How to evaluate $ \int_c(2x+y)\,ds$ where c is defined by $x^2+y^2=25$ from the point $(3,4)$ to $(4,3)$ why it gives me $-15$?

Question

If I parametrize the curve with $\begin{cases}x=5\cos t \\ y=5\sin t\end{cases}$ it gives me $-15$. Why if I parametrize the curve with $\begin{cases}x=t \\ y=\sqrt {25-t^2}\end{cases}$ the correct answer is $15$?

Answer 1

If you want to use the parametrization $\gamma(t) = (5 \cos t , 5 \sin t)$, you have to find out its domain as well. Your region is $$ C = \{(x , y) \in {\mathbb{R}}^2 : x^2 + y^2 = 25 , \quad 3 \leq x \leq 4 , \quad y > 0\}\mbox{.} $$ The module is clearly $5$, and the angle varies according to $\theta$, where this variables respects the two next inequalities: $$ 3 \leq 5 \cos \theta \leq 4 \qquad \mbox{ and } \qquad 5 \sin \theta > 0\mbox{.} $$ Then $\theta \in \left[\arccos \frac{4}{5} , \arccos \frac{3}{5}\right]$ and we are done: $$ \int_{C} (2 x + y) ds = 5 \int_{\arccos \frac{4}{5}}^{\arccos \frac{3}{5}} (2 \cos t + \sin t) \cdot 5 \sqrt{\cos^2 t + \sin^2 t} d t = 15\mbox{.} $$