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.
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.