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UPSC 2021 Maths Optional Paper 1 Q7a — Step-by-Step Solution

20 marks · Section B

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

Question

Verify Gauss divergence theorem for $\vec F=2x^2 y\hat\imath-y^2\hat\jmath+4xz^2\hat k$ taken over the first-octant region bounded by $y^2+z^2=9$ and $x=2$.

Technique

Compute both sides directly; first-octant cylinder region split into 5 boundary surfaces; cylindrical coordinates for volume integral.

Solution

Region $V$: $\{(x,y,z):0\le x\le 2,\;y\ge 0,\;z\ge 0,\;y^2+z^2\le 9\}$.

Boundary surface $S$ consists of:

• $S_1$: front face $x=2$, $0\le y,z$, $y^2+z^2\le 9$ (quarter disc).
• $S_2$: back face $x=0$, same region (quarter disc).
• $S_3$: bottom face $z=0$, $0\le x\le 2$, $0\le y\le 3$.
• $S_4$: side face $y=0$, $0\le x\le 2$, $0\le z\le 3$.
• $S_5$: curved cylindrical surface $y^2+z^2=9$, $0\le x\le 2$, $y\ge 0$, $z\ge 0$ (quarter cylinder).

Step 1 — Compute $\nabla\cdot\vec F$

$\partial_x(2x^2 y)=4xy$. $\partial_y(-y^2)=-2y$. $\partial_z(4xz^2)=8xz$.

$\nabla\cdot\vec F=4xy-2y+8xz$.

Step 2 — Volume integral

Use cylindrical (in $y,z$): $y=r\cos\phi$, $z=r\sin\phi$, $r\in[0,3]$, $\phi\in[0,\pi/2]$ (first octant), $x\in[0,2]$.

$dV=r\,dr\,d\phi\,dx$.

$\iiint_V\nabla\cdot\vec F\,dV=\int_0^2\!\int_0^{\pi/2}\!\int_0^3(4xr\cos\phi-2r\cos\phi+8xr\sin\phi)\,r\,dr\,d\phi\,dx$.

$=\int_0^2\!\int_0^{\pi/2}\!\int_0^3 r^2[(4x-2)\cos\phi+8x\sin\phi]\,dr\,d\phi\,dx$.

$\int_0^3 r^2\,dr=9$.

$=9\int_0^2\!\int_0^{\pi/2}[(4x-2)\cos\phi+8x\sin\phi]\,d\phi\,dx$.

$\int_0^{\pi/2}\cos\phi\,d\phi=1$. $\int_0^{\pi/2}\sin\phi\,d\phi=1$.

$=9\int_0^2[(4x-2)+8x]\,dx=9\int_0^2(12x-2)\,dx=9[6x^2-2x]_0^2=9(24-4)=9\cdot 20=180$.

Step 3 — Surface integral (sum over 5 faces)

$S_1$: $x=2$, $\hat n=\hat\imath$. $\vec F\cdot\hat\imath=2x^2 y|_{x=2}=8y$.

$\iint_{S_1}8y\,dA=8\int_0^{\pi/2}\!\int_0^3 r\cos\phi\cdot r\,dr\,d\phi=8\int_0^{\pi/2}\cos\phi\,d\phi\int_0^3 r^2\,dr=8\cdot 1\cdot 9=72$.

$S_2$: $x=0$, $\hat n=-\hat\imath$. $\vec F\cdot(-\hat\imath)=-2x^2 y|_{x=0}=0$. Contribution: 0.

$S_3$: $z=0$, $\hat n=-\hat k$. $\vec F\cdot(-\hat k)=-4xz^2|_{z=0}=0$. Contribution: 0.

$S_4$: $y=0$, $\hat n=-\hat\jmath$. $\vec F\cdot(-\hat\jmath)=-(-y^2)|_{y=0}=0$. Contribution: 0.

$S_5$: $y^2+z^2=9$, $0\le x\le 2$, first octant. $\hat n=(0,y,z)/3$ (outward radial in $yz$-plane).

$\vec F\cdot\hat n=(2x^2 y\cdot 0+(-y^2)\cdot y/3+4xz^2\cdot z/3)=(-y^3+4xz^3)/3$.

Parametrize: $y=3\cos\phi,z=3\sin\phi$, $\phi\in[0,\pi/2]$.

$dS=3\,d\phi\,dx$ (surface element on cylinder, radius 3).

$y^3=27\cos^3\phi$, $z^3=27\sin^3\phi$.

$\vec F\cdot\hat n=(-27\cos^3\phi+4x\cdot 27\sin^3\phi)/3=-9\cos^3\phi+36x\sin^3\phi$.

$\iint_{S_5}=\int_0^2\!\int_0^{\pi/2}(-9\cos^3\phi+36x\sin^3\phi)\cdot 3\,d\phi\,dx$.

$\int_0^{\pi/2}\cos^3\phi\,d\phi=2/3$. $\int_0^{\pi/2}\sin^3\phi\,d\phi=2/3$.

$=\int_0^2(-27\cdot 2/3+108x\cdot 2/3)\,dx=\int_0^2(-18+72x)\,dx=[-18x+36x^2]_0^2=-36+144=108$.

Step 4 — Total surface integral

$\oint_S=72+0+0+0+108=180$.

Step 5 — Compare

Volume integral $=180$ = Surface integral $=180$. Gauss divergence theorem verified. ✓

Answer