For the given vector field
$$\vec{H(r)} = rcos( \phi - \frac{\pi}{4} ) \vec{ a_{r} } +sin \phi \vec{ a_{ \phi }} $$
a) Calculate line integral of $\vec{H(r)} $ over the close path $ \Gamma $ with corners at ABCD on xy-plane shown in Figure.
b) Confirm the result of the line integral by using Stokes' theorem.-
i solved but I couldn't verified, I don't understand where I am doing wrong.
My solution:
$\vec{H(r)} = rcos( \phi - \frac{\pi}{4} ) \vec{ a_{r} } +sin \phi \vec{ a_{ \phi }} $
a) $ \int_A^B \vec{H(r)}.\vec{ a_{r}} .dr |_{\phi = \frac{ \pi }{4}} + \int_B^C \vec{H(r)}.r.\vec{ a_{ \phi }} d \phi |_{r=1} +\int_C^D \vec{H(r)}\vec{ a_{ r }} dr |_{ \phi = \frac{ 3\pi }{4}} +\int_D^A \vec{H(r)}r\vec{ a_{ \phi }} d \phi |_{r=2}$
then,
$\int_{r=2}^1 rcos(\phi - \frac{ \pi }{4} ) |_{ \phi = \frac{ \pi }{4}} + \left(\int_{\frac{ 3\pi }{2}}^\frac{ \pi }{4} sin\phi.d \phi.r |_{r=2}] + \int_{\frac{ 3\pi }{4}}^\frac{ 3\pi }{2} sin \phi. d \phi. r ) |_{ r=2}\right ) + \left( \int_{r=1}^2 rcos(\phi - \frac{ \pi }{4} ) |_{ \phi = \frac{ 3\pi }{4}} \right ) + [\int_{\frac{ 3\pi }{2}}^\frac{ 3\pi }{4} sin\phi.d\phi.r |_{r=1} + \int_{\frac{ \pi }{4}}^\frac{ 3\pi }{2} sin\phi.d\phi.r |_{r=1}] $
$\therefore $
= $ \frac{ r^2 }{2} |_{ 2}^1 + \left (2(-cos\phi)|_{ \frac{ 3\pi }{2}}^\frac{ \pi }{4} + 2(-cos\phi)|_{ \frac{ 3\pi }{4}}^\frac{ 3\pi }{2} \right ) + \left ( (-cos\phi)|_{ \frac{ 3\pi }{2}}^\frac{ 3\pi }{4} + (-cos\phi)|_{ \frac{ \pi }{4}}^\frac{ 3\pi }{2} \right )$
=$-\frac{ 3 }{2} - \sqrt{2}$
b) $\bigtriangledown \times \vec{H} = \vec{a_z} \left ( {\frac{ sin \phi }{r}} - sin( \frac{ \pi }{4}- \phi) \right )$
$\int_{S} \bigtriangledown \times \vec{H}.d\vec{s} = rd \phi dr \left ( {\frac{ sin \phi }{r}} - sin( \frac{ \pi }{4}- \phi) \right )$ $\Rightarrow $
$ = \left( \int_{r=1}^2 \int_{\frac{ 3\pi }{2}}^ \frac { \pi }{4}sin \phi d \phi dr - \int_{r=1}^2 \int_{\frac{ 3\pi }{2}}^ \frac { \pi }{4} sin (\frac{ \pi }{4} - \phi) r d \phi dr \right ) + \left( \int_{r=1}^2 \int_{\frac { 3\pi }{4}}^\frac{ 3\pi }{2} sin \phi d \phi dr - \int_{r=1}^2 \int_{\frac { 3\pi }{4}}^\frac{ 3\pi }{2} sin (\frac{ \pi }{4} - \phi) r d \phi dr \right )$
then,
$ \left( (-cos\phi)|_{ \frac{ 3\pi }{2}}^\frac{ \pi }{4}.r|_{1}^2 - cos(\frac{ \pi }{4} - \phi) |_{ \frac{ 3\pi }{2}}^\frac{ \pi }{4}.\frac{ r^2 }{2} |_{ 1}^2 \right ) + \left( (-cos\phi)|_{ \frac{ 3\pi }{4}}^\frac{ 3\pi }{2}.r|_{1}^2 - cos(\frac{ \pi }{4} - \phi) |_{ \frac{ 3\pi }{4}}^\frac{ 3\pi }{2}.\frac{ r^2 }{2} |_{ 1}^2 \right )$
$=-\frac{ 3 }{2} - \sqrt{2}$
Solution:
$ \int_{r=2}^1 r.cos(\phi-\frac{ \pi }{4}) .dr |_{\phi = \frac{ \pi }{4}} + \int_{\phi=\frac{ \pi }{4}}^\frac{ -5\pi }{4} sin\phi.d\phi.r |_{r=1} + \int_{\phi=\frac{ 3\pi }{4}}^\frac{ 9\pi }{4} sin\phi.d\phi.r |_{r=2} $
$=-\frac{ 3 }{2} - \sqrt{2}$ Line integral boundaries:
B $\Rightarrow $ C $\frac{ \pi }{4}, \frac{ -5\pi }{4}$
D $\Rightarrow $ A $\frac{ 3\pi }{4}, \frac{ 9\pi }{4}$
stokes' is verified, boundaries: $ \phi= \frac{ 3\pi }{4} \Rightarrow \frac{ 3\pi }{2}$ and $ \phi= \frac{ 3\pi }{2} \Rightarrow \frac{ \pi }{4}$