Q. Evaluate by Gauss divergence theorem $$\iint_S xz^2\,dy\,dz + \left(x^2y-z^3\right)\,dz\,dx + \left(2xy+y^2z\right)\,dx\,dy$$ where $S$ is the surface bounded by $z=0$ and $z=\sqrt{a^2-x^2-y^2}$.

I was not able to understand this question properly and when I solved this question I got the answer (2/3)π(a^5). Is this answer correct? In this question, I solve like this

∫∫∫ divF dv
div F =(x^2+y^2+z^2)
∫∫∫(x^2+y^2+z^2) dzdydx
a^2∫∫∫1 dzdydx from limit z=0 to z=(a^2-x^2-y^2)
               from limit y=-(a^2-x^2) to y=(a^2-x^2)        
                    limit x=-a to x=a

In this way, I got (2/3)π(a^5) as the answer.

  • 3
    $\begingroup$ Please use MathJax to format your question. It's very hard to read $\endgroup$
    – Andrei
    Mar 4 at 20:03
  • $\begingroup$ I can hardly read this. But you seem to be making the mistake of saying that the divergence is constant throughout the region. It is not. It is only constant on one piece of the boundary of the region. One last comment. You want to be working in spherical coordinates when you actually do the integral. $\endgroup$ Mar 4 at 20:14
  • $\begingroup$ Thank you for replying..Actually I 'm getting divergence= (x^2+ y^2+z^2) which I take as a^2.. I don't know how to proceed further..Can you solve it? $\endgroup$ Mar 4 at 20:19
  • 1
    $\begingroup$ Did you read my comment? The divergence is NOT $a^2$. $\endgroup$ Mar 4 at 20:20
  • 1
    $\begingroup$ The answer is that we expect you to integrate your divergence over half of the ball with radius $a\,.$ $\endgroup$
    – Kurt G.
    Mar 4 at 21:19

1 Answer 1


The integral in your post is over a closed volume bounded by half a sphere of radius $a$. By divergence theorem, $$\begin{aligned} \oint_{S}\langle xz^2,x^2y-z^3,2xy+y^2z\rangle\cdot\mathrm d\mathbf S&=\iiint_V\text{div}\langle xz^2,x^2y-z^3,2xy+y^2z\rangle \mathrm dV\\ &=\iiint_V(z^2+ x^2+y^2)\mathrm dV\\ &=\iiint_V r^2dV\\ &=\int_0^{2\pi}\int_0^{\pi/2}\int_0^a r^2r^2\sin\theta\mathrm dr\mathrm d\theta\mathrm d\Phi\\ &=2\pi\int_0^{\pi/2}\int_0^a r^4\sin\theta\mathrm dr\mathrm d\theta\\ &=2\pi\int_0^{\pi/2}\sin\theta\mathrm d\theta\int_0^a r^4\mathrm dr\\ &=2\pi\left[-\cos\theta\right]_0^{\pi/2}\left[\frac{r^5}{5}\right]_0^a=\frac{2\pi}{5}a^5 \end{aligned} $$ Notice how the divergence is $x^2+y^2+z^2=r^2\neq a^2$ because you need to integrate over $r\in[0,a]$. If not what you'd be doing is integrating over a surface and not a volume, and thus, divergence theorem wouldn't hold.

  • $\begingroup$ On youtube..I got answer as π/5 a^5..I don't know which one is correct.. $\endgroup$ Mar 4 at 22:58
  • $\begingroup$ @AnishBangia This answer here is crystal clear. If you think it is incorrect you should clarify why exactly. We have no idea what youtube videos you are watching. Don't even want to know them. $\endgroup$
    – Kurt G.
    Mar 5 at 12:12
  • 1
    $\begingroup$ $$\int _{-a}^{+a}\int _{-\sqrt{a^2-x^2}}^{+\sqrt{a^2-x^2}}\int _0^{\sqrt{a^2-x^2-y^2}}(x^2+y^2+z^2)dzdydx=\frac{2 \pi}{5}a^5$$ Same result. $\endgroup$
    – gpmath
    Mar 5 at 18:57
  • $\begingroup$ @KurtG. We are here to solve problems..okay....I'm not here to show you videos..even if you ask about videos..then also I will not show you..because you are a rude guy..I am nice to everyone here..you should be nice too..don't spread negativity here.. $\endgroup$ Mar 6 at 13:17
  • $\begingroup$ @AnishBangia that was not my intention but I stand by my statement that it is not helpful to claim that some youtube video got $\pi /5 a^5\,.$ You have received an answer with two upvotes an that is what we should work on if necessary. $\endgroup$
    – Kurt G.
    Mar 6 at 13:25

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