Multiple Integrals II

  • Introduction

In the previous post, evaluation of double and triple integrals was discussed. These integrals are sometimes referred to as surface integrals and volume integrals in specific cases; e.g. Gauss’ divergence theorem, Stokes’ theorem etc.

In this post, we will see some applications of multiple integration in physics.

  • Area and Volume

The area occupied by/ bounded by 2 curves is obtained by substituting the function f(x,y) as a constant function equal to 1. Thus,

{A = \iint \limits_R dxdy = \iint \limits_R dA}

Limits are to be put appropriately.

The volume occupied by a closed surface is obtained by substituting the function f(x,y,z) as a constant function equal to 1. Thus,

{V = \iiint \limits_V dxdydz = \iiint \limits_V dV}

Again, limits are to be put according to the data given in the problem.

NOTE THAT to get area/volume of a symmetrical shape, things get simplified when the area/volume is evaluated over 1st quadrant/octant and multiplied with appropriate no. of divisions. e.g. volume of an ellipsoid given by the equation

{\frac {x^2}{a^2} + \frac {y^2}{b^2} + \frac {z^2}{c^2} = 1}

will be equal to 8 times its volume in the first octant.

We know that

{density \ (\rho) = \frac {mass \ (M)}{volume \ (V)}}

So, if \rho is a function of x,y or x,y,z,

{M = \iint \limits_R \rho (x,y) dxdy}

OR

{M = \iiint \limits_V \rho (x,y,z) dxdydz}

  • Mean and Root Mean Square (RMS) Values

The mean value of f(x,y) over region R is given by

{\mu = \frac {\iint \limits_R f(x,y) dxdy}{\iint \limits_R dxdy}}

The mean value of f(x,y,z) over volume V is given by

{\mu = \frac {\iiint \limits_V f(x,y,z) dxdydz}{\iiint \limits_V dxdydz}}

The RMS values are useful in electrical circuits. For a periodic function y=f(x) with period T, the RMS value is given by

{y_{RMS} = \sqrt {\frac {\int \limits_{C}^{C+T} [f(x)]^2 dx}{ \int \limits_C^{C+T} dx}}}

  • Center of Mass, Center of Gravity

The concepts of center of mass and center of gravity have already been discussed in mechanics. So, only mathematical formulation is given.

Consider an elementary mass dm and density \rho as a function of position of the mass. Now, the C.G. is calculated as

{\bar r = \frac {\int r dm}{\int dm}}

I) If C.G. of a line or an arc is to be computed,

{dm = \rho ds ,}

where ds is the arc length given by

{ds = \sqrt {1 + \Big( \frac {dy}{dx} \Big)^2} dx}

II) If C.G. of a planar laminar is to be computed,

{dm = \rho(x,y) dA = \rho (x,y) dxdy }

So, the double integral over the lamina needs to be evaluated.

III) If C.G. of a solid object is to be computed,

{dm = \rho (x,y,z) dV = \rho (x,y,z) dxdydz}

  • Moment of Inertia

The moment of inertia, I, is a quantity analogous to mass, used in analysis of rotating bodies.

{force = mass \times linear \ acceleration}

{torque = moment \ of \ inertia \times angular \ acceleration}

Mathematically, the M.I. of a body about an axis is the second moment of mass about that axis. In other words,

{I = \int r^2 dm,}

where r is the perpendicular distance of the mass dm from the axis.

The substitutions for dm will be as explained earlier.

Parallel Axes Theorem states that, if I_G is the M.I. of the body through its centroidal axis, then M.I. of the body about a parallel axis, which is at a distance d from the centroidal axis is

{I = I_G + m d^2,}

where m is the mass of the body.

Perpendicular Axes Theorem states that if I_{XX} and I_{YY} are M.I.s of an object about 2 perpendicular axes XX and YY respectively, then, M.I. of the object about an axis, which perpendicular to both XX and YY is given by

{I = I_{XX } +I_{YY}}

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