Question

Let \(S\) be the surface defined by \[ \{(x, y, z) \in \mathbb{R}^3 : z = 1 - x^2 - y^2,\, z \ge 0\}. \] Let \[ \vec{F} = -y\hat{i} + (x - 1)\hat{j} + z^2\hat{k}, \] and let \(\hat{n}\) be the continuous unit normal field to the surface \(S\) with positive \(z\)-component. Then the value of \[ \frac{1}{\pi} \iint_S (\nabla \times \vec{F}) \cdot \hat{n}\, dS \] is _________.

Hint:

Always check if a vector field can be simplified using Stokes’ theorem — it often turns a difficult surface integral into a simple line integral.

Answer 1

Correct Answer: 2

Step 1: Apply Stokes’ theorem.
By Stokes’ theorem, \[ \iint_S (\nabla \times \vec{F}) \cdot \hat{n}\, dS = \oint_C \vec{F} \cdot d\vec{r}, \] where \(C\) is the boundary curve of \(S.\)
Step 2: Identify the boundary.
The surface \(z = 1 - x^2 - y^2\) meets \(z = 0\) when \(x^2 + y^2 = 1.\) So \(C\) is the circle \(x^2 + y^2 = 1,\ z = 0,\) traversed counterclockwise.
Step 3: Parameterize \(C.\)
Let \(x = \cos t,\ y = \sin t,\ 0 \le t \le 2\pi.\) Then \(d\vec{r} = (-\sin t\,\hat{i} + \cos t\,\hat{j})dt.\)
Step 4: Evaluate \(\vec{F}\) on \(C.\)
On \(z=0,\) \[ \vec{F} = -y\hat{i} + (x - 1)\hat{j}. \] So, \[ \vec{F} \cdot d\vec{r} = (-\sin t)(-\sin t) + (\cos t - 1)\cos t = \sin^2 t + \cos^2 t - \cos t = 1 - \cos t. \]
Step 5: Compute line integral.
\[ \oint_C (1 - \cos t)\,dt = \int_0^{2\pi} (1 - \cos t)\,dt = [t - \sin t]_0^{2\pi} = 2\pi. \]
Step 6: Compute required value.
\[ \frac{1}{\pi} \iint_S (\nabla \times \vec{F}) \cdot \hat{n}\, dS = \frac{1}{\pi}(2\pi) = 2. \]
Step 7: Conclusion.
The required value is \(\boxed{2}.\)