Evaluate the line integral int c f ds where f x y z ezi exyj...

Evaluate the line integral ∫C F · ds, where F(x, y, z) = (e^z)i + (e^(x−y))j + (e^y)k, and C is the path consisting of straight line segments from (0, 0, 0) to (0, 0, 1) and then from (0, 0, 1) to (0, 1, 1).

Transcript

00:01 In this question consider the path c1 which is line which is line from 0 0 0 to 0 0 1.

00:15 Now here x will be 0 y will be 0 z will be z and z is running from z limit is z is running from 0 to 1 therefore d r is d x i plus d y j plus d z k equals to 0 i plus 0 j plus d z k this will be equals to d z k.

00:52 So f of x comma y comma z will be e raise to the power x i plus e raise to the power x minus y j plus e raise to the power y k.

01:07 So f of 0 0 z will be equals to here it will be anything raise to the power 0 is the 1 into y that will be i plus j and here it will be k.

01:22 Now f of f of 0 0 z dot d r will be equals to d z will be equals to d z.

01:35 Now integral integral over curve c1 f dot d r so we have to find out that will be equals to integration from 0 to 1 d z.

01:49 So that will be equals to z and on substituting the limits we get 1 minus 0 which is equals to 1.

01:57 Now in next case we will be considering the path consider path consider path c2 the line c2 line from line from 0 0 1 to 0 1 1.

02:18 Now here the parametric equation will be here it will be x equals to 0 y equals to y and z equals to 1 where z is running from where z where y is running from 0 to 1.

02:41 So therefore d r will be equals to d x y plus d y j plus d z k and this is equals to 0 i plus d y j plus 0 k and this is equals to d y j.

03:00 Now in the next case we will get f of x comma y comma z will be e x i plus e x minus y j plus e raised to the power y k it is given then f of 0 y 1 will be equals to i plus e raised to the power minus y j and plus e raised to the power y k.

03:30 So from here we get f 0 y 1 dot d r that will be equals to e raised to the power minus y d y...

Question

Evaluate the line integral ∫C F · ds, where F(x, y, z) = (e^z)i + (e^(x−y))j + (e^y)k, and C is the path consisting of straight line segments from (0, 0, 0) to (0, 0, 1) and then from (0, 0, 1) to (0, 1, 1).

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