Question:
The value of the line integral
\[
\oint_{C} (-3y\,dx + 3x\,dy + z\,dz)
\]
along the circle \( C: x^{2} + y^{2} = 1,\ z = 1 \) oriented in the clockwise sense as seen from the origin, is
The value of the line integral
\[
\oint_{C} (-3y\,dx + 3x\,dy + z\,dz)
\]
along the circle \( C: x^{2} + y^{2} = 1,\ z = 1 \) oriented in the clockwise sense as seen from the origin, is
Show Hint
Clockwise orientation reverses the sign of a line integral.
Updated On: Dec 2, 2025
- \(2\pi\)
- \(4\pi\)
- \(6\pi\)
- \(8\pi\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Parametrize the circle.
The circle is \(x^{2}+y^{2}=1\) at \(z=1\). Parametric form (counter-clockwise) is:
\[ x=\cos t,\quad y=\sin t,\quad z=1 \] But the curve is oriented clockwise, so:
\[ x=\cos t,\quad y=-\sin t \]
Step 2: Compute differentials.
\[ dx=-\sin t\,dt,\qquad dy=-\cos t\,dt,\qquad dz=0 \]
Step 3: Substitute into the integral.
\[ -3y\,dx + 3x\,dy = -3(-\sin t)(-\sin t) + 3(\cos t)(-\cos t) \] \[ = -3(\sin^{2} t + \cos^{2} t) = -3 \]
Step 4: Integrate over \(0\) to \(2\pi\).
\[ \int_{0}^{2\pi} -3\,dt = -6\pi \]
Step 5: Account for clockwise orientation.
Clockwise reverses the standard orientation:
\[ -(-6\pi) = 6\pi \] Conclusion:
The value of the line integral is \(6\pi\).
The circle is \(x^{2}+y^{2}=1\) at \(z=1\). Parametric form (counter-clockwise) is:
\[ x=\cos t,\quad y=\sin t,\quad z=1 \] But the curve is oriented clockwise, so:
\[ x=\cos t,\quad y=-\sin t \]
Step 2: Compute differentials.
\[ dx=-\sin t\,dt,\qquad dy=-\cos t\,dt,\qquad dz=0 \]
Step 3: Substitute into the integral.
\[ -3y\,dx + 3x\,dy = -3(-\sin t)(-\sin t) + 3(\cos t)(-\cos t) \] \[ = -3(\sin^{2} t + \cos^{2} t) = -3 \]
Step 4: Integrate over \(0\) to \(2\pi\).
\[ \int_{0}^{2\pi} -3\,dt = -6\pi \]
Step 5: Account for clockwise orientation.
Clockwise reverses the standard orientation:
\[ -(-6\pi) = 6\pi \] Conclusion:
The value of the line integral is \(6\pi\).
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