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UPSC 2020 Maths Optional Paper 1 Q6b — Step-by-Step Solution

15 marks · Section B

Line integrals · Vector Analysis · asked 8× in 13 yrs · Read the full method →

Question

For the vector function $\vec A$, where $\vec A=(3x^2+6y)\hat i-14yz\,\hat j+20xz^2\,\hat k$, calculate $\int_C \vec A\cdot d\vec r$ from $(0,0,0)$ to $(1,1,1)$ along the following paths: (i) $x=t,\ y=t^2,\ z=t^3$ (ii) Straight lines joining $(0,0,0)$ to $(1,0,0)$, then to $(1,1,0)$ and then to $(1,1,1)$ (iii) Straight line joining $(0,0,0)$ to $(1,1,1)$ Is the result same in all the cases? Explain the reason.

Technique

Parameterize each path, reduce to a single-variable integral; test conservativeness via $\nabla\times\vec A$.

Solution

Path (i): $x=t,\ y=t^2,\ z=t^3$, $\;t:0\to1$

$dx=dt,\ dy=2t\,dt,\ dz=3t^2\,dt$. Substitute:

• $(3x^2+6y)dx=(3t^2+6t^2)\,dt=9t^2\,dt$,
• $-14yz\,dy=-14(t^2)(t^3)(2t)\,dt=-28t^6\,dt$,
• $20xz^2\,dz=20(t)(t^6)(3t^2)\,dt=60t^9\,dt$.

Path (ii): $(0,0,0)\to(1,0,0)\to(1,1,0)\to(1,1,1)$

Segment 1 $(0,0,0)\to(1,0,0)$: $y=0,z=0,\ dx=dx,\ dy=dz=0$, $x:0\to1$.

Segment 2 $(1,0,0)\to(1,1,0)$: $x=1,z=0$, only $dy$ term, $y:0\to1$.

Segment 3 $(1,1,0)\to(1,1,1)$: $x=1,y=1$, only $dz$ term, $z:0\to1$.

Total:

Path (iii): straight line $(0,0,0)\to(1,1,1)$: $x=y=z=t$, $t:0\to1$

$dx=dy=dz=dt$:

• $(3t^2+6t)\,dt$,
• $-14(t)(t)\,dt=-14t^2\,dt$,
• $20(t)(t^2)\,dt=20t^3\,dt$.

Answer