To find the value of the line integral \(\int_P^Q(z^2dx+3y^2dy+2xzdz)\) along the straight line joining the points \(P(1, 1, 2)\) and \(Q(2, 3, 1)\), we will first identify the vector field and then check if it is a conservative field. If it is conservative, we can use a potential function to simplify the calculation.

The given line integral can be written in the form \(\int_C \mathbf{F} \cdot d\mathbf{r}\), where the vector field \(\mathbf{F}\) is:

\[\mathbf{F} = z^2 \mathbf{\hat{i}} + 3y^2 \mathbf{\hat{j}} + 2xz \mathbf{\hat{k}}\]

From this, we identify the components \(P = z^2\), \(Q = 3y^2\), and \(R = 2xz\).

Conservative Field Verification

A vector field \(\mathbf{F} = P\mathbf{\hat{i}} + Q\mathbf{\hat{j}} + R\mathbf{\hat{k}}\) is considered conservative if its curl is zero, which translates to the following conditions on its partial derivatives:

• \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\)
• \(\frac{\partial P}{\partial z} = \frac{\partial R}{\partial x}\)
• \(\frac{\partial Q}{\partial z} = \frac{\partial R}{\partial y}\)

Let's calculate these partial derivatives for our given vector field components:

• For the first condition:
  • \(\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(z^2) = 0\)
  • \(\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(3y^2) = 0\)
  • Since \(0 = 0\), the first condition is satisfied.
• For the second condition:
  • \(\frac{\partial P}{\partial z} = \frac{\partial}{\partial z}(z^2) = 2z\)
  • \(\frac{\partial R}{\partial x} = \frac{\partial}{\partial x}(2xz) = 2z\)
  • Since \(2z = 2z\), the second condition is satisfied.
• For the third condition:
  • \(\frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}(3y^2) = 0\)
  • \(\frac{\partial R}{\partial y} = \frac{\partial}{\partial y}(2xz) = 0\)
  • Since \(0 = 0\), the third condition is satisfied.

As all three conditions are satisfied, the vector field \(\mathbf{F}\) is indeed a conservative vector field. This means the line integral's value depends only on the starting and ending points, not the path taken between them. We can therefore use a scalar potential function \(\phi(x, y, z)\) to evaluate the integral.

Potential Function Determination

For a conservative vector field \(\mathbf{F}\), there exists a scalar potential function \(\phi(x, y, z)\) such that \(\mathbf{F} = \nabla \phi\). This relationship implies:

• \(\frac{\partial \phi}{\partial x} = P = z^2\)
• \(\frac{\partial \phi}{\partial y} = Q = 3y^2\)
• \(\frac{\partial \phi}{\partial z} = R = 2xz\)

We can find \(\phi(x, y, z)\) by integrating each of these partial derivatives:

1. Integrating \(\frac{\partial \phi}{\partial x} = z^2\) with respect to \(x\): \[\phi(x, y, z) = \int z^2 \, dx = xz^2 + f(y, z)\]

   Here, \(f(y, z)\) represents any term that does not depend on \(x\).

2. Integrating \(\frac{\partial \phi}{\partial y} = 3y^2\) with respect to \(y\): \[\phi(x, y, z) = \int 3y^2 \, dy = y^3 + g(x, z)\]

   Here, \(g(x, z)\) represents any term that does not depend on \(y\).

3. Integrating \(\frac{\partial \phi}{\partial z} = 2xz\) with respect to \(z\): \[\phi(x, y, z) = \int 2xz \, dz = xz^2 + h(x, y)\]

   Here, \(h(x, y)\) represents any term that does not depend on \(z\).

By comparing the results from these three integrations, we can construct the unique potential function \(\phi(x, y, z)\) (ignoring the constant of integration, which cancels out when evaluating the definite integral):

The term \(xz^2\) appears in the first and third integrations. The term \(y^3\) appears in the second integration. Combining these, we get:

\[\phi(x, y, z) = xz^2 + y^3\]

To confirm, we can take the gradient of this potential function to see if it matches \(\mathbf{F}\):

• \(\frac{\partial \phi}{\partial x} = \frac{\partial}{\partial x}(xz^2 + y^3) = z^2\)
• \(\frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y}(xz^2 + y^3) = 3y^2\)
• \(\frac{\partial \phi}{\partial z} = \frac{\partial}{\partial z}(xz^2 + y^3) = 2xz\)

This matches the original components of \(\mathbf{F}\), so our potential function is correct.

Line Integral Calculation

For a conservative vector field \(\mathbf{F}\) and its potential function \(\phi\), the line integral from a starting point \(P\) to an ending point \(Q\) is given by the Fundamental Theorem of Line Integrals:

\[\int_P^Q \mathbf{F} \cdot d\mathbf{r} = \phi(Q) - \phi(P)\]

The given points are \(P(1, 1, 2)\) and \(Q(2, 3, 1)\).

First, evaluate the potential function at the starting point \(P(1, 1, 2)\):

\[\phi(P) = \phi(1, 1, 2) = (1)(2^2) + (1^3) = 1 \cdot 4 + 1 = 4 + 1 = 5\]

Next, evaluate the potential function at the ending point \(Q(2, 3, 1)\):

\[\phi(Q) = \phi(2, 3, 1) = (2)(1^2) + (3^3) = 2 \cdot 1 + 27 = 2 + 27 = 29\]

Finally, subtract the value at \(P\) from the value at \(Q\) to get the line integral:

\[\int_P^Q(z^2dx+3y^2dy+2xzdz) = \phi(Q) - \phi(P) = 29 - 5 = 24\]

The value of the line integral is \(24\).

Integral Solution Summary

Aspect	Description/Value
Vector Field \(\mathbf{F}\)	\(z^2 \mathbf{\hat{i}} + 3y^2 \mathbf{\hat{j}} + 2xz \mathbf{\hat{k}}\)
Starting Point \(P\)	(1, 1, 2)
Ending Point \(Q\)	(2, 3, 1)
Conservative Check	All conditions met, field is conservative.
Potential Function \(\phi(x, y, z)\)	\(xz^2 + y^3\)
Value of \(\phi(P)\)	5
Value of \(\phi(Q)\)	29
Result of Line Integral	24