Use the surface integral in​ Stokes' Theorem to calculate the circulation of the field Bold Upper F equals left parenthesis y squared plus z squared right parenthesis Bold i plus left parenthesis x squared plus y squared right parenthesis Bold j plus left parenthesis x squared plus y squared right parenthesis Bold k around the curve​ C: The square bounded by the lines x equals plus or minus 11 and y equals plus or minus 11 in the​ xy-plane, counterclockwise when viewed from above.
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Part 1
ModifyingBelow Contour integral With Upper CFtimesdrequals

To solve the problem using **Stokes' Theorem**, we need to follow these steps:

### **Step 1: State Stokes' Theorem**
Stokes' Theorem relates a surface integral of the curl of a vector field over a surface \( S \) to a line integral (circulation) of the vector field \( \mathbf{F} \) around the boundary curve \( C \). Mathematically, it is given by:

\[
\oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}
\]

Where:
- \( \mathbf{F} = (y^2 + z^2)\mathbf{i} + (x^2 + y^2)\mathbf{j} + (x^2 + y^2)\mathbf{k} \)
- \( C \) is the boundary of the square in the \( xy \)-plane, with vertices at \( (11, 11), (-11, 11), (-11, -11), (11, -11) \).

### **Step 2: Find the Curl of \( \mathbf{F} \)**
The curl of \( \mathbf{F} \) is computed as follows:

\[
\nabla \times \mathbf{F} = \begin{vmatrix} 
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
y^2 + z^2 & x^2 + y^2 & x^2 + y^2 
\end{vmatrix}
\]

Expanding this determinant:

\[
\nabla \times \mathbf{F} = \mathbf{i} \left( \frac{\partial (x^2 + y^2)}{\partial y} - \frac{\partial (x^2 + y^2)}{\partial z} \right) - \mathbf{j} \left( \frac{\partial (y^2 + z^2)}{\partial x} - \frac{\partial (x^2 + y^2)}{\partial z} \right) + \mathbf{k} \left( \frac{\partial (y^2 + z^2)}{\partial x} - \frac{\partial (x^2 + y^2)}{\partial y} \right)
\]

Computing the partial derivatives:

\[
\nabla \times \mathbf{F} = \mathbf{i} \left(2y - 0 \right) - \mathbf{j} \left(0 - 0 \right) + \mathbf{k} \left(0 - 2x\right)
\]

\[
\nabla \times \mathbf{F} = 2y\mathbf{i} - 2x\mathbf{k}
\]

Thus:

\[
\nabla \times \mathbf{F} = 2y\mathbf{i} + 0\mathbf{j} - 2x\mathbf{k}
\]

### **Step 3: Define the Surface \( S \)**
The surface \( S \) is the square in the \( xy \)-plane, so \( z = 0 \) on this surface. The vector normal to this surface is \( d\mathbf{S} = \mathbf{k} \, dA \), where \( dA \) is the area element in the \( xy \)-plane.

### **Step 4: Calculate the Surface Integral**
Now, we compute the surface integral:

\[
\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_S (2y\mathbf{i} - 2x\mathbf{k}) \cdot (\mathbf{k} \, dA)
\]

Since \( \mathbf{i} \cdot \mathbf{k} = 0 \) and \( \mathbf{k} \cdot \mathbf{k} = 1 \), we get:

\[
\iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_S -2x \, dA
\]

### **Step 5: Set up the Double Integral**
The area element \( dA \) in the \( xy \)-plane is \( dx \, dy \). The integral becomes:

\[
\iint_S -2x \, dA = -2 \iint_S x \, dx \, dy
\]

The limits of integration for the square are from \( -11 \) to \( 11 \) for both \( x \) and \( y \):

\[
\iint_S x \, dx \, dy = \int_{-11}^{11} \int_{-11}^{11} x \, dx \, dy
\]

### **Step 6: Compute the Integral**
Integrate with respect to \( x \):

\[
\int_{-11}^{11} x \, dx = \left[ \frac{x^2}{2} \right]_{-11}^{11} = \frac{11^2}{2} - \frac{(-11)^2}{2} = 0
\]

Since the integrand is an odd function over a symmetric interval, the result is 0.

Thus, the integral simplifies to:

\[
\oint_C \mathbf{F} \cdot d\mathbf{r} = -2 \times 0 = 0
\]

### **Final Answer:**
The circulation of the field \( \mathbf{F} \) around the curve \( C \) is 0.

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Do you want more details or have any questions?

**Related Questions:**
1. What is the physical interpretation of Stokes' Theorem?
2. How would the circulation change if the surface was tilted in space?
3. What if the vector field \( \mathbf{F} \) was different? How does that affect the outcome?
4. How does Stokes' Theorem relate to the Divergence Theorem?
5. Can Stokes' Theorem be used for non-planar curves?

**Tip:** When applying Stokes' Theorem, always check if the curve is oriented correctly (counterclockwise when viewed from above).

Learn how to calculate the circulation of the vector field \( \mathbf{F} = (y^2 + z^2)\mathbf{i} + (x^2 + y^2)\mathbf{j} + (x^2 + y^2)\mathbf{k} \) around the curve \( C \), a square bounded by the lines \( x = \pm 11 \) and \( y = \pm 11 \) in the \( xy \)-plane, counterclockwise when viewed from above. Follow the application of Stokes' Theorem, involving the curl of \( \mathbf{F} \) and surface integral calculations. Suitable for advanced undergraduate studies.

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