Step 1

Stoke's theorem given by

\(\displaystyle\int_{{C}}{F}.{d}{r}=\int\int_{{S}}{c}{u}{r}{l}{A}\times\hat{{{n}}}{d}{s}\)

According to the question,

A=−y i+x j

\(\displaystyle{C}{u}{r}{l}{A}={\left[\begin{array}{ccc} \hat{{{i}}}&\hat{{{j}}}&\hat{{{k}}}\\\frac{\partial}{{\partial{x}}}&\frac{\partial}{{\partial{y}}}&\frac{\partial}{{\partial{z}}}\\-{y}&{x}&{0}\end{array}\right]}\)

Step 2

Boundary condition of the curve C is,

\(\displaystyle\frac{{x}^{{2}}}{{a}^{{2}}}+\frac{{y}^{{2}}}{{b}^{{2}}}={1},{z}={0}\)

\(\displaystyle\hat{{{n}}}=\hat{{{k}}}\)

therefore,

\(\displaystyle{c}{u}{r}{l}{A}.{n}={2}\hat{{{k}}}.\hat{{{k}}}={2},{\left\langle.\hat{{{k}}}.\hat{{{k}}}={1}\right)}\)

Stoke's theorem given by

\(\displaystyle\int_{{C}}{F}.{d}{r}=\int\int_{{S}}{c}{u}{r}{l}{A}\times\hat{{{n}}}{d}{s}\)

\(\displaystyle=\int\int_{{S}}{2}{d}{s}\)

\(\displaystyle={2}{\left(\pi{a}{b}\right)}={2}\pi{a}{b}\)

Stoke's theorem given by

\(\displaystyle\int_{{C}}{F}.{d}{r}=\int\int_{{S}}{c}{u}{r}{l}{A}\times\hat{{{n}}}{d}{s}\)

According to the question,

A=−y i+x j

\(\displaystyle{C}{u}{r}{l}{A}={\left[\begin{array}{ccc} \hat{{{i}}}&\hat{{{j}}}&\hat{{{k}}}\\\frac{\partial}{{\partial{x}}}&\frac{\partial}{{\partial{y}}}&\frac{\partial}{{\partial{z}}}\\-{y}&{x}&{0}\end{array}\right]}\)

Step 2

Boundary condition of the curve C is,

\(\displaystyle\frac{{x}^{{2}}}{{a}^{{2}}}+\frac{{y}^{{2}}}{{b}^{{2}}}={1},{z}={0}\)

\(\displaystyle\hat{{{n}}}=\hat{{{k}}}\)

therefore,

\(\displaystyle{c}{u}{r}{l}{A}.{n}={2}\hat{{{k}}}.\hat{{{k}}}={2},{\left\langle.\hat{{{k}}}.\hat{{{k}}}={1}\right)}\)

Stoke's theorem given by

\(\displaystyle\int_{{C}}{F}.{d}{r}=\int\int_{{S}}{c}{u}{r}{l}{A}\times\hat{{{n}}}{d}{s}\)

\(\displaystyle=\int\int_{{S}}{2}{d}{s}\)

\(\displaystyle={2}{\left(\pi{a}{b}\right)}={2}\pi{a}{b}\)