Let's solve the line integral step-by-step as requested: --- ## **Given Vector Field** In cylindrical coordinates: \[ \mathbf{A} = 2r \sin\phi\, \mathbf{a}_r + r^2 \mathbf{a}_\phi + z \mathbf{a}_z \] ## **Path \( C \) Description** The path \( C \) is a closed loop comprising four segments: 1. **Segment 1:** \((0,0,0) \to (1,0,0)\) (straight line in \( r \)-direction) 2. **Segment 2:** \((1,0,0) \to (1,\pi/2,0)\) (arc at \( r=1, z=0 \)) 3. **Segment 3:** \((1,\pi/2,0) \to (1,\pi/2,1)\) (straight line in \( z \)-direction) 4. **Segment 4:** \((1,\pi/2,1) \to (0,0,0)\) (straight line back to the origin) --- # **Step 1: Compute Each Segment** We break the integral: \[ \oint_C \mathbf{A} \cdot d\mathbf{l} = \sum_{i=1}^4 \int_{C_i} \mathbf{A} \cdot d\mathbf{l} \] --- ## **Segment 1:** \((0,0,0) \to (1,0,0)\) - \( \phi = 0, z = 0 \), \( r \) varies from 0 to 1. - \( d\mathbf{l} = dr\,\mathbf{a}_r \) - At \( \phi = 0 \), \( \sin\phi = 0 \), so \( \mathbf{A} = r^2 \mathbf{a}_\phi \) - \( \mathbf{A} \cdot d\mathbf{l} = r^2 \mathbf{a}_\phi \cdot \mathbf{a}_r\, dr = 0 \) (since \( \mathbf{a}_\phi \perp \mathbf{a}_r \)) **Contribution:** \(\boxed{0}\) --- ## **Segment 2:** \((1,0,0) \to (1,\pi/2,0)\) (arc of a circle) - \( r = 1, z = 0 \), \( \phi \) varies from 0 to \( \pi/2 \) - \( d\mathbf{l} = r d\phi\,\mathbf{a}_\phi = d\phi\,\mathbf{a}_\phi \) - \( \mathbf{A} = 2\sin\phi\,\mathbf{a}_r + \mathbf{a}_\phi \) - \( \mathbf{A} \cdot d\mathbf{l} = \mathbf{a}_\phi \cdot \mathbf{a}_\phi\, d\phi = d\phi \) **Integral:** \[ \int_0^{\pi/2} d\phi = \left[ \phi \right]_0^{\pi/2} = \frac{\pi}{2} \] **Contribution:** \(\boxed{\frac{\pi}{2}}\) --- ## **Segment 3:** \((1,\pi/2,0) \to (1,\pi/2,1)\) - \( r = 1, \phi = \frac{\pi}{2} \), \( z \) varies from 0 to 1 - \( d\mathbf{l} = dz\,\mathbf{a}_z \) - At \( \phi = \frac{\pi}{2} \), \( \sin\phi = 1 \) - \( \mathbf{A} = 2\mathbf{a}_r + \mathbf{a}_\phi + z\mathbf{a}_z \) - \( \mathbf{A} \cdot d\mathbf{l} = z\, dz \) **Integral:** \[ \int_0^1 z\, dz = \left[ \frac{z^2}{2} \right]_0^1 = \frac{1}{2} \] **Contribution:** \(\boxed{\frac{1}{2}}\) --- ## **Segment 4:** \((1,\pi/2,1) \to (0,0,0)\) Let's parametrize this as: - \( r: 1 \to 0 \) - \( \phi: \frac{\pi}{2} \to 0 \) - \( z: 1 \to 0 \) Let \( t \) go from 0 to 1: - \( r = 1 - t \) - \( \phi = \frac{\pi}{2}(1 - t) \) - \( z = 1 - t \) Then, - \( dr = -dt \) - \( d\phi = -\frac{\pi}{2}dt \) - \( dz = -dt \) So, \[ d\mathbf{l} = dr\,\mathbf{a}_r + r\,d\phi\,\mathbf{a}_\phi + dz\,\mathbf{a}_z = (-dt)\,\mathbf{a}_r + (1-t)\left(-\frac{\pi}{2}dt\right)\mathbf{a}_\phi + (-dt)\mathbf{a}_z \] \[ \mathbf{A} = 2r\sin\phi\,\mathbf{a}_r + r^2\mathbf{a}_\phi + z\mathbf{a}_z \] Compute the dot product: \[ \mathbf{A} \cdot d\mathbf{l} = [2r\sin\phi](-dt) + [r^2]\, [(1-t)\left(-\frac{\pi}{2}dt\right)] + [z](-dt) \] \[ = -2r\sin\phi\, dt - r^2 (1-t) \frac{\pi}{2} dt - z\,dt \] Substitute \( r = 1-t, \phi = \frac{\pi}{2}(1-t), z = 1-t \): - \( r = 1-t \) - \( \sin\phi = \sin\left(\frac{\pi}{2}(1-t)\right) \) - \( r^2 = (1-t)^2 \) - \( z = 1-t \) So, \[ \mathbf{A} \cdot d\mathbf{l} = -2(1-t)\sin\left(\frac{\pi}{2}(1-t)\right)dt - (1-t)^2 (1-t)\frac{\pi}{2}dt - (1-t)dt \] \[ = -2(1-t)\sin\left(\frac{\pi}{2}(1-t)\right)dt - (1-t)^3\frac{\pi}{2}dt - (1-t)dt \] Integrate from \( t = 0 \) to \( t = 1 \): \[ I_4 = -2\int_0^1 (1-t)\sin\left(\frac{\pi}{2}(1-t)\right)dt - \frac{\pi}{2}\int_0^1 (1-t)^3 dt - \int_0^1 (1-t)dt \] Let's compute each term: --- ### **First Term:** Let \( u = 1-t \), so when \( t = 0, u = 1 \), \( t = 1, u = 0 \), \( dt = -du \) \[ \int_0^1 (1-t)\sin\left(\frac{\pi}{2}(1-t)\right)dt = \int_1^0 u \sin\left(\frac{\pi}{2}u\right)(-du) = \int_0^1 u \sin\left(\frac{\pi}{2}u\right) du \] Integrate by parts: Let \( v = u, dw = \sin\left(\frac{\pi}{2}u\right)du \) So, \( dv = du, w = -\frac{2}{\pi}\cos\left(\frac{\pi}{2}u\right) \) So: \[ \int u \sin\left(\frac{\pi}{2}u\right) du = -\frac{2u}{\pi}\cos\left(\frac{\pi}{2}u\right) + \frac{2}{\pi}\int \cos\left(\frac{\pi}{2}u\right) du \] But \( \int \cos(au) du = \frac{1}{a} \sin(au) \), so \( a = \frac{\pi}{2} \): \[ \int \cos\left(\frac{\pi}{2}u\right) du = \frac{2}{\pi} \sin\left(\frac{\pi}{2}u\right) \] So: \[ \int u \sin\left(\frac{\pi}{2}u\right) du = -\frac{2u}{\pi}\cos\left(\frac{\pi}{2}u\right) + \frac{4}{\pi^2} \sin\left(\frac{\pi}{2}u\right) \] Evaluate from \( u = 0 \) to \( u = 1 \): - At \( u = 1 \): \( \cos\left(\frac{\pi}{2}\right) = 0 \), \( \sin\left(\frac{\pi}{2}\right) = 1 \) - At \( u = 0 \): \( \cos(0) = 1 \), \( \sin(0) = 0 \) \[ \left[ -\frac{2u}{\pi}\cos\left(\frac{\pi}{2}u\right) + \frac{4}{\pi^2} \sin\left(\frac{\pi}{2}u\right) \right]_0^1 = \left( 0 + \frac{4}{\pi^2}(1) \right) - \left( 0 + 0 \right) = \frac{4}{\pi^2} \] So, the first term is: \[ -2 \times \frac{4}{\pi^2} = -\frac{8}{\pi^2} \] --- ### **Second Term:** \[ \int_0^1 (1-t)^3 dt = \int_0^1 u^3 du = \left[ \frac{u^4}{4} \right]_0^1 = \frac{1}{4} \] So, this term is: \[ - \frac{\pi}{2} \cdot \frac{1}{4} = -\frac{\pi}{8} \] --- ### **Third Term:** \[ \int_0^1 (1-t)dt = \left[ t - \frac{t^2}{2} \right]_0^1 = 1 - \frac{1}{2} = \frac{1}{2} \] So, this term is: \[ - \frac{1}{2} \] --- ### **Sum for Segment 4:** \[ I_4 = -\frac{8}{\pi^2} - \frac{\pi}{8} - \frac{1}{2} \] --- # **Step 2: Add All Segments** \[ \boxed{ \oint_C \mathbf{A} \cdot d\mathbf{l} = 0 + \frac{\pi}{2} + \frac{1}{2} + \left( -\frac{8}{\pi^2} - \frac{\pi}{8} - \frac{1}{2} \right) } \] Simplifying: \[ = \frac{\pi}{2} + \frac{1}{2} - \frac{8}{\pi^2} - \frac{\pi}{8} - \frac{1}{2} = \frac{\pi}{2} - \frac{\pi}{8} - \frac{8}{\pi^2} = \frac{4\pi - \pi}{8} - \frac{8}{\pi^2} = \frac{3\pi}{8} - \frac{8}{\pi^2} \] --- # **Final Answer** \[ \boxed{ \displaystyle \oint_C \mathbf{A} \cdot d\mathbf{l} = \frac{3\pi}{8} - \frac{8}{\pi^2} } \] --- **Summary of Steps:** 1. Broke the path into 4 segments and parametrized each. 2. Evaluated the integral for each segment. 3. Used substitution and integration by parts as needed. 4. Added all contributions for the final boxed answer.