Advanced Calculus fi..

Chapter 5 Vector Integral Calculus 3133. Let a wire be given as a curve C in space. Let its density (mass per unit length) beS = S(x, y, z), where (x, y, z) is a variable point in C. Justify the following formulas:a) length of wire = /, ds = L;b) mass of wire = Jc S ds = M;C) center of mass of the wire is (X, j, i), whered) moment of inertia of the wire about the z axis is 1JI, = (x2 + y2) S ds.C4. Formulate and justify the formulas analogous to those of Problem 3 for the surface area,mass, center of mass, and moment of inertia of a thin curved sheet of metal forming asurface S in space.5. Evaluate the following surface integrals:a) SSs x dy dz + y dz dx + z dx dy, where S is the triangle with vertices (1, 0, O),(0, 1, O), (0,0, 1) and the normal points away from (O,0,0);b) //, dy dz + dz dx + dx dy, where S is the hemisphere z = d m , x2 +y2 ( 1 , and the normal is the upper normal;c) SSs(x cos a + y cos + z cos y) du for the surface of part (b);d) SSs x2z do, where S IS the cylindrical surface x2 + y2 = 1 , 0 5 z 5 1.6. Evaluate the surface integrals of Problem 5, using the parametric representation:a)x=u+v, y=u-v, z=1-2ub) x = sin u cos v, y = sin u sin v, z = cos uC) same as (b)d) x =cosu, y =sinu, z = v7. Evaluate the surface integrals:a) SS, w . ndu, if w = xy2zi - 2x3j + yz2k, S is the surface z = 1 - x2 - y2, x2 +y2 5 1, and n is upper;b) SSs w . ndu, if w = i + 2j + 3k, S is the surface x = eU cosv, y = eu sinv,z = cos v sin v, 0 5 u 5 1 , 0 5 v 5 n/2, and n is given by (5.82) with the + sign; i *C) ISS 9 da if w = x ~ and ~ S and ~ n are z as in (a);d) SSs 2 da if w = n2 - y2 + z2 and S and n are as in (b);e) //,curlu . ndu if u = yzi - xzj + xzk, S is the triangle with vertices (1, 2, 8),(3, 1, 9). (2, 1, 7) and n is upper.8. a) Let a surface S: z = f (x, y) be defined by an implicit equation F(x, y, z) = 0. show^that the surface integral H da over S becomesrlIprovided that # 0.