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.) \(\mathbf{F}(x, y, z)=y z \mathbf{i}+x z \mathbf{j}+x y \mathbf{k}\) (a) \(\mathbf{r}_{1}(t)=t \mathbf{i}+2 \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 4\) (b) \(\mathbf{r}_{2}(t)=t^{2} \mathbf{i}+t \mathbf{j}+t^{2} \mathbf{k}, \quad 0 \leq t \leq 2\)

Short answer

The value of the line integral over path 1, \(C1\), is -16, and over path 2, \(C2\), is 6.

Step-by-step solution

Identify the vector field

The given vector field is \(\mathbf{F}(x, y, z)=y z \mathbf{i}+x z \mathbf{j}+x y \mathbf{k}\).

Determine whether the vector field is conservative

A vector field is conservative if it is the gradient of a scalar potential function. In other words, \(\mathbf{F} = \nabla f\). By inspection, it is clear that this vector field is not conservative because it cannot be written in the form \(\nabla f\). Hence, we can't use simple path based on the potential function and we will compute directly the line integral on each path.

Compute line integral on path 1

\( \mathbf{r}_{1}(t)=t \mathbf{i}+2 \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq 4 \). The bounds of the integral are \(0 \leq t \leq 4\). We find: \( \int_{C1} \mathbf{F} \cdot d\mathbf{r}_{1} = \int_{0}^{4} (t(2)t-t(2)-2t) dt = \int_{0}^{4} dt(-2t)=-16. \)

Compute line integral on path 2

\( \mathbf{r}_{2}(t)=t^{2} \mathbf{i}+t \mathbf{j}+t^{2} \mathbf{k}, \quad 0 \leq t \leq 2 \). The bounds of the integral are \(0 \leq t \leq 2\). We get: \( \int_{C2} \mathbf{F} \cdot d\mathbf{r}_{2} = \int_{0}^{2} (2t^{3}-t^{3}+t^{3}) dt = \int_{0}^{2} dt(2t^{3})=6. \)