use stokes theorem to evaluate the lone integral for c 4ydx+2zdy+7xdx where c is the intersection of the xy-plane and hemisphere z=sqrt1-x^2-y^2 travesed counterclockwise viewed from the top that is from the positive z axis toward the xy-plane

To use Stokes' theorem to evaluate the line integral \(\oint_C (4y \, dx + 2z \, dy + 7x \, dx)\) where \(C\) is the intersection of the \(xy\)-plane and the hemisphere \(z = \sqrt{1 - x^2 - y^2}\), we need to follow these steps:

1. **State Stokes' theorem:**
   \[
   \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S}
   \]
   where \(\mathbf{F} = (4y + 7x) \mathbf{i} + 2z \mathbf{j} + 0 \mathbf{k}\), \(S\) is the surface enclosed by \(C\), and \(d\mathbf{S}\) is the vector surface element.

2. **Parameterize the surface \(S\):**
   Here, \(S\) is the upper hemisphere \(z = \sqrt{1 - x^2 - y^2}\) above the circle \(x^2 + y^2 = 1\).

3. **Calculate the curl of \(\mathbf{F}\):**
   \[
   \mathbf{F} = (4y + 7x, 2z, 0)
   \]
   \[
   \nabla \times \mathbf{F} = \left| \begin{matrix}
   \mathbf{i} & \mathbf{j} & \mathbf{k} \\
   \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
   4y + 7x & 2z & 0
   \end{matrix} \right|
   = \left( \frac{\partial 0}{\partial y} - \frac{\partial 2z}{\partial z} \right) \mathbf{i}
   - \left( \frac{\partial 0}{\partial x} - \frac{\partial (4y + 7x)}{\partial z} \right) \mathbf{j}
   + \left( \frac{\partial (2z)}{\partial x} - \frac{\partial (4y + 7x)}{\partial y} \right) \mathbf{k}
   \]
   Simplifying the determinants:
   \[
   \nabla \times \mathbf{F} = (-2) \mathbf{i} + (0) \mathbf{j} + (2 - 4) \mathbf{k}
   = (-2, 0, -4)
   \]

4. **Calculate the surface element \(d\mathbf{S}\):**
   The surface element for the hemisphere can be written as \(d\mathbf{S} = \mathbf{n} \, dS\), where \(\mathbf{n}\) is the unit normal vector. For the upper hemisphere, \(\mathbf{n} = \mathbf{k}\) and \(dS = dA\) in the \(xy\)-plane.

5. **Evaluate the surface integral:**
   \[
   \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_S (-2, 0, -4) \cdot (0, 0, 1) \, dA
   = \iint_S (-4) \, dA
   \]
   Since \(S\) is the unit disk \(x^2 + y^2 \leq 1\) in the \(xy\)-plane:
   \[
   \iint_S dA = \text{Area of the unit disk} = \pi
   \]
   Therefore:
   \[
   \iint_S (-4) \, dA = -4 \pi
   \]

So, the value of the line integral is \(\boxed{-4\pi}\).

Would you like any additional details or have any questions?

Here are some other questions you might be interested in:

1. How to use Stokes' theorem for different vector fields and surfaces?
2. What is the physical significance of the curl of a vector field?
3. How can Green's theorem be applied to planar regions?
4. What are the differences between Green's theorem and Stokes' theorem?
5. How can surface integrals be used in physics and engineering problems?

**Tip:** When working with vector calculus theorems like Stokes' theorem, always ensure that the orientation of the surface and the curve are consistent with the right-hand rule for the curl of the vector field.

Learn how to use Stokes' theorem to evaluate the line integral \(\oint_C (4y \, dx + 2z \, dy + 7x \, dx)\), where \(C\) is the intersection of the \(xy\)-plane and the hemisphere \(z = \sqrt{1 - x^2 - y^2}\), traversed counterclockwise. Detailed step-by-step solution with calculations and concepts involved in vector calculus.

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