Find the volume of the solid that lies inside both of the spheres? #x^2 + y^2 + z^2 + 4x − 2y + 4z + 5 = 0# & #x^2 + y^2 + z^2 = 4#

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Feb 11, 2018

Answer:

#V = 11/12pi#

Explanation:

Given the sphere:

#x^2 + y^2 + z^2 + 4x − 2y + 4z + 5 = 0#

To obtain the form #(x-a)^2+(y-b)^2+(z-c)^2 = r^2#, we complete the squares:

#x^2 + 4x + a^2 + y^2 − 2y + b^2+ z^2 + 4z + c^2 = -5 +a^2+b^2+c^2#

#-2ax=4x#

#a = -2#

#-2by= -2y#

#b = 1#

#-2cz=4z#

#c = -2#

The standard equation of the first sphere is:

#(x- (-2))^2+(y-1)^2+(z-(-2))^2= 2^2#

The center is #(-2,1,-2)# and the radius, # r= 2#

The standard equation of second sphere is:

#(x-0)^2+(y-0)^2+(z-0)^2 = 2^2#

The center is #(0,0,0)# and the radius, #R = 2#

The distance between the two centers is:

#d = sqrt((-2-0)^2+(1-0)^2+(-2-0)^2)#

#d = 3#

This the same volume as between the two spheres:

#x^2+y^2+z^2=2^2#
#(x - 3)^2+ y^2+z^2= 2^2#

I found equations for this special case where #R = r# and d lies along the x axis at the following Wolfram Mathworld reference source; equation 17:

#V = 1/12pi(4R+d)(2R-d)^2#

#V = 1/12pi(4(2)+3)(2(2)-3)^2#

#V = 11/12pi#

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