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

5 marks · Section B

Stokes' theorem · Vector Analysis · asked 10× in 14 yrs · Read the full method →

Question

Evaluate by Stokes’ theorem $\displaystyle\oint_C e^x\,dx+2y\,dy-dz$ , where $C$ is the curve $x^2+y^2=4,\ z=2$ .

Technique

Recognise $\vec F=(e^x,2y,-1)$ has zero curl (each component a function of its own variable); Stokes’ theorem then gives $0$ immediately.

Solution

Step 1 — Identify the vector field

The line integral is $\displaystyle\oint_C\vec F\cdot d\vec r$ with

\vec F=e^x\,\hat i+2y\,\hat j+(-1)\,\hat k,

since $\vec F\cdot d\vec r=e^x\,dx+2y\,dy-1\,dz$ .

Step 2 — Stokes’ theorem

For a surface $S$ bounded by $C$ with unit normal $\hat n$ ,

\oint_C\vec F\cdot d\vec r=\iint_S(\nabla\times\vec F)\cdot\hat n\,dS.

Compute the curl:

\nabla\times\vec F= \begin{vmatrix}\hat i&\hat j&\hat k\\[2pt] \partial_x&\partial_y&\partial_z\\[2pt] e^x&2y&-1\end{vmatrix} =\Big(\partial_y(-1)-\partial_z(2y),\ \partial_z(e^x)-\partial_x(-1),\ \partial_x(2y)-\partial_y(e^x)\Big)=(0,0,0).

Each component of $\vec F$ depends only on its own variable, so the field is irrotational: $\nabla\times\vec F=\vec 0$ .

Step 3 — Evaluate

Since $\nabla\times\vec F=\vec0$ , by Stokes’ theorem the surface integral is zero regardless of the chosen cap (e.g. the disk $x^2+y^2\le4,\ z=2$ with $\hat n=\hat k$ ):

Answer

\boxed{\;\oint_C e^x\,dx+2y\,dy-dz=0.\;}