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I am stuck on the following problem:

Evaluate $\int_c xdx + ydy +zdz$ where $C$ is the line segment from $(4,1,1)$ to $(7,-2,4).$

I found the line equations (I believe that's what they're called) for $x, y,$ and $z$, getting $$x(t)=3t+4$$ $$y(t)=-3t+1$$ $$z(t)=3t+1$$ Then the derivatives equal $$x'(t)=z'(t)=3$$$$y'(t)=-3$$ And plugging it into $\int_c xdx + ydy +zdz$ gives: $$\int_c (3t+4)(3)dt$$ $$\int_c (-3t+1)(-3)dt$$ $$\int_c (3t+1)(3)dt$$ I got this question wrong so I'm probably way off the mark by now so I'll leave my thought process at that. I would appreciate if anyone can help, and also explain what the limits of the integral become (what to replace $C$ with).

mathjohnn
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1 Answers1

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First of all, the integral should be $\int_0^1$, since $t=0, 1$ corresponds to $(4, 1, 1)$ and $(7, -2, 4)$ respectively (this is why you this parametrization). Second, you should get

$$ \int_0^1 \left((3t+4) 3 + (-3t+1)(-3) + (3t+1) 3\right) dt$$

instead of three separate integrals. You should be able to find the answer easily.

Another way to tackle this is to notice that

$$ xdx + ydy+ zdz = \frac{1}{2} d(x^2 + y^2 + z^2)$$

is an exact form, so the line integral depends only on the endpoints

$$ \int_c (xdx + ydy + zdz) = \frac{1}{2} \left( 7^2 + (-2)^2 + 4^2 - (4^2 +1^2 + 1^2)\right).$$

Arctic Char
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  • Thank you, must have been too early in the morning for me to think straight! You made it look too easy haha – mathjohnn Jul 30 '20 at 16:28