Question

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(\mathbf{F}=y \mathbf{i}+x \mathbf{j}\) and \(C\) is given by \(\mathbf{r}(t)=(4 \sin t) \mathbf{i}+(3 \cos t) \mathbf{j}\) \(0 \leq t \leq \pi,\) then \(\int_{C} \mathbf{F} \cdot d \mathbf{r}=0\)

Short answer

The statement is false. The value of \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) is \(\pi\), not \(0\).

Step-by-step solution

Write the given vector field and curve

The vector field \(\mathbf{F}\) is given as \(y \mathbf{i}+x \mathbf{j}\), and the curve \(\mathbf{r}(t)\) is given as \((4 \sin t) \mathbf{i}+(3 \cos t) \mathbf{j}\), with \(0 \leq t \leq \pi\).

Find the derivative of the curve

Compute the derivative of \(\mathbf{r}(t)\), we have \(\mathbf{r}'(t)=(4 \cos t) \mathbf{i}-(3\sin t) \mathbf{j}\).

Express the vector field in terms of t

Using the curve to translate the vector field into function of \(t\), we obtain: \(y=3\cos t\) and \(x=4\sin t\). So, the vector field in terms of \(t\) is \(\mathbf{F}(t)=(3 \cos t) \mathbf{i}+(4 \sin t) \mathbf{j}\)

Calculate the dot product of vector field and curve's derivative

Compute \(\mathbf{F}(t) \cdot \mathbf{r}'(t)\), we have \(12\cos^{2}t-12\sin^{2}t\).

Integrate the dot product

Compute \(\int_{0}^{\pi} \mathbf{F}(t) \cdot \mathbf{r}'(t) dt\), the result is \(\pi\).

Conclusion

The given statement is false because \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) is equal to \(\pi\) rather than \(0\). Thus, the statement is not correct.