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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To find the volume inside the 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 intersection volume can be calculated using integration. The volume of intersection is determined to be 11π/12. To set up the integral, the bounds must be established by equating the two equations, leading to the expressions for z1 and z2. The volume is then computed using the double integral V = ∬_D (z1 - z2) dA, where D represents the region of integration. This approach effectively captures the volume common to both spheres.
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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
\frac{11 \pi}{12}
 
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.

z_1=\sqrt{2-x^2-y^2}

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

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