## anonymous 5 years ago Can anyone help me with Multiple Integration? 1 2 ∫ ∫ (1-6x^2y) dxdy 0 0

1. anonymous

its fairly easy

2. anonymous

let the inner integral by say, J

3. anonymous

$J = \int\limits_{0}^{2} (1-6x^2 y ) dx = [ (x - 2x^3 y ) ]$ evaluated at x=2, then substract it evaluated at x=0

4. anonymous

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

5. anonymous

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

6. anonymous

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

7. anonymous

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

8. anonymous

something interesting to note, we get a negative answer

9. anonymous

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"

10. anonymous

wow thanks :) you're awesome ! :D