Consider a vector field $F=xy^4 i +yz^2 j$. Its outward flux $\oint F.dS$ over the surface of a cube bounded by $|x|= 2,|y|= 2,|z|= 2 $ is nearest to

a) 410 b)-273 c)290 d)-300 e)0


I am confused here over the limits as I take the integral using divergence. So I did it taking limits to be from 0 to 2 as well as taking it from -2 to 2 and neither of them gets me the right answer. So This is how I did it:

$\oint F.dS$ = $\int \nabla.F\, dV$

$\nabla.F\ = y^4+z^2$

therefore $\int \nabla.F\, dV$ = $\int_V y^4+z^2 dx\, dy\, dz $

So now what should be the limits here? I do not understand what I should infer from |x|= 2 for the limits.

  • $\begingroup$ $V=\{-2\le x,y,z \le 2\}$ $\endgroup$
    – Nick
    May 2, 2017 at 7:10
  • $\begingroup$ Yes I got the answer. $\endgroup$
    – rahul rj
    May 2, 2017 at 7:19

1 Answer 1


The problem was trivial. Though I attempted with limits from -2 to 2 I didn't solve it completely thinking it might be wrong until it was affirmed to be right. So the solution is:

$\int \nabla.F\, dV$ = $\int_V y^4+z^2 dx\, dy\, dz$

$\int \nabla.F\, dV$ = $\iiint_{-2}^{2} y^4+z^2 dx\, dy\, dz$

$=\iint_{-2}^{2} 4y^4+4z^2 dy\, dz$

$=\int_{-2}^{2} 256/5+16z^2 dz$

= 290.13


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .