Integrate f(x, y, z)=x+√(y)-z^2 over the path from (0,0,0) to (1,1,1) (Figure 16.6 b ) given by

step 1 Hello, welcome to this lesson. In this last year, integrating F of X and the area C1, C2 and C3.

Integrate f(x, y, z)=x+√(y)-z^2 over the path from (0,0,0)...

Integrate $f(x, y, z)=x+\sqrt{y}-z^{2}$ over the path from $(0,0,0)$ to $(1,1,1)$ (Figure 16.6 $\mathrm{b} )$ given by $$ \begin{array}{ll}{C_{1} :} & {\mathbf{r}(t)=t \mathbf{k}, \quad 0 \leq t \leq 1} \\ {C_{2} :} & {\mathbf{r}(t)=t \mathbf{j}+\mathbf{k}, \quad 0 \leq t \leq 1} \\ {C_{3} :} & {\mathbf{r}(t)=t \mathbf{i}+\mathbf{j}+\mathbf{k}, \quad 0 \leq t \leq 1}\end{array} $$

Integrate $f(x, y, z)=x+\sqrt{y}-z^{2}$ over the path from $(0,0,0)$ to $(1,1,1)$ (Figure 16.6 $\mathrm{b} )$ given by
$$
\begin{array}{ll}{C_{1} :} & {\mathbf{r}(t)=t \mathbf{k}, \quad 0 \leq t \leq 1} \\ {C_{2} :} & {\mathbf{r}(t)=t \mathbf{j}+\mathbf{k}, \quad 0 \leq t \leq 1} \\ {C_{3} :} & {\mathbf{r}(t)=t \mathbf{i}+\mathbf{j}+\mathbf{k}, \quad 0 \leq t \leq 1}\end{array}
$$

Key Concepts

Piecewise Integration over Paths

When a path is divided into multiple segments, each segment is parameterized separately. The integral over the entire path is then found by computing the line integral over each piece and summing the results. This approach is essential when dealing with curves that are not smooth or have different segments defined by distinct parameterizations.

Line Integrals of Scalar Functions

A line integral of a scalar function involves integrating a scalar field over a curve. The procedure requires parameterizing the curve and then multiplying the function evaluated along the curve by the differential element of arc length. This method is used to aggregate the value of the scalar function along a path in space.

Curve Parameterization

Curve parameterization is the process of representing a curve using a parameter, typically denoted by t, in order to describe the coordinates as functions of that parameter. This converts a multi-dimensional integration problem into a single-variable integration problem, making it simpler to evaluate integrals along paths.

Scalar Fields

A scalar field assigns a single numerical value to every point in space. In the context of line integrals, a scalar field is integrated along a curve by evaluating the function at each point on the path. This process is used to measure quantities such as mass, charge, or potential distributed along a curve.

Transcript

00:01 Hello, welcome to this lesson.

00:01 In this last year, integrating f of x and the area c1, c2 and c3.

00:19 So the line integral for c is f of x, y, z, d x, okay.

00:31 And this can be split as.

00:41 Okay, so this has an input of zero, zero, t okay i mean from the parametric equation they see that we only have the cave we only have the cave component don't have the i energy so for that part then we look at the rt the t plus the second one that has components at there's nothing that there's no component of the i there's a t at the j and there's a one at the k okay the last one that has t at the i point one at the jth and the one at the kth so for us to get the length of our t for each for each for each of them would so find that for this okay and this is equal to the square root of the differential of the three parts the y the x the y and the z okay so at the z point we have you have t over there so when i differentiate i have just one squared okay and that gives me one okay so here to there's a one a coefficient of one here there's a zero over there so we have zero for i and i have one if i differentiate t i have over respect to t i have just one okay and if i differentiate one respect to t i have zero okay so another zero again so this gives me one.

04:09 The last but not least is this one.

04:20 Here, when i differentiate this respect to t, i have one.

04:27 There's a coefficient one there when i differentiate that respect to t i have zero, and i have zero.

04:34 So this is equal to one as well.

04:40 Okay.

04:42 So still you are using this.

05:05 Okay.

05:06 So here the integral becomes, so in the first one we substitute s for 0, then 0 for y, then z for t.

05:47 So we have from 0 to 1, the x is 0, the y is 0, z is t...

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