Given $\vec{F}=\left(x^2-2y\right)\vec{i}-4yz\vec{J}+4x{z}^2\vec{k},$ the value of the line integral _C^ F.dl along the straight line C from (0, 0, 0) to (1, 1, 1) is.

To find the value of the line integral _C^ F.dl along the given straight line C, we need to follow a series of steps. This involves parameterizing the path, calculating the dot product of the vector field and the differential displacement vector, and then performing the definite integral.

Line Integral Problem Analysis

We are given the vector field:

$$\vec{F}=\left(x^2-2y\right)\vec{i}-4yz\vec{j}+4x{z}^2\vec{k}$$

The path of integration C is a straight line segment connecting the point (0, 0, 0) to (1, 1, 1). This is a common type of line integral problem in vector calculus.

Parameterizing the Straight Line Path

To evaluate the line integral, we first need to parameterize the straight line C from the starting point $(0,0,0)$ to the end point $(1,1,1)$. A simple way to parameterize a straight line segment from $(x_0,y_0,z_0)$ to $(x_1,y_1,z_1)$ is using a parameter $t$ such that:

- $x=x_0+(x_1-x_0)t$
- $y=y_0+(y_1-y_0)t$
- $z=z_0+(z_1-z_0)t$

Here, $t$ varies from 0 to 1. Substituting our points:

- $x=0+(1-0)t=t$
- $y=0+(1-0)t=t$
- $z=0+(1-0)t=t$

So, the parametric equations for the path C are:

$$x=t,y=t,z=t$$

The differential displacement vector $\overrightarrow{dl}$ can be expressed as:

$$\overrightarrow{dl}=dx\vec{i}+dy\vec{j}+dz\vec{k}$$

Differentiating our parametric equations with respect to $t$:

- $dx=dt$
- $dy=dt$
- $dz=dt$

Therefore, $\overrightarrow{dl}$ becomes:

$$\overrightarrow{dl}=dt\vec{i}+dt\vec{j}+dt\vec{k}$$

Substituting into the Vector Field F

Next, we substitute the parametric equations of the path ($x=t,y=t,z=t$) into the given vector field $\vec{F}$:

$$\vec{F}=\left(x^2-2y\right)\vec{i}-4yz\vec{j}+4x{z}^2\vec{k}$$

Substitute $x=t,y=t,z=t$:

$$\vec{F}=\left(t^2-2t\right)\vec{i}-4(t)(t)\vec{j}+4(t)(t^2)\vec{k}$$

$$\vec{F}=\left(t^2-2t\right)\vec{i}-4t^2\vec{j}+4t^3\vec{k}$$

Calculating the Dot Product F · dl

Now, we compute the dot product $\vec{F}\cdot \overrightarrow{dl}$. This is a crucial step for evaluating the line integral of a vector field:

$$\vec{F}\cdot \overrightarrow{dl}=\left[\left(t^2-2t\right)\vec{i}-4t^2\vec{j}+4t^3\vec{k}\right]\cdot \left[dt\vec{i}+dt\vec{j}+dt\vec{k}\right]$$

Performing the dot product:

$$\vec{F}\cdot \overrightarrow{dl}=\left(t^2-2t\right)dt+\left(-4t^2\right)dt+\left(4t^3\right)dt$$

Combine the terms:

$$\vec{F}\cdot \overrightarrow{dl}=\left(t^2-2t-4t^2+4t^3\right)dt$$

$$\vec{F}\cdot \overrightarrow{dl}=\left(4t^3-3t^2-2t\right)dt$$

Evaluating the Definite Line Integral

Finally, we integrate the expression $\vec{F}\cdot \overrightarrow{dl}$ with respect to $t$ from the initial value $t=0$ to the final value $t=1$:

_C^ F.dl = _0^1 {≤ft( 4t^3 - 3t^2 - 2t )dt}

Perform the integration term by term:

$$={\left[\frac{4t^{3+1}}{3+1}-\frac{3t^{2+1}}{2+1}-\frac{2t^{1+1}}{1+1}\right]}_0^1$$

$$={\left[\frac{4t^4}{4}-\frac{3t^3}{3}-\frac{2t^2}{2}\right]}_0^1$$

$$={\left[t^4-t^3-t^2\right]}_0^1$$

Now, apply the limits of integration:

$$=\left(1^4-1^3-1^2\right)-\left(0^4-0^3-0^2\right)$$

$$=\left(1-1-1\right)-\left(0-0-0\right)$$

$$=\left(-1\right)-\left(0\right)$$

$$=-1$$Final Result of the Line Integral

The value of the line integral _C^ F.dl along the straight line C from (0, 0, 0) to (1, 1, 1) is -1.