00:01
In the first part we have function f is equal to minus y minus e to the power of y multiplied by cos x i cap plus y minus e to the power y multiplied by sine x j cap.
00:19
Now the line integral f d r over c is equal to integration over circle c minus y minus e to the power y multiplied by cos x multiplied by dx plus y minus e to the power y, sine x, d, y.
00:45
Now using b's theorem, we get double integration over c.
00:58
Now, del over del x, we have y minus e to 3 power y, sine x, sine x minus sine x, minus, sine x minus del over del y minus y minus e to the power y cost sets d a so at the end we get double integration over c d a now the limits are now d a can be transformed into r d r d theta now limit of theta is zero to pi by two and limit of r is zero to under root cos 2 -theta.
01:42
So if we integrate and evaluate we get at the end 0.
01:48
This is the answer of the first part.
01:51
Now in the second part we have function f is equal to 1 divided by 6 multiplied by x squared x squared plus y square 4 cube i cam.
02:03
Now the lined integral over c f d r is equal to line integral over c f d r is equal to line integral over c 1 divided by 6 multiplied by x square plus y square whole cube t x...