Answer to Question #347219 in Calculus for Nina

Question

Evalu‎ate ∫_C (x + 2y) ds, w‎here C is the cu‎rve defi‎ned by y = √(4 − x²), for x ∈ [0, 1].

Answer 1

ANSWER $\int_{C}(x+2y) ds=$ $8-2\sqrt{3} \cong4.5359$

EXPLANATION

To calculate the linear integral , we transform the curve $C$ ( it is a circle $x^{2}+y^{2}=4$ ) by parametric equation

$x(t)=2 \cdot \cos t, y(t)=2\cdot \sin t$

Changing the variable $x$ from to $1$ corresponds to $t\in [\frac {\pi}{3},\frac {\pi}{2}]$ .

Then the line integral is

$\int_{C}(x+2y) ds=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\left ( x(t)+2y(t) \right )\sqrt{\left ( x'\left ( t \right ) \right )^{2}+\left ( y'\left ( t \right ) \right )^{2}} dt$

Since $x'(t )=-2\cdot\sin t, y'(t)=2\cdot \cos t$ , then $\sqrt{\left ( x'\left ( t \right ) \right )^{2}+\left ( y'\left ( t \right ) \right )^{2}} = \sqrt{4 ( \sin t )^{2}+4 ( \cos t )^{2}} =2$ .

So,

$\int_{C}(x+2y) ds=2\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\left (2\cos t +4\sin t \right )dt=$ $=2 \left [2\sin t-4\cos t \right ] _{\frac{\pi}{3}}^{\frac{\pi}{2}}=2 \left [2\sin \frac{\pi }{2}-4\cos \frac{\pi }{2}- 2\sin \frac{\pi }{3}+4\cos \frac{\pi }{3} \right ]=\\=2\left ( 2-0-2\cdot \frac{\sqrt{3}}{2} +\frac{4}{2}\right )=8-2\sqrt{3} \cong4.5359$