Question #347219

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


1
Expert's answer
2022-06-02T08:51:33-0400

ANSWER C(x+2y)ds=\int_{C}(x+2y) ds= 8234.53598-2\sqrt{3} \cong4.5359

EXPLANATION

To calculate the linear integral , we transform the curve CC ( it is a circle x2+y2=4x^{2}+y^{2}=4 ) by parametric equation

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

Changing the variable xx from to 11 corresponds to t[π3,π2]t\in [\frac {\pi}{3},\frac {\pi}{2}] .



Then the line integral is

C(x+2y)ds=π3π2(x(t)+2y(t))(x(t))2+(y(t))2dt\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)=2sint,y(t)=2costx'(t )=-2\cdot\sin t, y'(t)=2\cdot \cos t , then (x(t))2+(y(t))2=4(sint)2+4(cost)2=2\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,

C(x+2y)ds=2π3π2(2cost+4sint)dt=\int_{C}(x+2y) ds=2\int_{\frac{\pi}{3}}^{\frac{\pi}{2}}\left (2\cos t +4\sin t \right )dt= =2[2sint4cost]π3π2=2[2sinπ24cosπ22sinπ3+4cosπ3]==2(20232+42)=8234.5359=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


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