00:01
Now, we have to calculate the work done by the force field f x y z equal to x square plus 10 z square i cap plus y square plus 14 x square j cap plus z square plus 12 y square k cap.
00:22
Now, when particle moves around the edge of the part of the sphere x square plus y square plus z square is equal to 16 that lies in the first octant.
00:34
So, now, r t parameterization will be r t is equal to 4 cos t i cap plus 4 sin t j cap plus 0 k cap t lying from 0 to pi by 2 since it is first octant.
00:49
So, r derivative t comes out to be minus 4 sin t i cap plus 4 cos t j cap.
00:56
Now, f of r t is equal to 16 cos square t plus 10 0 square i cap plus 16 sin square t plus 14 into 16 cos square t j cap plus 0 square plus 12 into 16 sin square t k cap.
01:24
So, this is simply equal to 16 cos square t i cap plus 16 sin square t plus 224 cos square t j cap plus 192 sin square t k cap.
01:40
So, work done is given by the formula integral a to b f r t r derivative t dt.
01:48
So, in this case this is simply equal to 0 to pi by 2 minus 64 sin t cos square t on taking the dot product plus 64 sin square t cos t plus 896 cos cube t dt.
02:14
So, this is simply equal to 64 by 3 cos cube t.
02:19
Now, over here we let cos t is equal to 0...