Question

Problem 4. (a) Let Co be the curve which forms the boundary of the shaded region Do below. Parametrize Co so that the curve is traversed once in a clockwise manner, starting from the point (-1,0) (if you need to parametrise different parts of the curve in different ways, adjust these partial parametrisations at the end so that they give you one continuous and piecewise smooth parametrisation of the entire curve Co). (b) Compute the line integral (x+1)ds (note that in part (b) you only need to compute the line integral; recall also that the value of the line integral is not affected by the parametrisation we use (as long as the curve, or part of the curve, along which the integral is considered, is traversed the same number of times according to each parametrisation we use); therefore here you could choose to use any parametrisation(s) that you find easiest to work with).

Answer 1

to on 40) co=e, w z ucz. .., := {(t,0) 1-1st sig C2:= {(t, 1-t) It goes from I to of = {(\-t, t) lost=1}... e = {(cost, sint) 1 • I3 t s} Ta = {(t+1, 0) 1-2 st of 9 = {1+t)) ostsi). es = {(Cos It, IL) *+ 2). This gives a continuos smooth spiecewise) of the curve paranetration c. [where t varies from -2 to 27. (6) ((x+1) as - Sa+as & Sixtilas + Axuds (61) ds. [In 9, x (t) = ++1, yet) = o]. a t +1+1) (a - (daj af

/ (+1) as [In 2, x (t) = 1-t, 46 47 = 47 1 x +Dds [In c3 x (t) = cost yet) = sin It = 3 (os Ft + i) / ( ( da je dit - cost + 1) FTSINFL) V + (1 cos Ft / dt = n cos It +) I dat 2 Adding 0, 0, @ We get, raudes 2+
When you parametrize a curve by (x(t),y(t)) then $ds=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^{2}}\ dt$ .

And limits of integration are the starting and ending values of t.