Mathematical Methods for Physics and Engineering - Matematica.NET

6.3 APPLICATIONS OF MULTIPLE INTEGRALSthe total volume of the tetrahedron is given byV =∫ a0dx∫ b−bx/a0dy∫ c(1−y/b−x/a)0dz, (6.8)which clearly gives the same result as above. This method is illustrated further inthe following example.◮Find the volume of the region bounded by the paraboloid z = x 2 + y 2z =2y.and the planeThe required region is shown in figure 6.4. In order to write the volume of the region inthe form (6.7), we must deduce the limits on each of the integrals. Since the integrationscan be performed in any order, let us first divide the region into vertical slabs of thicknessdy perpendicular to the y-axis, and then as shown in the figure we cut each slab intohorizontal strips of height dz, and each strip into elemental boxes of volume dV = dx dy dz.Integrating first with respect to x (adding up the elemental boxes to get a horizontal strip),the limits on x are x = − √ z − y 2 to x = √ z − y 2 . Now integrating with respect to z(adding up the strips to form a vertical slab) the limits on z are z = y 2 to z =2y. Finally,integrating with respect to y (adding up the slabs to obtain the required region), the limitson y are y =0andy = 2, the solutions of the simultaneous equations z =0 2 + y 2 andz =2y. So the volume of the region isV ==∫ 20∫ 20dy∫ 2yy 2∫ √ z−y 2dz dx =−√z−y 2dy [ 43 (z − y2 ) 3/2] z=2yz=y 2 =∫ 20∫ 20∫ 2ydy dz 2 √ z − y 2y 2dy 4 3 (2y − y2 ) 3/2 .The integral over y may be evaluated straightforwardly by making the substitution y =1+sinu, and gives V = π/2. ◭In general, when calculating the volume (area) of a region, the volume (area)elements need not be small boxes as in the previous example, but may be of anyconvenient shape. The latter is usually chosen to make evaluation of the integralas simple as possible.6.3.2 Masses, centres of mass and centroidsIt is sometimes necessary to calculate the mass of a given object having a nonuniformdensity. Symbolically, this mass is given simply by∫M = dM,where dM is the element of mass and the integral is taken over the extent of theobject. For a solid three-dimensional body the element of mass is just dM = ρdV,where dV is an element of volume and ρ is the variable density. For a laminarbody (i.e. a uniform sheet of material) the element of mass is dM = σdA,whereσ is the mass per unit area of the body and dA is an area element. Finally, fora body in the form of a thin wire we have dM = λds,whereλ is the mass per193