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anonymous
 5 years ago
Can anyone help me with Multiple Integration?
1 2
∫ ∫ (16x^2y) dxdy
0 0
anonymous
 5 years ago
Can anyone help me with Multiple Integration? 1 2 ∫ ∫ (16x^2y) dxdy 0 0

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let the inner integral by say, J

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[J = \int\limits_{0}^{2} (16x^2 y ) dx = [ (x  2x^3 y ) ] \] evaluated at x=2, then substract it evaluated at x=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0remember the inner integral, we have dx on the end, so x is the variable, and we treat y as a constant

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so J = ( [ 2  16y]  [ 0] )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now, let the outer integral be I \[I = \int\limits_{0}^{1} (216y) dy = [ 2y 8y^2 ] \] between the limits

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so final answer = ( [ 2(1) 8(1)]  [ 0] ) = 6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0something interesting to note, we get a negative answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0remember 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"

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow thanks :) you're awesome ! :D
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