00:11Definition of triple integrals: A triple integral is an extension of the concept of an integral.
00:16defined in three dimensions as well as the definite integral of a positive function x over a
00:22The interval a represents the area under the curve y equals the function of x from x equals to x.
00:28equal
00:28b and the double integral of a positive function of x and over a region r in the plane
00:34why
00:37represents the volume under the surface z equals a function of x and over that region the triple integral
00:43of a function x y z over a solid region v in three-dimensional space represents the hypervolume or
00:50an accumulated quantity in that volume weighted by the function f is represented by the volume of the
00:57integrating the function x and z in Cartesian coordinates can also be cylindrical and
01:04spherical
01:07here it is represented where the differential of v represents the infinitesimal volume element
01:12in three-dimensional space, which in Cartesian coordinates is the differential of v, is equal to the
01:18differential of x, y, z, or in any other order of integration, or in other coordinate systems
01:24as cylindrical, the differential of v is equal to r radius times the differential of r times the differential of
01:30the angular theta times the differential of zeta or spherical differential of the volume is equal to roto sine of phi times
01:37The differential of theta times the differential of phi is essentially a tool
01:43to sum the infinitesimal contributions of a function over a three-dimensional volume
01:51Use of triple integrals: Triple integrals have various applications in physics, engineering, and
01:59mathematics, among which the calculation of volume stands out: if the function f x y z equals 1 then the
02:07triple integral over a solid region v calculates the volume of that region volume equal to the integrand
02:13of v 1 of the triple integral of v mass calculation if rob x y z represents the density
02:21of an object
02:22three-dimensional at each point x y z then the total mass m of the object is calculated by mass equal to
02:29When integrating x and z, calculating moments and centroids, triple integrals are used to calculate the
02:37moments of inertia, center of mass and other static moments of three-dimensional objects
02:42which is crucial in mechanics and statics, total charge calculation in electromagnetism if sigma x y z is
02:51the volumetric charge density the total charge with a region is found with q equal to the triple
02:56integral of the sigma volume in x y z by the differential of the average value of a
03:03function in a volume the average value of a function f x y z over a volume b is defined
03:10as
03:10f equal volume 1 of the triple integral of the function x y z of the differential of the volume
03:20To define concepts of the topic I am going to do two exercises from the Piscuno differential calculus book
03:26Integral 2 from the defunct Mir Moscow publishing house. Here we have the first two exercises, not done.
03:38done with artificial intelligence, since it wouldn't make sense to do it with artificial intelligence
03:43This is where people are copying and not learning or developing anything; for this we have the following
03:48exercise done by hand and then we have the following presentation tells us to calculate the triple
03:54integral of the function 1 over x plus y plus z plus 1 times the differential of x y z
04:02if the domain of the
04:03function is limited by the coordinate planes and the x y y z plane plus 1 of answer we have
04:10logarithm
04:11natural of 2 over 2 minus 5 sixteenths first we find the limits for the exercise we equate to
04:20the function of x plus y plus y plus z plus 1, then we perform the differential of the function u
04:26we obtain
04:27The differential of u is equal to x plus y plus 1 times the differential of z of x plus y
04:34plus z plus 1 we clear z and obtain
04:381 minus x minus and we replace this in the function u the value of z simplifying we get 2
04:52For the limits we cleared and setting z equal to 0, we also do the same for x, we clear
04:59x and y
05:00setting x equal to 0 and y equal to 0
05:17the triple integral of u to the power of negative 3 with respect to z from x plus y plus
05:221 to 2 then with
05:23with respect to y from 0 and finally with respect to x from 0 to 1 and finally with respect to
05:27x from 0
05:28up to 1 as mentioned before we have to take the differential of the function of u obtaining
05:35x plus y plus 1 of the differential of z, where z is the variable and the rest are constants, now we do
05:45the
05:45triple integral of u to the power of negative 3 with respect to z from x plus y plus 1
05:50up to 2 integrating
05:51We obtain minus one half times u raised to the power of minus 2 of limits from x plus y plus
05:571 to 2
05:57Replacing and simplifying, we get negative one-eighth minus one-half times u to the power of negative 2
06:04being u x plus y plus 1 with respect to z by the differential of y x we make the following
06:12integral definite
06:13with respect to y from 0 to 1 minus x of the previously found result and we have minus one eighth minus
06:20a
06:20half times u raised to the power of -1 with limits 1 minus x to 0 times the differential of
06:26x and finally
06:28the definite integral with respect to x from 0 to 1 gives us one-eighth times x squared.
06:34minus three-eighths x plus one-half of the natural logarithm of x plus 1 simplifying and replacing its
06:42limits we get minus 5 16ths plus one half times the natural logarithm of 2 the following exercise gives us
06:51It says to calculate the triple integral of x and z with respect to z from 0 to y and then with respect to
06:57and from 0 to
06:58x and finally with respect to x from 0 to your answer is a raised to the sixth over
07:0548 we performed
07:07Integration with respect to z: we find x and z squared over 2 with limits 0 to y.
07:15the
07:16Substituting its limits, we get the result of x and y squared over 2, that is x
07:23and cubed
07:24over 2 we then integrate with respect to y and obtain the important thing here is to separate factors so x over 2 that
07:33Multiply x to the fourth over 4, this gives us a constant of one-eighth and the integrand is
07:39x to the
07:39Fifth, finally solving and substituting its limits from 0 to a, we obtain a raised to the sixth power
07:46about 48 now as an example to solve in matlab I'm going to take a triple integral from the open page
07:54star tells us to find the triple integral of x plus y z squared with respect to x from
08:00minus 1 up to
08:015 then with respect to y from 2 to 4 and finally with respect to z from 0 to 1
08:0736
08:10It is optional; we comment with the percent sign; within this we write triple integral.
08:18Next, we write that gm s and establish the variables x, y, and z.
08:25We write the function f equals x plus y z squared. Once the function is established, we have
08:34We can find solutions in a single line of code or in parts.
08:40We perform this in a single line of code: y equals y in parentheses and y in parentheses
08:48This is integral, we open parentheses, we write the function of f(x) with limits from -1 to 5, we close
08:55We place a comma in parentheses. Now it is with respect to y with limits 2 to 4 and finally z with limits 0
09:03up to 1 we close parentheses
09:09We press enter and get the solution to 36
09:17The following exercise in Matlab is performed in parts; that is, for each integral.
09:23triple integral of the function x squared y z your answer is 162
09:31We make the comment, we open the percent sign (this is optional), we write triple integral
09:39We designate the variables with that gm s x y z
09:48We perform the function f equal to x squared and z; we establish the function once the
09:56We integrated the function in parts in the previous exercise, we did it with a single line of code, now we
10:03We will do it in parts, we write and 1 is equal to int, we open parentheses, inside we write f
10:12comma function and limits 0 to 3, press enter
10:26Next we write y 2 is equal to int we open parentheses inside this we write y 1 comma
10:33From limits minus 2 to 1, press enter. Note: here goes the integral and 1, not the function f.
10:49Next we write y 3 is equal to int we open parentheses inside this we write y 2 comma of limits
10:58minus 1 to 5, press enter, note: here goes the integral and 2 not the function f, press enter, we have
11:04of answer 162
11:15That's all! If you found the video helpful, don't forget to subscribe.
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