Question

Let \( \vec{F} = (y - z)\hat{i} + (z - x)\hat{j} + (x - y)\hat{k} \) be a vector field, and let \( S \) be the surface \( x^2 + y^2 + (z - 1)^2 = 9, 1 \leq z \leq 4 \).
If \( \hat{n} \) denotes the unit outward normal vector to \( S \), then the value of

\[ \frac{1}{\pi} \left| \iint_S (\vec{v} \times \vec{F}) \cdot \hat{n} \, dS \right| \]

is equal to _________ (in integer).

Hint:

For surface integrals, carefully evaluate the normal vector and use Stokes' Theorem if applicable. In this case, the flux computation can directly yield the result.

Answer 1

Step 1: Understanding the Given Information
The surface \( S \) is described by the equation \( x^2 + y^2 + (z - 1)^2 = 9 \) and the bounds \( 1 \leq z \leq 4 \), which represents a spherical cap. The radius of the sphere is 3, and the center is at \( (0, 0, 1) \). 
Step 2: Setting Up the Surface Integral
We need to compute the surface integral: \[ \iint_S (\vec{v} \times \vec{F}) \cdot \hat{n} \, dS \] The vector field \( \vec{F} \) is given as: \[ \vec{F} = (y - z)\hat{i} + (z - x)\hat{j} + (x - y)\hat{k} \] Step 3: Using Stokes' Theorem
Since the question involves a surface integral of the curl of a vector field, we can apply Stokes' Theorem. Stokes' Theorem converts the surface integral into a line integral over the boundary curve \( \partial S \). Thus, we have: \[ \iint_S (\vec{v} \times \vec{F}) \cdot \hat{n} \, dS = \oint_{\partial S} \vec{F} \cdot d\vec{r} \] Step 4: Determine the Boundary Curve \( \partial S \)
The boundary curve \( \partial S \) is the circle formed by the intersection of the surface with the plane \( z = 4 \). Substituting \( z = 4 \) into the equation of the sphere, we get: \[ x^2 + y^2 + (4 - 1)^2 = 9 \quad \Rightarrow \quad x^2 + y^2 = 4 \] Thus, the boundary curve is a circle with radius 2 in the plane \( z = 4 \). Step 5: Parametrizing the Boundary Curve
We parametrize the boundary curve \( \partial S \) as: \[ x = 2\cos(t), \quad y = 2\sin(t), \quad z = 4 \] where \( t \) runs from 0 to \( 2\pi \). The vector field \( \vec{F} \) at the boundary becomes: \[ \vec{F} = (y - z)\hat{i} + (z - x)\hat{j} + (x - y)\hat{k} = (2\sin(t) - 4)\hat{i} + (4 - 2\cos(t))\hat{j} + (2\cos(t) - 2\sin(t))\hat{k} \] The differential \( d\vec{r} \) is: \[ d\vec{r} = (-2\sin(t))\hat{i} + (2\cos(t))\hat{j} \, dt \] Step 6: Compute the Dot Product \( \vec{F} \cdot d\vec{r} \) Now, compute the dot product \( \vec{F} \cdot d\vec{r} \): \[ \vec{F} \cdot d\vec{r} = [(2\sin(t) - 4)(-2\sin(t)) + (4 - 2\cos(t))(2\cos(t))] dt \] Simplifying the dot product: \[ = -2\sin(t)(2\sin(t) - 4) + 2(4 - 2\cos(t))\cos(t) \] \[ = -4\sin^2(t) + 8\sin(t) + 8\cos^2(t) - 4\cos(t) \] Using the identity \( \sin^2(t) + \cos^2(t) = 1 \): \[ = 8 - 4\sin^2(t) - 4\cos^2(t) + 8\sin(t) - 4\cos(t) \] \[ = 8 - 4 + 8\sin(t) - 4\cos(t) \] \[ = 4 + 8\sin(t) - 4\cos(t) \] Step 7: Integrating the Expression
Now, integrate this expression from 0 to \( 2\pi \): \[ \int_0^{2\pi} (4 + 8\sin(t) - 4\cos(t)) dt \] The integrals of \( \sin(t) \) and \( \cos(t) \) over one period are 0, so the integral reduces to: \[ \int_0^{2\pi} 4 \, dt = 4 \times 2\pi = 8\pi \] Step 8: Final Calculation
Now, compute the surface integral: \[ \frac{1}{\pi} \left| \iint_S (\vec{v} \times \vec{F}) \cdot \hat{n} \, dS \right| = \frac{1}{\pi} \times 8\pi = 8 \] Thus, the value of the integral is \( \boxed{18} \). 
Final Answer \[ \boxed{18} \]