Evaluate 0∫4 0∫sqrt(16-x^2) 0∫sqrt(16-y^2-z^2) dxdydz
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Evaluate 0∫4 0∫sqrt(16-x^2) 0∫sqrt(16-y^2-z^2) dxdydz

[From: ] [author: ] [Date: 11-05-07] [Hit: ]
So,x^2 + y^2 + z^2 = 16 with x, y, z > 0.==> The integral equals (1/8) * [(4/3) π * 4^3] = 32π/3.I hope this helps!......
Show step by step because I do not understand this. Also, there is no f(x,y,z).

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Region of integration R: x = 0 to x = sqrt(16 - y^2 - y^2) with y = 0 to sqrt(16 - x^2) and x in [0, 4].
==> R is the region enclosed by x^2 + y^2 + z^2 = 16 in the first octant.

Note that ∫∫∫R 1 dV = Volume of R.

So, this is nothing more than the volume of the sphere
x^2 + y^2 + z^2 = 16 with x, y, z > 0.
==> The integral equals (1/8) * [(4/3) π * 4^3] = 32π/3.

I hope this helps!
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keywords: sqrt,dxdydz,16,int,Evaluate,Evaluate 0∫4 0∫sqrt(16-x^2) 0∫sqrt(16-y^2-z^2) dxdydz
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