Consider F\left ( x, \ y, \ z \right ) = -\frac{y}{x^2 + y^2}i + \frac{x}{x^2 + y^2}j + zk a....

Question:

Consider {eq}F\left ( x, \ y, \ z \right ) = -\frac{y}{x^2 + y^2}i + \frac{x}{x^2 + y^2}j + zk {/eq}

a. Compute {eq}\int_{C}F\cdot dr {/eq}

C: is the circle {eq}x^2 + y^2 = 1 {/eq} in the xy plane oriented counterclockwise.

b. Compute {eq}\iint_{S}curlF\cdot dS {/eq}

For the surface {eq}z = 1 - x^2 - y^2 {/eq} bounded by the plane z = 0.

c. Explain why Stokes? Theorem does not work in this case.

Checking Stokes' Theorem:

Stokes' Theorem tells us that if we have a smooth surface {eq}S {/eq} with an outward pointing unit normal vector {eq}\mathbf{N}, {/eq} where {eq}S {/eq} is bounded by a piecewise continuous curve {eq}C, {/eq} and a vector field {eq}\mathbf{F}(x, y, z) = f(x, y, z) \mathbf{i} + g(x, y, z) \mathbf{j} + h(x, y, z) \mathbf{k}, {/eq} we have the identity

{eq}\displaystyle\int_C \mathbf{F} \cdot d\mathbf{r} = \iint_S \mathbf{curl \: F} \cdot \mathbf{N} \: dS {/eq}

There is one additional condition: The components of {eq}\mathbf F {/eq} must have continuous partial derivatives on an open region that contains {eq}S. {/eq} If this condition is not satisfied, then the identity does not necessarily hold.

Answer and Explanation:

a. We are given the vector field {eq}\mathbf F\left ( x, \ y, \ z \right ) = -\displaystyle\frac{y}{x^2 + y^2} \: \mathbf i + \frac{x}{x^2 + y^2} \: \mathbf j + z \: \mathbf k {/eq} and the circle {eq}x^2 + y^2 = 1 {/eq} in the xy plane oriented counterclockwise. We can parametrize the circle as {eq}x(t) = \cos t, \: y(t) = \sin t, \: z(t) = 0, \: 0 \leq t \leq 2\pi. {/eq} We can now evaluate the line integral:

{eq}\begin{eqnarray*}\int_{C} \mathbf F \cdot d \mathbf r & =& \int_0^{2\pi} \left( -\displaystyle\frac{y}{x^2 + y^2} \: \mathbf i + \frac{x}{x^2 + y^2} \: \mathbf j + z \: \mathbf k \right) \cdot \left( -\sin t \: \mathbf{i} + \cos t \: \mathbf j \right) \: dt \\ \\ & =& \int_0^{2\pi} \displaystyle\frac{\sin^2 t}{\cos^2 t + \sin^2 t} + \frac{\cos^2 t}{\cos^2 t + \sin^2 t} \: dt \\ \\ & =& \int_0^{2\pi} dt \\ \\ & =& 2 \pi \end{eqnarray*} {/eq}

b. The boundary of the surface {eq}z = 1 - x^2 - y^2, \: z \geq 0 {/eq} is the unit circle from part (a). The curl of the vector field is

{eq}\begin{eqnarray}\mathbf{curl \: F} & = & \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \displaystyle\frac{\partial}{\partial x} & \displaystyle\frac{\partial}{\partial y} & \displaystyle\frac{\partial}{\partial z} \\ -\displaystyle\frac{y}{x^2 + y^2} & \displaystyle\frac{x}{x^2 + y^2} & 0 \end{vmatrix} \\ \\ & = & \left( \displaystyle\frac{(1)(x^2 + y^2) - x(2x)}{(x^2 + y^2)^2} + \frac{(1)(x^2 + y^2) - (-y)(2y)}{(x^2 + y^2)^2} \right) \mathbf{k} \\ \\ & = & \mathbf 0 \end{eqnarray} {/eq}

Therefore, {eq}\displaystyle\iint_{S} \mathbf{curl \: F} \cdot \mathbf N \: dS = 0. {/eq}

c. Stokes' Theorem does not apply because the vector field {eq}\mathbf F {/eq} is not defined at the origin, which is inside the boundary curve {eq}C. {/eq}


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