# Evaluation of the given line integral

Question: Evaluate $$\int_{C}$$B.dr along the curve $$x^{2}$$+$$y^{2}$$=1,$$z$$= 1 in the positive direction from (0,1,2) to (1,0,2);given B= (xz²+y)i+(z-y)j+(xy-z)k

The question itself is easy,but I don't know how to handle z=1

Here's my attempt:-

$$\int_{C}$$B.dr= $$\int_{C}$$(xz²+y)i+(z-y)j+(xy-z)k.(dxi+dyj+dzk)

=$$\int_{C}$$(xz²+y)dx+$$\int_{C}$$(z-y)dy+$$\int_{C}$$(xy-z)dz

Should I put z=1 in the above integral?after this I will integrate all the three integrals and put up the values given in the question.

• You should use the parametrisation $x=sin(t),y=cos(t),z=2, 0\leq t \leq \frac{\pi}{2}$ ( I have written "$z=2$" as your coordinates suggest that we are on the curve $x^2+y^2=1, z=2$, but it contradicts your curve written in the question, so you may want to check you have written the question correctly).
– J.D
Commented May 24 at 14:37
• The question is correct. Commented May 24 at 14:58
• The question as it stands cannot be correct as (0,1,2) and (1,0,2) do not lie on the curve.
– J.D
Commented May 24 at 15:21

As J.D pointed out, yes, it looks like the coordinates tell $$z = 2$$ while your question says $$z = 1$$. So there might be something to check in your notes.
In either case, this does not matter much since $$z = \text{cst}$$. So the answer to your question is: yes. You can replace $$z$$ by its constant value in your above integrals (and you will have $$\mathrm{d}z = 0$$, so that the third integral vanishes, since $$z \equiv$$ constant).
The reason for that is that your curve is defined on a slice of constant height ($$z = 1$$ or $$z = 2$$, depending on what your question says). So in a way, your curve $$x^2 + y^2 = 1, z = 1$$ is simply a curve in the plane $$\{z = 1\}$$. And as such, you can forget about the third dimension, and do everything like it were on the plane. (By setting $$z = 1$$ wherever $$z$$ appears, and $$\mathrm{d}z = 0$$).