Math Problem Statement

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

Solution

To use Stokes' theorem to evaluate the line integral C(4ydx+2zdy+7xdx)\oint_C (4y \, dx + 2z \, dy + 7x \, dx) where CC is the intersection of the xyxy-plane and the hemisphere z=1x2y2z = \sqrt{1 - x^2 - y^2}, we need to follow these steps:

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

  2. Parameterize the surface SS: Here, SS is the upper hemisphere z=1x2y2z = \sqrt{1 - x^2 - y^2} above the circle x2+y2=1x^2 + y^2 = 1.

  3. Calculate the curl of F\mathbf{F}: F=(4y+7x,2z,0)\mathbf{F} = (4y + 7x, 2z, 0)

    \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 dSd\mathbf{S}: The surface element for the hemisphere can be written as dS=ndSd\mathbf{S} = \mathbf{n} \, dS, where n\mathbf{n} is the unit normal vector. For the upper hemisphere, n=k\mathbf{n} = \mathbf{k} and dS=dAdS = dA in the xyxy-plane.

  5. Evaluate the surface integral:

    = \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 4π\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.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Stokes' Theorem
Surface Integrals

Formulas

Stokes' Theorem: \( \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \)

Theorems

Stokes' Theorem

Suitable Grade Level

Undergraduate