem are presented without agreement on the final result.

Contextual Notes

There are unresolved aspects regarding the specific bounds for the integral and the method of setting up the volume calculation, as well as potential assumptions about the intersection region.

real10
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x^2+y^2+z^2=4
(x+2)^2+(y-1)^2+(z+2)^2=4

find volume inside both.
thanks,
 
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Just draw and integrate.You will get
[tex]\frac{11 \pi}{12}[/tex]
 
i meant the volume of intersection due to those spheres.(common to both)
any details like what are the bounds for the integral u set up and which one..

thanks again,
 
I may be wrong, but try to find the region first by setting both equations equal. The "height" of each point in the intersection will be equal to the first equation minus the second.

[tex]z_1=\sqrt{2-x^2-y^2}[/tex]

[tex]z_2=\sqrt{4-(x+2)^2-(y-1)^2}-2[/tex]

[tex]V=\iint\limits_{D}\left(z_1-z_2\right)dA[/tex]
 

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