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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.

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  • $\begingroup$ 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). $\endgroup$
    – J.D
    Commented May 24 at 14:37
  • $\begingroup$ The question is correct. $\endgroup$ Commented May 24 at 14:58
  • $\begingroup$ The question as it stands cannot be correct as (0,1,2) and (1,0,2) do not lie on the curve. $\endgroup$
    – J.D
    Commented May 24 at 15:21

1 Answer 1

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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 $\mathrm{d}z = 0$).

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  • $\begingroup$ Made my life easier, thanks $\endgroup$ Commented May 24 at 14:57
  • $\begingroup$ It's a question from Calcutta honours examination (1985) $\endgroup$ Commented May 24 at 14:59

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