ANSWER ∫C(x+2y)ds= 8−23≅4.5359
EXPLANATION
To calculate the linear integral , we transform the curve C ( it is a circle x2+y2=4 ) by parametric equation
x(t)=2⋅cost,y(t)=2⋅sint
Changing the variable x from to 1 corresponds to t∈[3π,2π] .
Then the line integral is
∫C(x+2y)ds=∫3π2π(x(t)+2y(t))(x′(t))2+(y′(t))2dt
Since x′(t)=−2⋅sint,y′(t)=2⋅cost , then (x′(t))2+(y′(t))2=4(sint)2+4(cost)2=2 .
So,
∫C(x+2y)ds=2∫3π2π(2cost+4sint)dt= =2[2sint−4cost]3π2π=2[2sin2π−4cos2π−2sin3π+4cos3π]==2(2−0−2⋅23+24)=8−23≅4.5359
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