# Evaluate line integral

I've been trying to work out this problem - the answer's $28$, but I can't understand how my textbook gets to that.

I have Vector $\vec F(x,y,z) = yz^2\hat i + xz^2\hat j + 2xyz\hat k$ and $C$ is the path from $(-1,2,-2)$ to $(1,5,2)$ that consists of three line segments parallel to the $z$-axis, the $x$-axis and the $y$-axis.

This implies that I should first integrate the whole thing by parametrizing for $z$ and holding $x$ and $y$ at $0$, then setting $z$ to $2$ and parametrizing for $x$ and then finally parametrizing for $y$ with the values for $x$ and $z$ set.

In all cases this ultimately comes down to an integral of $xz^2$ over some parametrization of $y$ - but how do I select which parametrization? I don't think that this Force field is conservative, so path independence does not occur.

Thanks

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I am going to write out a solution here, since some incorrect conclusions were drawn by the OP. The vector field $\overrightarrow{F} = \ <yz^2, xz^2, 2xyz>$ is in fact conservative, as the curl calculation yields

$$\overrightarrow{\nabla} \times \overrightarrow{F} \ = \ < 2xz-2xz\ , \ -2yz - (-2yz) \ , \ z^2 - z^2> \ = \ \overrightarrow{0} .$$

So the line integral $\int_C \ (yz^2) dx \ + \ (xz^2) dy \ + \ (2xyz) dz \$ from $A(-1,2,-2)$ to $B(1,5,2)$ is path-independent. $$\\$$

Following the path $C$ specified in the problem statement, we have, for the three segments,

$$\int_{C_1} \ (yz^2) dx \ + \ (xz^2) dy \ + \ (2xyz) dz \ \ [dx = 0 \ , \ dy = 0 ]$$

$$+ \ \int_{C_2} \ (yz^2) dx \ + \ (xz^2) dy \ + \ (2xyz) dz \ \ [dy = 0 \ , \ dz = 0 ]$$

$$+ \ \int_{C_3} \ (yz^2) dx \ + \ (xz^2) dy \ + \ (2xyz) dz \ \ [dx = 0 \ , \ dz = 0 ]$$

$$= \ \int_{-2}^2 \ (2xyz) dz \ + \ \int_{-1}^1 \ (yz^2) dx \ + \ \int_{2}^5 \ (xz^2) dy$$

$$= \ 0^{*} + \ (xyz^2) \ |_{-1, y=2,z=2}^1 \ + \ (xz^2y) \ |_{2, x=1,z=2}^5$$

[with the first integral equal to zero since an odd function of $z$ is being integrated over a path symmetric about the xy-plane]

$$= \ [ \ (1 \cdot 2 \cdot 2^2) \ - \ (-1 \cdot 2 \cdot 2^2) \ ] \ + \ [ \ (1 \cdot 2^2 \cdot 5) \ - \ (1 \cdot 2^2 \cdot 2) \ ]$$

$$= \ \ (8) \ - \ (-8) \ + \ (20) \ - \ (8) \ = \ 28 \ . \\$$

Because the integral is path-independent, we can verify this result by choosing a path $C'$ directly from $A$ to $B$ along the parametrized line segment

$$x \ = \ -1 + 2t \ \ [dx = 2 dt] \ , \ y \ = \ 2 + 3t \ \ [dy = 3 dt] \ , \ z \ = \ -2 + 4t \ \ [dz = 4 dt] \ ,$$

over the range $t: \ 0 \rightarrow 1 \ .$ Our line integral becomes

$$\int_0^1 \ (2 + 3t) \cdot (-2 + 4t)^2 \cdot 2 \ dt \ + \ (-1 + 2t) \cdot (-2 + 4t)^2 \cdot 3 \ dt \$$

$$+ \ 2 \cdot (-1 + 2t) \cdot (2 + 3t) \cdot (-2 + 4t) \cdot 4 \ dt \$$

[passing over a lot of tedious algebra]

$$= \ \int_0^1 \ 36 \ - \ 48t \ - \ 240t^2 \ + \ 384t^3 \ dt \ = \ ( 36t \ - \ 24t^2 \ - \ 80t^3 \ + \ 96t^4 ) \ |_0^1 \ = \ 28 \ .$$

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A late upvote is still an upvote, thanks for answering an old question. +1 – Shuhao Cao Aug 5 '13 at 20:27

Because the line integral is independent of the variable used to parametrize, you can use a different variable for each of the line segments, and then add the integrals from each together.

Assuming you solve the x segment, then the y segment, then the z segment:

For the x segment, $y=0, z=0 \Rightarrow dy=0, dz=0$, which should make the integral easy to evaluate from -1 to 1.

For the y segment, $x=x_1, z=0 \Rightarrow dx=0, dz=0$ which should create an integral in terms of y.

Do the same thing for z.

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I've tried this, but I don't get 28 – praks5432 Nov 13 '12 at 1:02
also, your method doesn't work because that will evaluate to 0 - what I tried was to move z with x = 0 and y = 0, then move x, with z held at 2 and with y = 0, then move y with z and x at their end values – praks5432 Nov 13 '12 at 1:03