00:06This video will study double and integrated integrals; for this we have the
00:10The following concepts, double integral, is called the double integral of a function of two variables
00:15or equal to the function of x, y, extended to a region s, is designated by, the integral
00:21sub s, of the function x, y, by the differential of s, implies the double integral sub s, surface
00:27or region of the integrand function of x, y, for the geometric locus some are established
00:32Under these conditions, region s, figure 1, is arbitrarily divided into n, elementary regions,
00:382, inside or on the border, 'a' of each elementary region is chosen arbitrarily
00:43a point, y, x, y, y, y, 3, is considered u as a function of a point that can be given
00:49not only in a Cartesian coordinate system, 4, the value of the function or at this point function
00:55of x, y, y, y, is multiplied by the value of the differential area of s, obtaining the
01:01sum, 5, the limit of the sum obtained is found, a sigma from equal to 1, up to n, values of
01:07the function x, y, y, y, by the differential of s, when each region contracts towards
01:13a point, and therefore n implies infinity, here we can see the different n regions
01:18in 3D, as well as the behavior of a single region s, 6, tends to infinity,
01:24if this limit does not depend on the procedure of the region s in elementary regions and of
01:29The choice of points, y, x, y, y, y, then this is called the double integral of the function
01:35u, extended to region s, in this case s is called region, enclosure, domain or field
01:40From integration, we have the following integration equality, of the integral sub s, of the
01:45function x, y, is equal, when the limit of the increment of s tends to zero, and when
01:51the limit n tends to infinity, of the summation from n equal to 1, to n, as a function of
01:56x, y, by the differential of s, if the function x, y, is continuous throughout the region of integration
02:03closed, that is, including the points of the contour, then there exists an integral in
02:081. Cylindrical coordinates, in addition to the classic rectangular x, y, and z coordinate axes, we have
02:14Cylindrical coordinates are a three-dimensional coordinate system that generalizes
02:18the polar coordinates of the plane, adding a third coordinate to represent the
02:23height, its use is widespread in exact sciences, such as physics, mathematics, among other disciplines,
02:30Cylindrical coordinates definition: a point in space is represented in cylindrical coordinates
02:35by an ordered triple r, theta, z, where, r, is the radial distance from the point to the z-axis,
02:42that is, the distance from the point to a perpendicular projection onto the z-axis, theta,
02:48is the angle formed by the radius vector, measured counterclockwise, z, that is,
02:56is the height of the point above the x, y plane, conversion formulas, to convert between
03:01Cartesian coordinates, x, y, z, and cylindrical coordinates, r, theta, z, are used as follows
03:09formulas, from Cartesian to cylindrical, for r, equal to the square root of x squared,
03:15plus and square, so also theta is equal to the inverse tangent of x on y, z is equal to z.
03:22From cylindrical to Cartesian, we have x equals r times cosine of theta, and y equals r,
03:28by sine of theta, and also here z is equal to z, spherical coordinates, are a system of
03:34three-dimensional coordinates that use a distance from the origin, r, an angle from
03:39the positive z-axis, pi, and an angle from the positive x-axis in the x, y plane, theta, for
03:44specify
03:45the position of a point in space, spherical coordinate elements, r, radial distance from
03:52from the origin to the point, is always a non-negative value, r greater than or equal to zero, pi, angle
03:59polar or with latitude, measures the angle from the positive z-axis to the segment joining
04:05The origin with the point varies between zero and pi, zero less than or equal to pi less than or equal to pi,
04:10theta, asimuthal angle, measures the angle from the positive x-axis in the x-y plane to the projection
04:17of the segment that joins the origin with the point on that plane, varies between zero and two pi,
04:23zero less than or equal to theta less than or equal to two pi, conversion formulas, to convert
04:27Between spherical and Cartesian coordinates, the following formulas are used, from Cartesian coordinates
04:34For cylindrical functions, x equals rho times sine of phi times cosine of theta, so too y equals
04:41a ro times sine of phi times cosine of theta, in z, we have, z is equal to ro times cosine
04:46of phi,
04:47From cylindrical to Cartesian, we have, ro is equal to the square root of the summation
04:52The square of x, y, z, in phi is equal to the inverse cosine of z over ro, in the tangent
04:58We have the inverse theta and y over x, in mathematical terms, an iterated integral is an integral
05:04multiple that is calculated by successively evaluating integrals in one or more variables, that is,
05:10A multiple integral is decomposed into a sequence of simple integrals, let's suppose
05:15We want to calculate the volume under a surface z equal to the function x, y, over a region
05:21r, o s, in the x, y plane, if the region r, o s, can be described as a set
05:27of
05:27rectangles, we can evaluate this double integral as an iterated integral, which we do
05:33Here we integrate first with respect to a variable, for example, and, while holding
05:38The other variable, x, is a constant, then we integrate the result with respect to the other variable,
05:45Here we have an example in GeoGebra; what the iterated integral does is calculate the area.
05:49Under the curves of the functions established in the example, let's imagine we want to calculate
05:54the volume of an object with an irregular shape, instead of trying to find a single formula
05:59For the entire volume, we can slice it into many smaller pieces and calculate the
06:04volume of each slice individually, then we add the volume of all the slices to
06:10To obtain the total volume, this is, in essence, the concept behind iterated integrals.
06:19Thus we have the following exercise: calculate the iterated integral in spherical coordinates,
06:23as a function of ρ and θ, this is the double integral from the inverse tangent of 1.5 to 0,
06:28and another definite integral from secant of θ to 0, by the differentials of ρ and θ.
06:34The important thing here is to integrate first with respect to one variable, and the other can be constant.
06:40If it exists, in which case we integrate directly, with the variable ρ, and develop.
06:47But we have secant squared of θ which is equal to tangent θ plus e, and simplifying we get,
06:532 by the definite integral from the inverse tangent of 1.5 to 0, and finally we obtain,
06:592 that multiplies tangent of θ of limits of 56 degrees, to 0, of the definite integral,
07:05The answer obtained is 2.9, which is equivalent to 3. The following exercise tells us to calculate the
07:11iterated integral in spherical coordinates, of the double integral from b to b over 2,
07:16and of the definite integral of limits, and means, up to 0. Here we have two variables ρ and θ,
07:22We take ρ as a constant and integrate with respect to θ. Expanding, we obtain the definite integral that goes
07:28from b to b means of the integrand ρ times π means of the differential of ρ. Developing
07:33This integral gives us 3π/16, times b squared. The next exercise tells us,
07:39Calculate the iterated integral, where s is the region in the first quadrant, bounded by the
07:44semicubic parabola, and y squared equals x cubed and the line y equals x. For this
07:50we calculate
07:50First, the intersection of the axes, initially from the semicubic parabola,
07:55Solving for x, we get x equal to 2/3. We also have that y equals x, and the parabola
08:01semicubic in x, we equate x squared equal to x cubed, solving the equation and factoring
08:07We have x squared multiplying x minus 1. Solving, we obtain the roots of the intersection
08:13of the axes, and taking into account the conditions in x, that is, that x is equal to y, also
08:19that x is
08:20equal to y raised to 2/3. Now we develop the iterated integral of the definite integral of
08:26limits 1 to 0, and the definite integral of limits and 2 thirds up to and taking into account
08:30The conditions in x, by the differential of x and then y. We perform the integration for x,
08:36and replacing values we obtain, and raised to 2 thirds, minus y. We perform the following integral
08:43for y, from the definite integral of limits from 1 to 0, expanding and replacing we obtain
08:48One tenth. Now we'll use GeoGebra to determine the intersection of the axes graphically.
08:55We enter the first equation, that is, q1, where x equals y, then the next equation
09:00eq2 in x equal to y raised to 2 thirds, we obtain the following intersections of 1,1 and the origin
09:080.0. That's all. If you found the video helpful, don't forget to subscribe.
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