Understanding the Problem

We are asked to find the value of the integral \( \int \int Curl \vec F. d\vec r \) for the given vector field \( \vec F(x, y, z) = (y + z, z + x, x + y) \). The notation \( \int \int Curl \vec F. d\vec r \) typically represents a surface integral of the curl of the vector field over some surface S, which is often written as \( \iint_S (\nabla \times \vec F) \cdot d\vec S \). According to Stokes' Theorem, this surface integral is equal to the line integral of the vector field over the boundary curve of the surface. However, the problem asks for the value of the integral of the curl itself, which can be calculated by first finding the curl and then evaluating the integral.

Calculating the Curl of the Vector Field \( \vec F \)

The vector field is given by \( \vec F(x, y, z) = (y + z) \hat{i} + (z + x) \hat{j} + (x + y) \hat{k} \). We can write this as \( \vec F = P \hat{i} + Q \hat{j} + R \hat{k} \), where:

• \( P = y + z \)
• \( Q = z + x \)
• \( R = x + y \)

The curl of the vector field \( \vec F \) is denoted by \( \nabla \times \vec F \) and is calculated using the determinant formula:

Let's find the necessary partial derivatives:

• \( \frac{\partial R}{\partial y} = \frac{\partial}{\partial y}(x + y) = 1 \)
• \( \frac{\partial Q}{\partial z} = \frac{\partial}{\partial z}(z + x) = 1 \)
• \( \frac{\partial R}{\partial x} = \frac{\partial}{\partial x}(x + y) = 1 \)
• \( \frac{\partial P}{\partial z} = \frac{\partial}{\partial z}(y + z) = 1 \)
• \( \frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(z + x) = 1 \)
• \( \frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(y + z) = 1 \)

Now, substitute these partial derivatives into the curl formula:

The curl of the vector field \( \vec F \) is the zero vector \( \vec 0 \).

Evaluating the Integral

The integral we need to evaluate is \( \int \int Curl \vec F. d\vec r \), which we interpret as the surface integral \( \iint_S (\nabla \times \vec F) \cdot d\vec S \) over some surface S. Since we found that \( \nabla \times \vec F = \vec 0 \), the integral becomes:

The dot product of the zero vector with any vector \( d\vec S \) is zero:

Therefore, the integral is:

The value of the integral \( \int \int Curl \vec F. d\vec r \) is 0.

Final Answer

Based on the calculation, the curl of the given vector field is the zero vector. The surface integral of the zero vector over any surface is zero.

The value of \( \int \int Curl \vec F. d\vec r \) is 0.

Revision Table: Key Vector Calculus Concepts

Additional Information on Conservative Vector Fields

A vector field \( \vec F \) is called conservative if its curl is the zero vector, i.e., \( \nabla \times \vec F = \vec 0 \). For a simply connected domain, a vector field is conservative if and only if it is the gradient of a scalar potential function \( \phi \), i.e., \( \vec F = \nabla \phi \). If a vector field is conservative, the line integral \( \int_C \vec F \cdot d\vec r \) is path-independent, meaning it only depends on the start and end points of the curve C, not the path taken between them.

In this specific problem, since \( \nabla \times \vec F = \vec 0 \), the vector field \( \vec F(x, y, z) = (y + z, z + x, x + y) \) is conservative. This means we can find a potential function \( \phi(x, y, z) \) such that \( \nabla \phi = \vec F \). Let's find it:

Integrating the first equation with respect to x, we get \( \phi(x, y, z) = xy + xz + g(y, z) \). Differentiating with respect to y:

Comparing with the second equation, \( x + \frac{\partial g}{\partial y} = z + x \), which implies \( \frac{\partial g}{\partial y} = z \). Integrating with respect to y, we get \( g(y, z) = yz + h(z) \). So, \( \phi(x, y, z) = xy + xz + yz + h(z) \). Differentiating with respect to z:

Comparing with the third equation, \( x + y + \frac{\partial h}{\partial z} = x + y \), which implies \( \frac{\partial h}{\partial z} = 0 \). This means \( h(z) = C \), where C is a constant.

Thus, a potential function is \( \phi(x, y, z) = xy + xz + yz + C \).

This confirms that the vector field is conservative. A direct consequence of a vector field being conservative (in a simply connected region) is that its curl is zero. This aligns with our calculation and the integral result.