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### Problem 40
Given \(\mathbf{v} = (2x + yz, 2y + zx, 2z + xy)\), you need to show that \(\text{curl} \, \mathbf{v} = 0\) and find a function \(f(\mathbf{r})\) such that \(\mathbf{v} = \nabla f\).

1. **Compute the Curl**: The curl of a vector field \(\mathbf{v} = (v_1, v_2, v_3)\) is given by:

\[
\text{curl} \, \mathbf{v} = \nabla \times \mathbf{v} = \left( \frac{\partial v_3}{\partial y} - \frac{\partial v_2}{\partial z}, \frac{\partial v_1}{\partial z} - \frac{\partial v_3}{\partial x}, \frac{\partial v_2}{\partial x} - \frac{\partial v_1}{\partial y} \right)
\]

Let's compute each component:
- \( \frac{\partial v_3}{\partial y} = \frac{\partial (2z + xy)}{\partial y} = x \)
- \( \frac{\partial v_2}{\partial z} = \frac{\partial (2y + zx)}{\partial z} = x \)
- \( \frac{\partial v_1}{\partial z} = \frac{\partial (2x + yz)}{\partial z} = y \)
- \( \frac{\partial v_3}{\partial x} = \frac{\partial (2z + xy)}{\partial x} = y \)
- \( \frac{\partial v_2}{\partial x} = \frac{\partial (2y + zx)}{\partial x} = z \)
- \( \frac{\partial v_1}{\partial y} = \frac{\partial (2x + yz)}{\partial y} = z \)

Thus:
\[
\text{curl} \, \mathbf{v} = (x - x, y - y, z - z) = (0, 0, 0)
\]

This shows that \(\text{curl} \, \mathbf{v} = 0\), meaning \(\mathbf{v}\) is conservative, so there exists a scalar potential \(f\) such that \(\mathbf{v} = \nabla f\).

2. **Finding the Potential Function \(f\)**: 

Since \(\mathbf{v} = \nabla f\), we have:
\[
\frac{\partial f}{\partial x} = 2x + yz, \quad \frac{\partial f}{\partial y} = 2y + zx, \quad \frac{\partial f}{\partial z} = 2z + xy
\]

Integrate \(\frac{\partial f}{\partial x} = 2x + yz\) with respect to \(x\):
\[
f(x, y, z) = x^2 + xyz + g(y, z)
\]
where \(g(y, z)\) is a function of \(y\) and \(z\) only.

Next, differentiate this expression with respect to \(y\) and compare with \(\frac{\partial f}{\partial y}\):
\[
\frac{\partial f}{\partial y} = xz + \frac{\partial g}{\partial y}
\]
\[
2y + zx = xz + \frac{\partial g}{\partial y}
\]
Thus, \( \frac{\partial g}{\partial y} = 2y \), integrate with respect to \(y\):
\[
g(y, z) = y^2 + h(z)
\]
where \(h(z)\) is a function of \(z\) only.

Differentiate \(f(x, y, z)\) with respect to \(z\) and compare with \(\frac{\partial f}{\partial z}\):
\[
\frac{\partial f}{\partial z} = xy + \frac{\partial h}{\partial z}
\]
\[
2z + xy = xy + \frac{\partial h}{\partial z}
\]
Thus, \( \frac{\partial h}{\partial z} = 2z \), integrate with respect to \(z\):
\[
h(z) = z^2 + C
\]
where \(C\) is a constant.

Thus, the potential function \(f\) is:
\[
f(x, y, z) = x^2 + y^2 + z^2 + xyz + C
\]

### Problem 41
You need to verify the identity:
\[
\nabla \times (f \mathbf{v}) = f (\nabla \times \mathbf{v}) + (\nabla f) \times \mathbf{v}
\]
for \(f(\mathbf{r}) = x^3 - y\) and \(\mathbf{v}(\mathbf{r}) = (z, 0, -x)\).

1. **Compute Each Term Separately**:
   
   a. **Calculate \(\nabla f\)**:
   \[
   \nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = \left( 3x^2, -1, 0 \right)
   \]

   b. **Compute \(\nabla \times \mathbf{v}\)**:
   \[
   \nabla \times \mathbf{v} = \nabla \times (z, 0, -x) = \left( \frac{\partial (-x)}{\partial y} - \frac{\partial 0}{\partial z}, \frac{\partial z}{\partial z} - \frac{\partial (-x)}{\partial x}, \frac{\partial 0}{\partial x} - \frac{\partial z}{\partial y} \right) = (0 - 0, 1 - (-1), 0 - 0) = (0, 2, 0)
   \]

   c. **Calculate \(\nabla \times (f \mathbf{v})\)**:
   First, \(f \mathbf{v} = (f \cdot z, 0, -f \cdot x)\).
   \[
   \nabla \times (f \mathbf{v}) = \nabla \times \left( fz, 0, -fx \right)
   \]
   Compute this using the definition of curl.

   d. **Compute \(f (\nabla \times \mathbf{v})\)**:
   \[
   f \cdot (0, 2, 0) = (0, 2f, 0)
   \]

   e. **Compute \((\nabla f) \times \mathbf{v}\)**:
   \[
   (\nabla f) \times (z, 0, -x) = \left| \begin{array}{ccc}
   \mathbf{i} & \mathbf{j} & \mathbf{k} \\
   3x^2 & -1 & 0 \\
   z & 0 & -x
   \end{array} \right| = \left( 1x, 3x^2(-x) - 0, -(3x^2 \cdot 0 + z) \right) = (x, -3x^3, -z)
   \]

Now, sum up \(f (\nabla \times \mathbf{v})\) and \((\nabla f) \times \mathbf{v}\) and compare it with \(\nabla \times (f \mathbf{v})\).

If you compare all, you'll verify the identity.

Would you like further details or explanations on any of these steps?

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### Follow-up Questions:
1. How do you compute the gradient of a scalar function?
2. What is the significance of the curl of a vector field in physics?
3. How do you verify if a vector field is conservative?
4. Can you find a potential function \(f\) for any vector field?
5. How do you calculate the cross product of two vectors?

### Tip:
Always verify vector field identities step by step, starting from the basic definitions of gradient, curl, and cross product.

Show that if v = (2x + yz, 2y + zx, 2z + xy) then curl v = 0, and find f(r) such that v = grad f. Verify the identity ∇ × (f∇ × v) = f(∇ × v) + (∇f) × v for f(r) = x³ - y and v(r) = (z, 0, -x).

Explore the verification of key vector calculus identities including curl, gradient, and cross product. This detailed solution covers curl calculation, gradient determination, and the vector triple product identity. Suitable for university-level vector calculus studies.

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