Evaluate ∮C 𝐅 ·d 𝐫 for the following vector fields and...

Evaluate $\oint_{C} \mathbf{F} \cdot d \mathbf{r}$ for the following vector fields and closed oriented curves $C$ by parameterizing $C .$ If the integral is not zero, give an explanation. $$\mathbf{F}=\langle x, y, z\rangle ; \quad C: \mathbf{r}(t)=\langle\cos t, \sin t, 2\rangle, \text { for } 0 \leq t \leq 2 \pi$$

Evaluate $\oint_{C} \mathbf{F} \cdot d \mathbf{r}$ for the following vector fields and closed oriented curves $C$ by parameterizing $C .$ If the integral is not zero, give an explanation.
$$\mathbf{F}=\langle x, y, z\rangle ; \quad C: \mathbf{r}(t)=\langle\cos t, \sin t, 2\rangle, \text { for } 0 \leq t \leq 2 \pi$$

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Key Concepts

Evaluation of the Integral

After parameterizing the curve and computing the derivative, one substitutes these into the dot product. The resulting expression is then integrated over the specified interval. In problems where symmetry or cancellation occurs, the computed dot product may be identically zero, leading to a zero integral.

Line Integral of a Vector Field

A line integral of a vector field along a curve computes the work done by the field when moving along the curve. It is defined by integrating the dot product of the vector field and the differential element of the curve (often obtained via parameterization) over the parameter interval.

Parameterization of Curves

Parameterizing a curve means expressing its coordinates as functions of a parameter, typically t. This technique transforms the line integral into an ordinary integral by substituting the parameterized expressions for the coordinates and their differentials.

Dot Product in the Integral

The dot product is used in line integrals to project the vector field onto the direction of the curve. By taking the dot product of the vector field evaluated along the parameterized path and the derivative of the position vector, one obtains the integrand which represents the component of the field acting in the tangent direction of the curve.

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