Let f(x, y) = x³ + y³. Consider the line segment from (0, 0)to (2, 2) using two different parametrizations:C1 : x = t, y = t, 0

1. The given function is: f(x,y)=x3+y3 Formula: Let C be a curve with parametric equation r(t), where a≤t≤b. The line integration of the function f(x,y) over the curve C is: ∫Cf(x,y) ds=∫abfx(t), y(t) r'(t) dt The parametric equation for the curve C1 is: r(t)=x(t), y(t)=t, t where 0≤t≤2. Find r'(t): r(t)=t, tr'(t)=(1, 1) Formula: (a,b)=a2+b2 Find r'(t): r'(t)=1+1=2f(x,y)=x3+y3 x(t)=t and y(t)=t Hence, f(x(t),y(t))=t3+t3=2t3∫Cf(x,y) ds=∫abfx(t), y(t) r'(t) dt f(x(t),y(t))=2t3, r'(t)=2 and r(t)=t, t, 0≤t≤2 Substitute the values in the integral: ∫C1f(x,y) ds=∫022t3 2 dt=22∫02t3 dt Formula: ∫abtn dt=tn+1n+! ∫C1f(x,y) ds=22t4402=22416-0=224=82 2. The parametric equation r1(t) of the curve C2 is: r1(t)=x(t), y(t)=(2t, 2t) where 0≤t≤1. Find r1'(t): r1(t)=(2t, 2t)r1'(t)=(2, 2) Find r1'(t): r1'(t)=22+22=8=22f(x,y)=x3+y3 x(t)=2t and y(t)=2t Hence, f(x(t),y(t))=(2t)3+(2t)3 =8t3+8t3=16t3∫Cf(x,y) ds=∫abfx(t), y(t) r'(t) dt f(x(t),y(t))=16t3, r1'(t)=22 and r1(t)=(2t, 2t), 0≤t≤1. ∫C2f(x,y) ds=∫0116t3 22 dt=1622∫01t3 dt=1622t4401=82(1)∫C2f(x,y) ds=82 and ∫C1f(x,y) ds=82 Hence ∫C1f(x,y) ds=∫C2f(x,y) ds1. ∫C1f(x,y) ds=82 2. ∫C1f(x,y) ds=∫C2f(x,y) ds