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UPSC 2019 Maths Optional Paper 1 Q8c-i — Step-by-Step Solution

15 marks · Section B

Gauss divergence theorem · Vector Analysis · asked 10× in 14 yrs · Read the full method →

Question

State Gauss divergence theorem. Verify this theorem for $\vec F=4x\hat i-2y^2\hat j+z^2\hat k$, taken over the region bounded by $x^2+y^2=4,\ z=0$ and $z=3$.

Technique

Compute $\nabla\cdot\vec F=4-4y+2z$ and integrate over the cylinder (odd $y$-term vanishes by symmetry); compute the flux through the three faces separately and sum.

Solution

Statement (Gauss divergence theorem)

Let $V$ be a closed bounded region in $\mathbb R^3$ with piecewise-smooth boundary surface $S$ and outward unit normal $\hat n$. If $\vec F$ is continuously differentiable on $V$, then

Here $V$ is the solid cylinder $x^2+y^2\le4,\ 0\le z\le3$, and $S$ is its surface (top disk $z=3$, bottom disk $z=0$, lateral surface $x^2+y^2=4$). The radius is $2$.

Step 1 — Volume integral of the divergence

In cylindrical coordinates ($x=r\cos\theta,\ y=r\sin\theta$, $dV=r\,dr\,d\theta\,dz$), the $-4y$ term integrates to zero over a full revolution. So

(the $-4y=-4r\sin\theta$ term vanishes on $\int_0^{2\pi}\sin\theta\,d\theta=0$). The remaining integral factorises:

so

Step 2 — Surface integral (three faces)

Top disk $z=3$, $\hat n=+\hat k$: $\vec F\cdot\hat n=z^2=9$. Area $=\pi(2)^2=4\pi$:

Bottom disk $z=0$, $\hat n=-\hat k$: $\vec F\cdot\hat n=-z^2=0$:

Lateral surface $x^2+y^2=4$: outward normal $\hat n=\dfrac{(x,y,0)}{2}=(\cos\theta,\sin\theta,0)$, with $x=2\cos\theta,\ y=2\sin\theta$, surface element $dS=2\,d\theta\,dz$ (radius $\times d\theta\times dz$):

Over a full period $\int_0^{2\pi}\cos^2\theta\,d\theta=\pi$ and $\int_0^{2\pi}\sin^3\theta\,d\theta=0$, so the $\theta$-integral is $8\pi$. Then

Total flux:

Step 3 — Conclusion

Answer