Evaluate the line integral ∫cxyds Where c is the given curve c:x=t2,y=2t,0≤t≤5

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Evaluate the line integral  ∫cxyds  Where c is the given curve  c:x=t2,y=2t,0≤t≤5

yagombyeR · Accepted Answer

ExplanationLine integrals can be used to eavaluate integrals of two or three dimensioal curvesThe curve c is expressed by the following parametric equations given byx=t2andy=2tThus, to evaluate the following integral given byI=∫cxydswe use the following line integral formula given by∫cf(x,y)ds=∫abf(x(t),y(t))|r′|dt=∫abf(x(t),y(t))|drdt|dt=∫abf(x(t),y(t))(dxdt)2+(dydt)2dtSince x(t)=t2 and y(t)=2tWe want to evaluate ∫cxyds as followsFirstly, we must find the integrand f(x(t),y(t)) as followsf(x(t),y(t))=xy=t2(2t)=2t3Also, the derivatives of x(f) and y(t) must beddtx(t)=ddt(t2)=2tandddty(t)=ddt(2t)=2Since 0≤t≤5, so that the upper and lower limits of the integral must bea=0andb=5Thus, using the line integral formula, we get∫cxyds=∫abf(x(t))(dxdt)2+(dydt)2dt=∫05(2t3)(2t)2+(2)2dt=2∫05t34t2+4dt=2∫05t34(t2+1)dt=2∫052t3(t2+1)dt=4∫05t3(t2+1)dtTo evaluate the following integral ∫05t3(t2+1)dt, we use the method of substitution as followsLet u=t2+1, so thatDifferentiate both sides of the following equation u=t2+1 implies thatdu=2tdtand thustdt=du2Also, we havet2=u−1Also, we find that the upper and lower limits of the integral must bet=0⇒u=02+1=1andt=5⇒u=52+1=26Thus, the line integral becomes∫cxyds=4∫t=05t2t2+1tdt=4∫u=126(u−1)u(du2)=42∫u=126(u−1)udu=2∫u=126(u32−u12)duThus, evaluating the following line integral implie

Evaluate the line integral \int_c xyds Where c is the given curve c:x=t^2,y=2t,0\leq t\leq5

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