Question

A vector field a is given by $(z^2+2xy)\mathbf{i}+(x^2+2yz)\mathbf{j}+(y^2+2zx)\mathbf{k}$. Show that $\mathbf{a}$ is conservative and that the line integral $\int \mathbf{a}\cdot d\mathbf{r}$ along any line joining $(1,1,1)$ and $(1,2,2)$ has the value 11 .

Short answer

The vector field is conservative, and the line integral value is 11.

Step-by-step solution

Define Vector Field

Given vector field $\mathbf{a}=(z^2+2xy)\mathbf{i}+(x^2+2yz)\mathbf{j}+(y^2+2zx)\mathbf{k}$.

Check for Conservative Field

A vector field $\mathbf{a}$ is conservative if it can be written as the gradient of some potential function $\varphi$, i.e., $\mathbf{a}=abla\varphi$. This means we need to find a function $\varphi$ such that: $\frac{\partial \varphi }{\partial x}=z^2+2xy$ $\frac{\partial \varphi }{\partial y}=x^2+2yz$ $\frac{\partial \varphi }{\partial z}=y^2+2zx$

Integrate with Respect to x

Integrate $\frac{\partial \varphi }{\partial x}=z^2+2xy$ with respect to $x$: $\varphi (x,y,z)=x{z}^2+x^{2}y+h(y,z)$, where $h(y,z)$ is an arbitrary function of $y$ and $z$.

Integrate with Respect to y

Take the partial derivative of $\varphi$ with respect to $y$: $\frac{\partial \varphi }{\partial y}=x^2+\frac{\partial h(y,z)}{\partial y}$. Set this equal to the given $\frac{\partial \varphi }{\partial y}$ from step 2: $x^2+\frac{\partial h(y,z)}{\partial y}=x^2+2yz$. Therefore, $\frac{\partial h(y,z)}{\partial y}=2yz$, which implies $h(y,z)=y^{2}z+g(z)$, where $g(z)$ is an arbitrary function of $z$.

Integrate with Respect to z

Take the partial derivative of $\varphi$ with respect to $z$: $\frac{\partial \varphi }{\partial z}=xz+y^2+\frac{dg(z)}{dz}$. Set this equal to the given $\frac{\partial \varphi }{\partial z}$ from step 2: $y^2+2zx=y^2+xz+\frac{dg(z)}{dz}$. Therefore, $2zx=zx+\frac{dg(z)}{dz}$ which implies $\frac{dg(z)}{dz}=zx$ and hence $g(z)=z{x}^2$.

Form the Potential Function

Combine results from steps 3, 4, and 5 to obtain the potential function: $\varphi (x,y,z)=x{z}^2+x^{2}y+y^{2}z+x^{2}z$.

Evaluate the Line Integral

Since $\mathbf{a}$ is conservative, the line integral ${\int }_{(1,1,1)}^{(1,2,2)}\mathbf{a}\cdot d\mathbf{r}$ equals the difference in potential function values at the endpoints: $\varphi (1,2,2)-\varphi (1,1,1)$.

Calculate Potential Function at Endpoints

Calculate $\varphi (1,1,1)$: $\varphi (1,1,1)=(1)(1^2)+(1^2)(1)+(1^2)(1)+(1^2)(1)=4$. Calculate $\varphi (1,2,2)$: $\varphi (1,2,2)=(1)(2^2)+(1^2)(2)+(2^2)(2)+(1)(2^2)=15$.

Find the Line Integral Value

Compute the line integral value: $\varphi (1,2,2)-\varphi (1,1,1)=15-4=11$.

Key concepts

Vector Field

A vector field is a function that assigns a vector to every point in space. You can think of it as a map that tells you which direction and how strongly something (like wind or force) is pushing at each point. In our exercise, we are given the vector field:
$\mathbf{a}=(z^2+2xy)\mathbf{i}+(x^2+2yz)\mathbf{j}+(y^2+2zx)\mathbf{k}$. This notation means that at every point $(x,y,z)$ in space, the vector field has three components:
* along the $x$-axis (represented by $\mathbf{i}$), * along the $y$-axis (represented by $\mathbf{j}$), * along the $z$-axis (represented by $\mathbf{k}$).
Each component points in a distinct direction and is a function of $x,y,$ and $z$. If you visualize it, it looks like arrows spread across the space, each indicating the direction and magnitude of the vector field at that point.

Gradient of a Potential Function

The gradient of a potential function is a vector field representing the rate of change of that function in space. If a vector field is the gradient of some potential function $\varphi$, it is called conservative.
For our vector field to be conservative, we need to find a function $\varphi$ such that $\mathbf{a}=abla\varphi$. This means:
$\frac{\partial \varphi }{\partial x}=z^2+2xy$, $\frac{\partial \varphi }{\partial y}=x^2+2yz$, and $\frac{\partial \varphi }{\partial z}=y^2+2zx$.
These equations represent how the potential function $\varphi$ changes along each axis. By solving them, we can construct $\varphi$.

Line Integral

A line integral computes the sum of a vector field along a curve or path. It helps in evaluating how much work is done by a force field in moving an object along a specific path.
For a conservative vector field, the line integral from point $A$ to point $B$ depends only on the values of the potential function at the endpoints. So we only need to find $\varphi$ at these points.
In our problem, the line integral ${\int }_{(1,1,1)}^{(1,2,2)}\mathbf{a}\cdot d\mathbf{r}$ is computed by calculating the values of $\varphi$ at points $(1,1,1)$ and $(1,2,2)$, then taking the difference $\varphi (1,2,2)-\varphi (1,1,1)$. This considerably simplifies the process as opposed to integrating along the actual path.

Potential Function

The potential function $\varphi$ is a scalar function whose gradient equals a given vector field. Finding this function shows the field is conservative.
To find $\varphi$, we integrate partial derivatives step by step. First, we integrate $\frac{\partial \varphi }{\partial x}=z^2+2xy$:
$\varphi (x,y,z)=x{z}^2+x^{2}y+h(y,z)$, where $h(y,z)$ is an arbitrary function.
Then, we find $h(y,z)$ by differentiating with respect to $y$ and setting it equal to the given partial derivative:
$\frac{\partial \varphi }{\partial y}=x^2+2yz$:
$h(y,z)=y^{2}z+g(z)$.
Finally, we find $g(z)$ by differentiating with respect to $z$:
$\frac{\partial \varphi }{\partial z}=y^2+xz$.
Combining results, we obtain:
$\varphi (x,y,z)=x{z}^2+x^{2}y+y^{2}z+x^{2}z$.