Let's go through your problem step by step: --- ## (a) **Describe and Parameterize the Surface** The surface is \[ z = 4 - x^2 - y^2, \quad z \ge 0 \] This is an **upper hemisphere** (paraboloid cap) opening downward, cut off where \( z = 0 \). The boundary occurs when \( z = 0 \): \[ 0 = 4 - x^2 - y^2 \implies x^2 + y^2 = 4 \] So the domain in the \(xy\)-plane is the disk of radius 2 centered at the origin. ### **Diagram** This is a "cap" of the paraboloid \( z = 4 - x^2 - y^2 \) above the disk \( x^2 + y^2 \le 4 \) in the \(xy\)-plane. ### **Parameterization** Use polar coordinates: \[ \begin{align*} x &= r\cos\theta \\ y &= r\sin\theta \\ z &= 4 - r^2 \\ \end{align*} \] where \( 0 \le r \le 2 \), \( 0 \le \theta < 2\pi \). So the parameterization is: \[ \vec{r}(r, \theta) = \langle r\cos\theta,\, r\sin\theta,\, 4 - r^2 \rangle \] --- ## (b) **Compute \( \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} \) using the parameterization** Given: \[ \vec{F} = \langle y,\, 2z,\, x^2 \rangle \] First, compute the **curl**: \[ \nabla \times \vec{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y & 2z & x^2 \end{vmatrix} \] Compute each component: ### \( \mathbf{i} \) component: \[ \frac{\partial}{\partial y}(x^2) - \frac{\partial}{\partial z}(2z) = 0 - 2 = -2 \] ### \( \mathbf{j} \) component: \[ \frac{\partial}{\partial z}(y) - \frac{\partial}{\partial x}(x^2) = 0 - 2x = -2x \] ### \( \mathbf{k} \) component: \[ \frac{\partial}{\partial x}(2z) - \frac{\partial}{\partial y}(y) = 0 - 1 = -1 \] So, \[ \nabla \times \vec{F} = \langle -2,\, -2x,\, -1 \rangle \] --- ### **Surface element \( d\vec{S} \)** For a surface \( \vec{r}(r,\theta) \), the oriented surface element is: \[ d\vec{S} = \vec{r}_r \times \vec{r}_\theta \, dr\, d\theta \] Compute: \[ \vec{r}_r = (\cos\theta, \sin\theta, -2r) \] \[ \vec{r}_\theta = (-r\sin\theta, r\cos\theta, 0) \] Cross product: \[ \vec{r}_r \times \vec{r}_\theta = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \cos\theta & \sin\theta & -2r \\ - r\sin\theta & r\cos\theta & 0 \end{vmatrix} \] Expand: - \( \mathbf{i} \): \( \sin\theta \cdot 0 - (-2r) \cdot r\cos\theta = 2r^2\cos\theta \) - \( -\mathbf{j} \): \( \cos\theta \cdot 0 - (-2r)(-r\sin\theta) = -2r^2\sin\theta \) - \( \mathbf{k} \): \( \cos\theta \cdot r\cos\theta - \sin\theta \cdot (-r\sin\theta) = r(\cos^2\theta + \sin^2\theta) = r \) So, \[ \vec{r}_r \times \vec{r}_\theta = \langle 2r^2\cos\theta,\, -2r^2\sin\theta,\, r \rangle \] --- ### **Dot with the curl** Recall: \[ \vec{C} = \nabla \times \vec{F} = \langle -2,\, -2x,\, -1 \rangle \] Substitute \( x = r\cos\theta \): \[ \vec{C} = \langle -2,\, -2r\cos\theta,\, -1 \rangle \] Dot product: \[ \vec{C} \cdot (\vec{r}_r \times \vec{r}_\theta) = (-2) (2r^2\cos\theta) + (-2r\cos\theta) (-2r^2\sin\theta) + (-1) (r) \] Simplify: - First term: \( -4r^2\cos\theta \) - Second term: \( [ -2r\cos\theta ] \cdot [ -2r^2\sin\theta ] = 4r^3 \cos\theta \sin\theta \) - Third term: \( -r \) So: \[ \vec{C} \cdot d\vec{S} = \left[ -4r^2 \cos\theta + 4r^3 \cos\theta \sin\theta - r \right] dr d\theta \] --- ### **Set up the integral** \[ \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \int_0^{2\pi} \int_0^{2} \left[ -4r^2 \cos\theta + 4r^3 \cos\theta \sin\theta - r \right] dr d\theta \] Integrate term by term. #### **First term:** \[ \int_0^{2\pi} \int_0^2 -4r^2\cos\theta \, dr d\theta \] Integrate \( r \) first: \[ \int_0^2 r^2 dr = \left[\frac{1}{3} r^3\right]_0^2 = \frac{8}{3} \] So: \[ -4 \frac{8}{3} \int_0^{2\pi} \cos\theta d\theta = -\frac{32}{3} \left[ \sin\theta \right]_0^{2\pi} = -\frac{32}{3} (0 - 0) = 0 \] #### **Second term:** \[ \int_0^{2\pi} \int_0^2 4r^3 \cos\theta \sin\theta dr d\theta \] Integrate \( r \) first: \[ \int_0^2 r^3 dr = \left[ \frac{1}{4} r^4 \right]_0^2 = \frac{16}{4} = 4 \] So: \[ 4 \cdot 4 \int_0^{2\pi} \cos\theta \sin\theta d\theta = 16 \int_0^{2\pi} \cos\theta\sin\theta d\theta \] But: \[ \int_0^{2\pi} \cos\theta\sin\theta d\theta = 0 \] (as the integrand is an odd function over a full period) #### **Third term:** \[ \int_0^{2\pi} \int_0^2 -r dr d\theta \] Integrate \( r \): \[ \int_0^2 r dr = \left[\frac{1}{2} r^2\right]_0^2 = 2 \] So: \[ -2 \int_0^{2\pi} d\theta = -2 (2\pi) = -4\pi \] --- ### **Sum up** Only the last term is nonzero: \[ \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = -4\pi \] **Boxed Answer:** \[ \boxed{-4\pi} \] --- ## (c) **Find the circulation around the boundary curve** The circulation is \[ \oint_C \vec{F} \cdot d\vec{r} \] By **Stokes' Theorem**: \[ \oint_C \vec{F} \cdot d\vec{r} = \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} \] So the answer is also \[ \boxed{-4\pi} \] --- ### **Alternatively, compute directly using the boundary parameterization** The boundary curve is the circle \( x^2 + y^2 = 4, z=0 \): \[ \vec{r}_C(\theta) = \langle 2\cos\theta,\, 2\sin\theta,\, 0 \rangle, \quad 0 \le \theta < 2\pi \] Compute \( d\vec{r}_C = \langle -2\sin\theta,\, 2\cos\theta,\, 0\rangle d\theta \) Evaluate \( \vec{F} \) on the curve (\( z = 0 \), \( x = 2\cos\theta \), \( y = 2\sin\theta \)): \[ \vec{F} = \langle 2\sin\theta,\, 0,\, 4\cos^2\theta \rangle \] Dot product: \[ \vec{F} \cdot d\vec{r}_C = [2\sin\theta] \cdot [-2\sin\theta] + [0] \cdot [2\cos\theta] + [4\cos^2\theta] \cdot 0 \] \[ = -4\sin^2\theta \, d\theta \] Integrate over \( 0 \le \theta < 2\pi \): \[ \oint_C \vec{F} \cdot d\vec{r} = \int_0^{2\pi} -4\sin^2\theta d\theta = -4 \int_0^{2\pi} \sin^2\theta d\theta \] But \[ \int_0^{2\pi} \sin^2\theta d\theta = \pi \] So, \[ \oint_C \vec{F} \cdot d\vec{r} = -4\pi \] This matches the result from Stokes' theorem. --- ## (d) **OPTIONAL: Compute the surface integral using the implicitly defined function** The surface is given by \( f(x, y, z) = x^2 + y^2 + z - 4 = 0 \). For a surface defined by \( f(x, y, z) = 0 \): \[ d\vec{S} = \frac{\nabla f}{|\nabla f|} dS \] where \( dS \) is the area element, and \( \nabla f = \langle 2x, 2y, 1 \rangle \). Alternatively, the flux through the surface is: \[ \iint_S \vec{A} \cdot d\vec{S} = \iint_D \vec{A}(x, y, z(x, y)) \cdot \left( -\frac{\partial f}{\partial x}, -\frac{\partial f}{\partial y}, 1 \right) \frac{dx\,dy}{\left| \nabla f \cdot \mathbf{k} \right|} \] But since \( \frac{\partial f}{\partial z} = 1 \), and \( \left| \nabla f \cdot \mathbf{k} \right| = 1 \), this simplifies to: \[ d\vec{S} = (-f_x, -f_y, 1) dx dy = (-2x, -2y, 1) dx dy \] So, \[ \iint_S (\nabla \times \vec{F}) \cdot d\vec{S} = \iint_{x^2 + y^2 \le 4} \langle -2, -2x, -1 \rangle \cdot \langle -2x, -2y, 1 \rangle dx dy \] Compute the dot product: \[ (-2)(-2x) + (-2x)(-2y) + (-1)(1) = 4x + 4xy - 1

-4π (surface integral of curl over the paraboloid cap) -4π (circulation around the boundary circle)