Let's solve the problem **step by step** using **Stokes's Theorem**. --- ### **Step 1: Understanding the Surface \( S \) and its Boundary \( C \)** - The surface \( S \) is the upper hemisphere \( z = \sqrt{4 - x^2 - y^2} \) (since \( z \geq 0 \)), which is the upper half of the sphere \( x^2 + y^2 + z^2 = 4 \). - The boundary \( C \) of \( S \) is the circle in the \( xy \)-plane where \( z = 0 \): \[ x^2 + y^2 = 4, \quad z = 0 \] This is a circle of radius 2 centered at the origin in the \( xy \)-plane. --- ### **Step 2: Stokes's Theorem** Stokes's theorem states: \[ \iint_{S} (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_{C} \mathbf{F} \cdot d\mathbf{r} \] So, instead of evaluating the surface integral, we can evaluate the line integral around \( C \). --- ### **Step 3: Parametrize the Boundary Curve \( C \)** Let \[ C: \begin{cases} x = 2\cos t \\ y = 2\sin t \\ z = 0 \\ \end{cases}, \quad t \in [0, 2\pi] \] --- ### **Step 4: Compute \( \mathbf{F} \) on \( C \)** Given \( \mathbf{F}(x, y, z) = (-y, x, 1) \): On \( C \): \[ \mathbf{F}(2\cos t, 2\sin t, 0) = (-2\sin t, 2\cos t, 1) \] --- ### **Step 5: Compute \( d\mathbf{r} \) on \( C \)** \[ d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = (-2\sin t, 2\cos t, 0) dt \] --- ### **Step 6: Evaluate the Dot Product \( \mathbf{F} \cdot d\mathbf{r} \)** \[ \mathbf{F} \cdot d\mathbf{r} = (-2\sin t, 2\cos t, 1) \cdot (-2\sin t, 2\cos t, 0) dt \] \[ = (-2\sin t)(-2\sin t) + (2\cos t)(2\cos t) + (1)(0) \] \[ = 4\sin^2 t + 4\cos^2 t \] \[ = 4(\sin^2 t + \cos^2 t) = 4 \] --- ### **Step 7: Evaluate the Line Integral** \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} 4 \, dt = 4[ t ]_{0}^{2\pi} = 4(2\pi - 0) = 8\pi \] --- ## **Final Answer** \[ \boxed{8\pi} \] This is the value of the flux integral using **Stokes's theorem**.