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I need to calculate the following line integral over the curve C:

$$\int_C e^{\sqrt{x^2+y^2}} ds $$

where $C$ is the circuit bounded by $r = a, \varphi = 0$ and $\varphi = \pi/4$. I tried separating the integral into these three parts and got an answer of $(\pi/4)ae^a$ However, the answer the book gives (the problem is from Demidovich's Problems in Analysis) is

$$(\pi/4)ae^a + 2(e^a - 1)$$

The $e^a - 1$ term comes from evaluating the integral over the bounds of the curve $\varphi = 0$ and $\varphi= \pi/4$. But I thought these would cancel out, and not add up.

Jean Marie
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Nacho
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  • Please check that my Latex-ification (placing formulas between dollar signs etc.) of your text is correct. – Jean Marie Mar 07 '21 at 11:14
  • @nacho just noticed your field is a scalar field. So pls check that in your integral. – Math Lover Mar 07 '21 at 11:22
  • @MathLover are you saying the book's answer is actually right? It's a scalar field, yes, but I still think what I said holds true. – Nacho Mar 07 '21 at 11:55
  • Yes given it is a scalar field, it should not matter the direction. Regardless of direction, you are traversing through the same set of points. So integral should be same. That is why I said to check how you set up your integral. – Math Lover Mar 07 '21 at 12:20
  • You are evaluating the path length. If you go from A to B and then back, your path length will be $L=2|AB|$ – Svyatoslav Mar 07 '21 at 12:27
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    You should accept the result of Math Lover. – Jean Marie Mar 07 '21 at 23:08

1 Answers1

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a) Between point $(0,0)$ and $(a,0)$, $\vec r(t) = (t,0), 0 \leq t \leq a$.

$I_1 = \displaystyle \int_C f(\vec r(t)) \ |r'(t)| \ dt = \int_0^a e^t \ dt = e^a - 1$

b) Between point $(\frac a {\sqrt2},\frac a {\sqrt2})$ and $(0,0)$, $\vec r(t) = (\frac a {\sqrt2} - t, \frac a {\sqrt2} - t), 0 \leq t \leq \frac a {\sqrt2}$.

$I_2 = \displaystyle \int_0^{a / \sqrt2} \sqrt2 \ e^{(a - \sqrt2 t)} \ dt = e^a - 1$

c) over circular segment, as you calculated.

Math Lover
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