00:01
Hello friends, in this question we are given with f vector which is equal to 2 root z i vector minus x j vector plus 2 root y k vector.
00:09
So, for the first question the c1 which is r of t that is t i vector t j vector plus t k vector.
00:15
So, t lies between 0 and 1.
00:17
So, from this we get x equal to t, y is equal to t, z equal to t and dx equal to dt, dy which is equal to dt and dz equal to dt.
00:25
So, integration over the region c1 f vector dot dr vector which is equal to product of that is dot product of the given f vector into dx i vector plus dy j vector plus dz k vector.
00:41
So, from this we get the integral to the limit 0 to 1 2 root t dt minus t dt plus 2 root t dt.
00:48
So, after integrating and substituting the limit we get the value for this question the over the c1 which is the answer for this is 13 by 16.
00:57
The second question the c2 is r of t from r of t we are given with t i vector plus t square j vector plus t to the power 4 k vector, where t lies between 0 and 1.
01:08
From this we get dx equal to dt, dy which is equal to 2t and dz which is equal to 4 t square.
01:16
So, we can write this into integral 0 to 1 2t square dt minus t 2t dt plus 2t square into 4t square into t dt...