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\begin{equation} =(0-0) \vec{i}-(0-0) \vec{j}+\left(e^{x} \sin y-\left(-e^{x} \sin y\right)\right) \vec{k} \end{equation}

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If \(\vec F = \left\langle {P,Q,R} \right\rangle \) is a vector field on \({\mathbb{R}^3}\) and the partial derivatives of \(P\), \(Q\), and \(R\) exist, then the curl of \(\vec F\) is the vector field on \({\mathbb{R}^3}\) defined by:

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\begin{equation} \operatorname{curl} \vec{F}=\nabla \times \vec{F}=\left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{array}\right| \end{equation}

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Well, guess what. The water spreading out from the faucet is an example of divergence, and the act of scrubbing is your curl!

So, when you scrub your dinner plate with a sponge, the soap and water begin to swirl around the plate to make it clean, which is the curl. And the water dispersing out of the faucet and subsequently leaving through the drain is divergence.

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The divergence of a vector field measures the fluid flow “out of” or “into” a given point. The curl indicates how much the fluid rotates or spins around a point.

Find the curl and divergence of the vector field \(\vec F\left( {x,y,z} \right) = \left\langle {{e^x}\cos y,{e^x}\sin y,z} \right\rangle \).

If \(\vec F = \left\langle {P,Q,R} \right\rangle \) is a vector field on \({\mathbb{R}^3}\) and \(\frac{\partial }{{\partial x}},\frac{\partial }{{\partial y}},\) and \(\frac{\partial }{{\partial z}}\) exist, then the divergence of\(\vec F\) is the function of three variables defined by:

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Okay, so now that we have a general idea about curl and divergence, let’s define both of these differentiation operators, along with some necessary notation, that will help us to make sense of some essential theorems and properties like Green’s theorem, surface integrals, and Stokes’ theorem.

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\begin{equation} \operatorname{div} \vec{F}=\nabla \cdot \vec{F}=\left\langle\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right\rangle \cdot\left\langle e^{x} \cos y, e^{x} \sin y, z\right\rangle \end{equation}

\begin{equation} =\left(\frac{\partial}{\partial y}(z)-\frac{\partial}{\partial z}\left(e^{x} \sin y\right)\right) \vec{i}-\left(\frac{\partial}{\partial x}(z)-\frac{\partial}{\partial z}\left(e^{x} \cos y\right)\right) \vec{j}+\left(\frac{\partial}{\partial x}\left(e^{x} \sin y\right)-\frac{\partial}{\partial y}\left(e^{x} \cos y\right)\right) \vec{k} \end{equation}

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\begin{equation} \begin{aligned} &=\frac{\partial}{\partial x}\left(e^{x} \cos y\right)+\frac{\partial}{\partial y}\left(e^{x} \sin y\right)+\frac{\partial}{\partial z}(z) \\ &=e^{x} \cos y+e^{x} \cos y+1 \\ &=2 e^{x} \cos y+1 \end{aligned} \end{equation}

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\begin{equation} =\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right) \vec{i}+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right) \vec{j}+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) \vec{k} \end{equation}

If \(\nabla = \vec i\frac{\partial }{{\partial x}} + \vec j\frac{\partial }{{\partial y}} + \vec k\frac{\partial }{{\partial z}}\) is a vector comprised of \(\frac{\partial }{{\partial x}},\frac{\partial }{{\partial y}},\) and \(\frac{\partial }{{\partial z}}\) components, then the curl of a vector field can be computed using the cross product.

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\begin{equation} \operatorname{div} \vec{F}=\frac{\partial P}{\partial x}+\frac{\partial Q}{\partial y}+\frac{\partial R}{\partial z} \end{equation}

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\begin{equation} =\left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ e^{x} \cos y & e^{x} \sin y & z \end{array}\right| \end{equation}

\begin{equation} \operatorname{curl} \vec{F}=\nabla \times \vec{F}=\left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ P & Q & R \end{array}\right| \end{equation}

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But this formula seems a bit difficult to remember. Thankfully we can simplify things using the nabla operator \(\nabla \), or as we like to call it, the del or gradient operator.

We will look more closely at these two versions of writing Green’s theorem in our lecture and work through an example to demonstrate its ability to find the work done in moving an object about a curve.

You turn on the faucet to let water pour out of the tap, and then you proceed to scrub your dinner plate with a soapy sponge to clean your dishes.

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So together, we will learn all about how if the vector represents the velocity of a fluid, the curl of the vector field measures the tendency of an object immersed in that fluid to rotate or swirl around a point. In contrast, the divergence of the vector field measures the tendency for fluid to gather or disperse at a point. And how these two operators help us in representing Green’s theorem.

\begin{equation} \operatorname{curl} \vec{F}=\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right) \vec{i}+\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right) \vec{j}+\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) \vec{k} \end{equation}

\begin{equation} \begin{array}{|c|c|} \hline \text { Vector form of Green’s Theorem using Curl } & \text { Vector form of Green’s Theorem using Divergence } \\ \hline \oint_{C} \vec{F} \cdot d \vec{r}=\iint_{D}(\operatorname{curl} \vec{F}) \cdot \vec{k} d A & \oint_{C} \vec{F} \cdot \vec{n} d s=\iint_{D} \operatorname{div} \vec{F}(x, y) d A \\ \hline \end{array} \end{equation}

First, we will compute the curl using our cross-product formula replacing P, Q, and R from our vector field and taking the respective partial derivatives.