Find the line integral of f(x, y, z)=x+y+z over the straightline segment from (1,2,3) to (0,-1,1

Find the line integral of f(x, y, z)=x+y+z over the...

Key Concepts

Arc Length Differential

The arc length differential (often denoted as ds) represents an infinitesimal element of length along the curve. It is computed as the norm (magnitude) of the derivative of the parametrized curve with respect to its parameter. This differential is essential for weighting the contributions of the scalar field during integration.

Integral Evaluation Techniques

Evaluating a line integral typically involves substituting the parametrized form of the curve into the scalar function, finding the derivative of the parametrization to determine ds, and then integrating the resulting expression with respect to the parameter over its specified interval. This systematic approach simplifies complex spatial integrations into standard single-variable integrals.

Line Integral of a Scalar Field

A line integral over a scalar field involves integrating a scalar function along a curve, taking into account the varying value of the function along the path. This type of integral accumulates the contributions of the function multiplied by an element of length along the curve, effectively summing the function's influence along that path.

Curve Parametrization

Parametrization is the process of expressing a curve by defining a vector-valued function that maps an interval of a parameter to points on the curve. This allows one to represent the path in a manageable form, transforming the integration over a curve into an integration over a parameter interval.

Transcript

00:01 Okay, folks, so in this video, we're going to take a look at this line integral.

00:05 This is problem number 13.

00:08 We have this function f, which is a function of x, y, and z, and it's equal to the sum of the three variables, x, y, and z.

00:19 And we're integrating from 1 to 3 via a straight line segment to 0, negative 1, and 1.

00:29 So, in other words, we want the line integral from 1, 2, 3, to 0, negative 1, 1 of this function, of this function x, y, and z multiplied by the usual ds, which is an infinitesimal line segment.

00:52 So, unfortunately, we're not given the parameterization, so we're going to have to find the parameterization ourselves.

00:59 So the way we're going to find the parameterization of x, y, and z with respect to t, is by using the condition.

01:08 I hope you remember this condition, x minus x0 of x final minus x0 equals y minus y 0 of y final minus y 0 is equal to z minus z zero, z final minus z zero.

01:30 This condition we have this condition right here that every point on a line segment needs to satisfy so this is this is a set of three expressions that are all equal to each other and we're going to use them to obtain our parameterization by that i mean i'm going to set all of them to equal to t so um so from this we can get three separate equations for x and y and z each as a function of time so let's do it so for for x we're going to use this is equal to t we're going to use this to get what's x0 well x0 is just um this number right here that's x 0 x minus 1 so we have um what about x final well x final is 0 and x 0 is 1 again that is equal to t so as you can see x minus 1 must be equal to next t therefore we have x equals 1 minus t so we have obtained our first function which is a which is as a function of time okay so that's x let's do the same thing for y we have y minus y not y not is 2 as you can see here so y minus 2 over y final which is negative 1 negative 1 minus y 2 is just um excuse me negative 1 minus 2 that's equal to t so we get y minus two over minus three equal t so that means that we have y equals minus three t plus two okay um so we have obtained our function for y what about z well z we're going to do the same thing again so z minus z zero z zero is three as you can see above and then 1 minus 3 that's equal to t so from this we're going to get z minus 3 over minus 2 is equal to t okay so we have this z is looking like a 2 here okay so we have z equals 3 minus 2 t so as you can see we have obtained all three of our functions now what we can do is we can just plug it in using the usual you know the way you the way you do a line integral is by first parameterizing it and then rewriting the ds as a function of of t.

04:19 The way you write that is by first of all you say that ds is equal to x prime squared plus y prime squared plus z prime squared the whole thing multiplied by t and then x prime is just the derivative the time derivative of the function x so as you can see here the time derivative of this function is just negative 1.

04:49 So negative 1 squared is just 1.

04:51 So we have 1 plus we're going to do the same thing for y.

04:56 You take the derivative of y with respect to t you get negative 3.

05:00 So you square that, you get 9 plus plus 4, okay, multiplied by dt.

05:07 And then when you add these three things up, 10 plus 4, just 14, dt, which is 2 times 7.

05:17 You can't do anything about 2 times 7 inside of the square root.

05:20 So i'm going to leave this right here.

05:22 Okay, i'm not going to do anything about it.

05:26 So we have ds as an expression of dt.

05:30 And then now what we're going to do is we're going to plug it back into this line integral here.

05:36 So we have what's f again? f is x plus y plus z, but now we're not going to write x, y, and z.

05:42 We're going to write all of them as a function of t.

05:45 So we have 1 minus t.

05:47 That's for x plus y is 2 minus 3 t okay um and z is going to be 3 minus 2 t so we have these three guys multiplied by d s but the s is just a constant times d t square root of 14 times d t let me copy this down so that i have more room i'm running on a room right now okay so we have we have this thing...

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