Let S be the cylinder x squared plus y squared equals a squared​, 0 less than or equals z less than or equals h​, together with its​ top, x squared plus y squared less than or equals a squared​, zequalsh. Let Bold Upper F equals negative 2 y Bold i plus 2 x Bold j plus 2 x squared Bold k. Use​ Stokes' Theorem to find the flux of gradienttimesF through S in the direction away from the interior of the cylinder.
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Part 1
The flux of gradienttimesF outward through S is 
  
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To find the flux of \( \nabla \times \mathbf{F} \) through the surface \( S \) using Stokes' Theorem, we follow these steps:

### **Step 1: State Stokes' Theorem**
Stokes' Theorem states that:

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

However, since we are asked to find the flux of \( \nabla \times \mathbf{F} \) through the surface \( S \), the problem already provides the surface integral directly:

\[
\text{Flux of } \nabla \times \mathbf{F} \text{ through } S = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}
\]

### **Step 2: Compute the Curl of \( \mathbf{F} \)**
Given the vector field:

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

We need to calculate \( \nabla \times \mathbf{F} \):

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

Expanding the determinant:

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

Calculating the partial derivatives:

\[
\nabla \times \mathbf{F} = \mathbf{i} \left(0 - 0\right) - \mathbf{j} \left(4x - 0 \right) + \mathbf{k} \left(2 - (-2)\right)
\]

This simplifies to:

\[
\nabla \times \mathbf{F} = -4x \mathbf{j} + 4 \mathbf{k}
\]

Thus:

\[
\nabla \times \mathbf{F} = -4x \mathbf{j} + 4 \mathbf{k}
\]

### **Step 3: Set Up the Surface Integral**
The surface \( S \) consists of the cylindrical surface and its top. We will compute the flux through each part separately.

#### **Part 1: Flux through the Top Surface**
The top surface is the disk \( x^2 + y^2 \leq a^2 \) at \( z = h \). The surface normal vector \( d\mathbf{S} \) points in the \( \mathbf{k} \) direction, so \( d\mathbf{S} = \mathbf{k} \, dA \), where \( dA \) is the area element on the \( xy \)-plane.

The flux through the top surface is:

\[
\iint_{\text{top}} (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_{\text{top}} 4 \, dA
\]

The area of the top surface is:

\[
\text{Area} = \pi a^2
\]

So the flux through the top surface is:

\[
\text{Flux}_{\text{top}} = 4 \times \pi a^2
\]

#### **Part 2: Flux through the Cylindrical Surface**
For the cylindrical surface, the surface normal is radial and lies in the \( xy \)-plane, i.e., \( d\mathbf{S} \) is in the direction of \( \mathbf{r} \) and has a magnitude \( d\mathbf{S} = a \, dz \, d\theta \). The flux through the cylindrical surface is:

\[
\iint_{\text{cylinder}} (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_{\text{cylinder}} (-4x \mathbf{j} + 4 \mathbf{k}) \cdot (r \, dz \, d\theta)
\]

Given that \( \mathbf{r} \) is normal to the cylindrical surface, the \( \mathbf{k} \) component does not contribute to the flux through the cylindrical surface because it is orthogonal to \( d\mathbf{S} \). Hence:

\[
\text{Flux}_{\text{cylinder}} = \iint_{\text{cylinder}} 0 \, dz \, d\theta = 0
\]

### **Final Answer:**
The total flux through the surface \( S \) is the flux through the top plus the flux through the cylindrical surface:

\[
\text{Total Flux} = \text{Flux}_{\text{top}} + \text{Flux}_{\text{cylinder}} = 4 \pi a^2 + 0 = 4 \pi a^2
\]

Thus, the flux of \( \nabla \times \mathbf{F} \) outward through \( S \) is:

\[
\boxed{4 \pi a^2}
\]

---

Do you want more details or have any questions?

**Related Questions:**
1. How does the flux change if the height \( h \) of the cylinder is doubled?
2. What would happen if the vector field \( \mathbf{F} \) had a different \( z \)-component?
3. Can Stokes' Theorem be applied if the surface \( S \) was open instead of closed?
4. How does the direction of the flux change if the orientation of the surface is reversed?
5. How is Stokes' Theorem related to the Divergence Theorem in this scenario?

**Tip:** Always verify the orientation of the surface and the direction of the normal vector when applying Stokes' Theorem to ensure correct results.

Learn how to calculate the flux of gradient times F through a cylindrical surface using Stokes' Theorem. Detailed step-by-step solution provided for vector field F = -2y i + 2x j + 2x^2 k over the specified surface S.

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