Question

Consider the line integral \( \int_C \mathbf{F}(\mathbf{r}) \cdot d\mathbf{r \), with \( \mathbf{F}(\mathbf{r}) = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \), where \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors in the \( (x, y, z) \) Cartesian coordinate system. The path \( C \) is given by \( \mathbf{r}(t) = \cos(t) \mathbf{i} + \sin(t) \mathbf{j} + t \mathbf{k} \), where \( 0 \leq t \leq \pi \). The value of the integral, rounded off to 2 decimal places, is:}

Hint:

For line integrals, parametrize the path carefully, compute \( \mathbf{F}(\mathbf{r}) \cdot \frac{d\mathbf{r}}{dt} \), and simplify the resulting integral for accurate evaluation.

Answer 1

Given: (i) \[ F(r) = x\hat{i} + y\hat{j} + z\hat{k}, \] where \( \hat{i}, \hat{j}, \hat{k} \) are unit vectors in the \((x, y, z)\) Cartesian coordinate system. (ii) The path \( C \) is given by: \[ r(t) = \cos(t)\hat{i} + \sin(t)\hat{j} + t\hat{k}. \] Line Integral: \[ \int_C F(r) \cdot dr = \int_C (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (dx\hat{i} + dy\hat{j} + dz\hat{k}). \] \[ = \int_C (x dx + y dy + z dz). \] Parameters: \[ x = \cos(t), \quad y = \sin(t), \quad z = t. \] \[ dx = -\sin(t)dt, \quad dy = \cos(t)dt, \quad dz = dt. \] Substitute these values: \[ \int_C F(r) \cdot dr = \int_0^\pi \left[\cos(t)(-\sin(t)dt) + \cos(t)(\sin(t)dt) + t(dt)\right]. \] Simplify: \[ \int_C F(r) \cdot dr = \int_0^\pi t dt. \] Solve the integral: \[ \int_0^\pi t dt = \left[\frac{t^2}{2}\right]_0^\pi = \frac{\pi^2}{2}. \] Final value of the integral: \[ \int_C F(r) \cdot dr = \frac{\pi^2}{2} \approx 4.93. \] Conclusion: The value of the integral is \( 4.93 \).