00:05This video will discuss the field due to a continuous distribution of electric charge
00:10volumetric, in relation to the topic of triple integral, here it will be in cylindrical coordinates,
00:15Also in MATLAB. Well, it's a field due to a continuous distribution of electric charge
00:21Volumetric force is the region around a charged object where the force manifests itself.
00:26The electrical force that this object exerts on other charges. As a central idea, the integral is performed
00:33triple, infinitesimal fields, differential of E, produced by each charge element
00:39differential load Q. We have the following types of load: linear, surface, volumetric.
00:51What is an infinitesimal field? It is a tiny portion of an electric field.
00:55often used in calculations to determine the total field of a continuous distribution of
01:00Charge, here we're analyzing a small portion of the electric field. And the total field, is the effect.
01:06from the vector combination of the electric fields that each charge creates in the area. It can be calculated
01:12using spherical and cylindrical coordinates. It is important to know the symmetry in order to perform
01:19Integration. Cylindrical coordinates; here, Cartesian coordinates are usually used.
01:24However, some problems present complicated symmetry, and it is more convenient to develop
01:29exercises using cylindrical coordinates, because it offers us an angular direction and
01:34radial. These are a three-dimensional system that denotes the point of the point in 3D using
01:39three coordinates.
01:43Here we have the integral of the charge Q, it is the integral of the volume, of ρ sub b of the
01:48The differential of V, the differential of E, signifies an integration over the entire volume e
01:53This implies a triple integration; however, it is usually indicated with a single symbol.
01:58integration. Thus ρ sub b, whose units are Coulomb per cubic meter, c over cubic meters,
02:05The problem is to find the volumetric charge of 4XYZ squared.
02:13Of limits in intervals of inequalities, ρ less than 2, 0 less than or equal to phi, less than or equal to pi means,
02:19or 90 degrees, also from 0, less than or equal to Z less than or equal to 3. It is observed that we have a value
02:26angular
02:27of 90 degrees, we perform them in cylindrical coordinates, where X is equal to ρ cosine of phi,
02:32And equal to ρ sine of phi, Z equal to Z. Now we solve the triple integral, replacing the
02:38limits, from 2 to 0, phi, pi means to 0, 3 to 0, of the volumetric density by the differential
02:46B.
02:47We substitute into the density function to find, that is, 4XYZ squared, to 4 ρ cosine of phi sine of
02:54phi
02:54by Z squared. We perform the integral, performing the equality, in cylindrical coordinates, the integral
03:00triple of 4 times ρ times cosine of phi times sine of phi Z squared, this equals 4XYZ,
03:06this is
03:06valid, since by integrating we can develop the substitution of the problem, and we carry out the
03:11integration by Z, and the rest as constants. Importantly, be careful here, the limits of
03:17Integration, why not substitute with zero? Because when we multiply by zero, we get zero.
03:25There are cases in the definite integral when substituting zero, in angles, and trigonometric ratios.
03:30fundamental, if the substitution by zero must be made. Performing the integral, for Z, gives us Z cubed over 3,
03:39From limits 3 to zero, this gives us 36, as a constant, continuing and replacing, since
03:46We have phi in the differential, we perform, ρ times cosine of phi times sine of phi, of the differential.
03:54Now, in the integrand, factoring out ρ, from cosine of phi times sine of phi, we have a
04:00Trigonometric equality, that is, the sine of twice the angle is equal to the cosine of the angle times
04:05sine of the angle. Now we have, 36 or half as a constant, we perform the sine integral, for phi
04:12by the differential of ρ, we obtain, minus cosine, of limits pi mean, up to zero, by the differential
04:18of ρ. Integrating and substituting the limits, we obtain, in the integrand, negative cosine
04:25two times pi halves, minus, in parentheses minus cosine of two times zero, we obtain, the sum of
04:31One plus one gives us two. Now 36 or half of the definite integral from two to zero of two
04:38differential ρ, important here, we treat it as a constant, and perform the integration for ρ by the
04:44differential of ρ, operating and substituting we obtain 72 coulombs. In MATLAB, for the general graph in
04:53spherical coordinates, not cylindrical, taking as an example we have the equalities for x and z,
04:58This in turn implies that the radius, r, will be the same in each of the conversion equations.
05:03Cartesian to spherical coordinates. Thus we have the following general code where we type,
05:08expression, expression, expression, expression, parameter variable 1, initial value, final value,
05:15variable with parameter 2, initial value, final value, this will give us a sphere, in general,
05:21Here you can vary the intervals of phi and theta. In MATLAB, we type the expression,
05:31expression, expression, goes the equalities in x, y, z, in that order, but in spherical, variable of
05:38Parameter 1 corresponds to phi, the initial value; this is where the limits go, telling us that phi goes from 0 to
05:44pi, u is for the angle phi and b is for the values of theta, initial value, final value, va
05:50from 0 to
05:50360 degrees.
06:04We press enter and we get a sphere.
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