Can anyone help me with Multiple Integration? 1 2 ∫ ∫ (1-6x^2y) dxdy 0 0

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Can anyone help me with Multiple Integration? 1 2 ∫ ∫ (1-6x^2y) dxdy 0 0

Mathematics
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its fairly easy
let the inner integral by say, J
\[J = \int\limits_{0}^{2} (1-6x^2 y ) dx = [ (x - 2x^3 y ) ] \] evaluated at x=2, then substract it evaluated at x=0

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Other answers:

remember the inner integral, we have dx on the end, so x is the variable, and we treat y as a constant
so J = ( [ 2 - 16y] - [ 0] )
now, let the outer integral be I \[I = \int\limits_{0}^{1} (2-16y) dy = [ 2y -8y^2 ] \] between the limits
so final answer = ( [ 2(1) -8(1)] - [ 0] ) = -6
something interesting to note, we get a negative answer
remember the geometric meaning of the double integral, "the volume between the surface and the xy plane", well , in this case our surface is below the xy plane over our domain of integration , thus why we get a "negative volume"
wow thanks :) you're awesome ! :D

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