Question

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

Answer 1

its fairly easy

Answer 2

let the inner integral by say, J

Answer 3

\[J = \int\limits_{0}^{2} (1-6x^2 y ) dx = [ (x - 2x^3 y ) ] \]

evaluated at x=2, then substract it evaluated at x=0

Answer 4

remember the inner integral, we have dx on the end, so x is the variable, and we treat y as a constant

Answer 5

so J = ( [ 2 - 16y] - [ 0] )

Answer 6

now, let the outer integral be I \[I = \int\limits_{0}^{1} (2-16y) dy = [ 2y -8y^2 ] \]

between the limits

Answer 7

so final answer = ( [ 2(1) -8(1)] - [ 0] ) = -6

Answer 8

something interesting to note, we get a negative answer

Answer 9

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"

Answer 10

wow thanks :) you're awesome ! :D