Question:medium

The circulation of \(\vec{F}=y\hat{i}+z\hat{j}+x\hat{k}\) around the circle \(x^2+y^2=1,\ z=0\) is ____.

Show Hint

For circulation around a closed curve, parametrize the curve and calculate \(\oint_C \vec{F}\cdot d\vec{r}\).
  • \(\pi\)
  • \(\dfrac{\pi}{2}\)
  • \(\dfrac{\pi}{4}\)
  • \(-\pi\)
Show Solution

The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
Circulation is defined as the line integral of a vector field around a closed curve, $\oint_C \vec{F} \cdot d\vec{r}$. We can compute this either by direct parameterization or by using Stokes' Theorem. Stokes' Theorem is often simpler if the curl of $\vec{F}$ is easy to compute.

Step 2: Key Formula or Approach (using Stokes' Theorem):

Stokes' Theorem states: $\oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S}$.
The curve $C$ is the circle $x^2+y^2=1$ in the $z=0$ plane. The surface $S$ is the disk $x^2+y^2 \le 1$ in the $z=0$ plane.
The standard (counter-clockwise) orientation of the curve corresponds to an upward-pointing normal vector for the surface, so $d\vec{S} = \hat{k} \,dA$.

Step 3: Detailed Explanation:

First, calculate the curl of $\vec{F}$:
\[ \nabla \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} y & z & x \end{vmatrix} \] \[ = \hat{i}\left(\frac{\partial}{\partial y}(x) - \frac{\partial}{\partial z}(z)\right) - \hat{j}\left(\frac{\partial}{\partial x}(x) - \frac{\partial}{\partial z}(y)\right) + \hat{k}\left(\frac{\partial}{\partial x}(z) - \frac{\partial}{\partial y}(y)\right) \] \[ = \hat{i}(0 - 1) - \hat{j}(1 - 0) + \hat{k}(0 - 1) = -\hat{i} - \hat{j} - \hat{k} \] Now, compute the dot product $(\nabla \times \vec{F}) \cdot d\vec{S}$:
\[ (\nabla \times \vec{F}) \cdot d\vec{S} = (-\hat{i} - \hat{j} - \hat{k}) \cdot (\hat{k} \,dA) = -1 \,dA \] Finally, integrate over the surface S (the unit disk):
\[ \iint_S (-1) \,dA = -1 \iint_S dA = -1 \times (\text{Area of the unit disk}) \] The area of a disk with radius $r=1$ is $\pi r^2 = \pi$.
\[ \text{Circulation} = -1 \times \pi = -\pi \]

Step 4: Final Answer:

The circulation of the vector field is $-\pi$.
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