Question

Question 2 (30 points) Integrate f(x, y,2) xzv2-z2 - y2 over the path C, which consists of two curves, C1 and C2 from (1, 0, 0) to (1,0, 0), then to (-1,3, 0). Curve C1 is only half of the circle2 Curve C2 is a straight-line segment. The parametric equation for G is G: r! (t)-cos t î + sin t k, 0 π Find the line integral: Jcf(x. y,z)ds - (25 points) C2 (-1,3,0)

Answer 1

The given function in question is , The given path of integration is C , which C, and C, consists of the two given curves , First , we consi der the path C Here ,C is the semi - circle x4 + z4-1 , and it has the parametric equations as , with , In Cartesian coordinates , we can write , (iv) , J , £ , are the unit vectors along the Here , positive directions of the x , y , z axes respectively , in the three - dimensional Cartesian coordinate system From (ii) , (iv) , we get , after equating the coefficients of i , j, k, x=cos t , y=0 , z=sin t Using (v) , it is seen that , t=0:x=1,y=0,z=0 at and , at x=-1 , y=0 , z = 0 t=π: gain gives , dz d a (sint) (0) dt dt dt dt dt dt (vi) dx =-sin t dt , dy=0 , dz = cos t dt Then , we have , by formula , dt V(- sin t)* +(0) + (cos t) dt using (vi) -' (vii) sin 2 t + cos 2 t dt-J 1 dt = dt ds-

Again , using (v) in (i) . we get t ) = (cos t) (sin t) 2-(sin อั -0 (vi) →f(t)=J2-sin 2 tsin t cos t ; V111 Now , along the curve C1, from (1,0,0 1,0,0 , we get , (ix) 2 - sin t sin t cos t dt; 12 (viii) . (vii) . (ii) . and noting (*) . (**) after using 2-sin 2 t = u Let , Then dt dt →0-2sin t cos t =- dt (zi) →sin t cos t dt =-- Also . (x) gives (xi) (xi) Then , using (xiii) . (xii) . (xi). (x). in (ix) , we get t=0,u=2-sin 2 0-2-0-2 At X11 sin 2 π = 2-0-2 At t=7,11 = 2- X111 (xiv) ter using property of definite integrals at Now , we consider the path C , which is clearly a straight (xv) This ine jeins the points x)-(1.0) and (x2 , Уг)-(-1,3) in the xy-plane , for which z-0 line in the xy-plane , for which z=0

(xi , y.) Now , the equation of this line joining the points (, . У.) and 1S given by ; using the above data HV1 x+1 0 Now , using (xvi) . (xv). in (i) . we get (xvii) And , we get , by formula , dy using (xvi) dy (xvii) -1,0,0) to Now , along the path C3 , from -1,3,0) we get . using (xvi) , , XV111 33 03 X1X

Thus , the value of the given integral Jcf(x. y.z) ds . with usual notations , is given by =04 (-9) (xiv), (xix) using ; 0-9 (x , y ,z) is and this answers the given question