Let's solve the problem step-by-step using **Stokes' Theorem**: ### **Step 1: Stokes' Theorem** Stokes' Theorem states: \[ \iint_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} = \oint_C \mathbf{F} \cdot d\mathbf{r} \] Where \( C \) is the boundary curve of surface \( S \). ### **Step 2: Identify the Boundary Curve \( C \)** The surface \( S \) is the part of the paraboloid \( z = 9 - x^2 - y^2 \) above the xy-plane (\(z\geq 0\)). The boundary occurs when \(z = 0\): \[ 0 = 9 - x^2 - y^2 \implies x^2 + y^2 = 9 \] So, \( C \) is the circle of radius 3 in the \(xy\)-plane. ### **Step 3: Parameterize \( C \)** Let: \[ \mathbf{r}(t) = \langle 3\cos t, 3\sin t, 0 \rangle, \quad t \in [0, 2\pi] \] ### **Step 4: Evaluate \( \mathbf{F} \) on \( C \)** \[ \mathbf{F}(x, y, z) = x^2 \sin(z)\,\mathbf{i} + y^2\,\mathbf{j} + xy\,\mathbf{k} \] On \(C\): \(z = 0\), \(x = 3\cos t\), \(y = 3\sin t\): - \(x^2 \sin(z) = (3\cos t)^2 \cdot 0 = 0\) - \(y^2 = (3\sin t)^2 = 9\sin^2 t\) - \(xy = (3\cos t)(3\sin t) = 9\sin t \cos t\) So, \[ \mathbf{F}(\mathbf{r}(t)) = \langle 0, 9\sin^2 t, 9\sin t \cos t \rangle \] ### **Step 5: Compute \( d\mathbf{r} \)** \[ d\mathbf{r} = \frac{d\mathbf{r}}{dt} dt = \langle -3\sin t, 3\cos t, 0 \rangle dt \] ### **Step 6: Calculate the Line Integral** \[ \mathbf{F} \cdot d\mathbf{r} = \langle 0, 9\sin^2 t, 9\sin t \cos t \rangle \cdot \langle -3\sin t, 3\cos t, 0 \rangle dt \] \[ = (0)(-3\sin t) + (9\sin^2 t)(3\cos t) + (9\sin t \cos t)(0) \] \[ = 27\sin^2 t \cos t \, dt \] ### **Step 7: Integrate over \( t \) from 0 to \( 2\pi \)** \[ \oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} 27\sin^2 t \cos t \, dt \] Let’s use substitution \(u = \sin t\), \(du = \cos t dt\): When \(t = 0\), \(u = 0\). When \(t = 2\pi\), \(u = 0\). So, \[ \int_0^{2\pi} \sin^2 t \cos t dt = \int_0^{0} u^2 du = 0 \] Thus, \[ \boxed{0} \] --- **Final Answers:** **First box:** \(\int_0^{2\pi} 27\sin^2 t \cos t \, dt\) **Second box:** \(0\) --- If you need the explicit form for the integral or more explanation, let me know!