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# Lecture 24: Curl ## Curl The curl of a vector field $\mathbf{F} = \langle P, Q, R \rangle$ is the vector field: $\qquad \operatorname{curl} \mathbf{F} = \left\langle \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \fr...

# Lecture 24: Curl ## Curl The curl of a vector field $\mathbf{F} = \langle P, Q, R \rangle$ is the vector field: $\qquad \operatorname{curl} \mathbf{F} = \left\langle \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right\rangle$ or $\qquad \operatorname{curl} \mathbf{F} = \nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{vmatrix}$ ## Example 1 Find the curl of $\mathbf{F}(x, y, z) = \langle xz, x y z, -y^2 \rangle$ ### Solution $\qquad \operatorname{curl} \mathbf{F} = \nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ xz & xyz & -y^2 \end{vmatrix} = (-2y - xy)\mathbf{i} - (0 - x)\mathbf{j} + (yz - z)\mathbf{k} = \langle -2y - xy, x, yz - z \rangle$ ## Example 2 Find the curl of $\mathbf{F}(x, y, z) = \langle xz, x y z, -x y \rangle$ ### Solution $\qquad \operatorname{curl} \mathbf{F} = \nabla \times \mathbf{F} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ xz & xyz & -xy \end{vmatrix} = (-x - 0)\mathbf{i} - (-y - x)\mathbf{j} + (yz - z)\mathbf{k} = \langle -x, y + x, yz - z \rangle$ ## Theorem Suppose that $\Sigma$ is an oriented surface with a unit normal vector $\mathbf{n}$, bounded by an oriented curve $C$ with positive orientation. If $\mathbf{F}$ is a vector field whose component functions have continuous partial derivatives on an open region in $\mathbb{R}^3$ containing $\Sigma$, then $\qquad \oint_C \mathbf{F} \cdot d\mathbf{r} = \iint_\Sigma \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = \iint_\Sigma \operatorname{curl} \mathbf{F} \cdot \mathbf{n} \, dS$ ## Example 3 Evaluate $\iint_\Sigma \operatorname{curl} \mathbf{F} \cdot d\mathbf{S}$ where $\mathbf{F}(x, y, z) = \langle xz, -xy, yz \rangle$ and $\Sigma$ is the surface $x^2 + y^2 + z^2 = 4$, $z \geq 0$. ### Solution Let $C$ be the boundary curve of $\Sigma$, that is, $x^2 + y^2 = 4$ and $z = 0$. Parameterize $C$ by $\qquad \mathbf{r}(t) = \langle 2\cos t, 2\sin t, 0 \rangle, \quad 0 \leq t \leq 2\pi$ Then $\qquad \mathbf{r}'(t) = \langle -2\sin t, 2\cos t, 0 \rangle$ $\qquad \mathbf{F}(\mathbf{r}(t)) = \langle (2\cos t)(0), -(2\cos t)(2\sin t), (2\sin t)(0) \rangle = \langle 0, -4\cos t \sin t, 0 \rangle$ Thus $\qquad \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) = \langle 0, -4\cos t \sin t, 0 \rangle \cdot \langle -2\sin t, 2\cos t, 0 \rangle = -8 \cos^2 t \sin t$ and $\qquad \oint_C \mathbf{F} \cdot d\mathbf{r} = \int_0^{2\pi} \mathbf{F}(\mathbf{r}(t)) \cdot \mathbf{r}'(t) \, dt = \int_0^{2\pi} -8 \cos^2 t \sin t \, dt = \left[ \frac{8}{3} \cos^3 t \right]_0^{2\pi} = \frac{8}{3} - \frac{8}{3} = 0$ Therefore $\qquad \iint_\Sigma \operatorname{curl} \mathbf{F} \cdot d\mathbf{S} = 0$