Question

Find the value of the line integral $$\int_{C} \mathbf{F} \cdot d \mathbf{r}$$ (Hint: If \(\mathbf{F}\) is conservative, the integration may be easier on an alternative path.) \(\mathbf{F}(x, y, z)=(2 y+x) \mathbf{i}+\left(x^{2}-z\right) \mathbf{j}+(2 y-4 z) \mathbf{k}\) (a) \(\mathbf{r}_{1}(t)=t \mathbf{i}+t^{2} \mathbf{j}+\mathbf{k}, \quad 0 \leq t \leq 1\) (b) \(\mathbf{r}_{2}(t)=t \mathbf{i}+t \mathbf{j}+(2 t-1)^{2} \mathbf{k}, \quad 0 \leq t \leq 1\)

Short answer

Since the field is not conservative, you cannot use the fundamental theorem of line integrals, so simply compute each integral separately.

Step-by-step solution

Check if the field is conservative

Compute the curl of the vector field \( \mathbf{F} \), which is \( \nabla \times \mathbf{F} \). If \( \nabla \times \mathbf{F} = \mathbf{0} \), then \( \mathbf{F} \) is conservative. Note that, curl \( \mathbf{F} = \left( \frac{\partial}{\partial y} (2y-4z) - \frac{\partial}{\partial z} (x^{2}-z) \right) \mathbf{i} - \left( \frac{\partial}{\partial x} (2y-4z) - \frac{\partial}{\partial z} (2y+x) \right) \mathbf{j} + \left( \frac{\partial}{\partial x} (x^{2}-z) - \frac{\partial}{\partial y} (2y+x) \right) \mathbf{k} = 2\mathbf{i} + 0\mathbf{j} + 0 \mathbf{k} \neq 0 \)}. Hence the field is not conservative.

Parameterization of paths and compute \( \mathbf{F} \cdot d \mathbf{r} \)

The given paths are \( \mathbf{r}_1(t) \) and \( \mathbf{r}_2(t) \). Compute the derivative of these paths and calculate \( \mathbf{F}( \mathbf{r}_1(t)) \cdot \mathbf{r}_1'(t) \) and \( \mathbf{F}( \mathbf{r}_2(t)) \cdot \mathbf{r}_2'(t) \).

Compute the line integral over \( \mathbf{r}_1 \) and \( \mathbf{r}_2 \)

Then, perform the line integral which is \(\int_{0}^{1} \mathbf{F}( \mathbf{r}_1(t)) \cdot \mathbf{r}_1'(t) dt \) and \(\int_{0}^{1} \mathbf{F}( \mathbf{r}_2(t)) \cdot \mathbf{r}_2'(t) dt \). Simplify the integral and find the solution.