Formulas to memorise

1The del operator¶

Given $\vec F = Q\hat i + P \hat j + R \hat k$ and $f(x, y, z)$:

1.1The curl¶

Turns a vector into a vector (lol)

$$\text{curl } \vec F = \nabla \times \vec F = \det \begin{pmatrix} \hat i & \hat j & \hat k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial }{\partial z} \\ Q & P & R \end{pmatrix}$$

Uses: Stokes' theorem, showing that a vector field is conservative

1.2The divergence¶

Turns a vector into a scalar

$$\text{div } \vec F = \nabla \cdot \vec F = \frac{\partial Q}{\partial x} + \frac{\partial P}{\partial y} + \frac{\partial R}{\partial z}$$

Uses: Divergence theorem, showing that a vector field is solenoidal

1.3The gradient¶

Turns a scalar into a vector

$$\nabla f = \frac{\partial f}{\partial x} \hat i + \frac{\partial f}{\partial y} \hat j + \frac{\partial f}{\partial z} \hat k$$

Uses: You can use this to check that a potential function is indeed a potential function for the given vector field - take the potential function's gradient, and it should be equal to the vector field.

2The theorems¶

2.1Stokes' theorem¶

$$\int_C \vec F \cdot d\vec r = \iint_D \text{curl }\vec F \,d\vec S$$

where $C$ is a simple, closed, smooth, boundary curve in a oriented counter-clockwise direction and $S$ is the surface bounded by that curve. $\vec F$ is a vector field.

Uses: later

2.2Green's theorem¶

$$\oint_C P\,dx + Q \,dy = \iint_D \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \,dA$$

where $C$ is a simple, closed, smooth, boundary curve oriented in a counter-clockwise direction and $D$ is the region enclosed by that curve. Also, $P$ and $Q$ must have continuous first order partial derivatives over $D$.

Another version of this theorem gives us a way of computing the area enclosed by a curve:

$$\text{Area } = \oint_C x\,dy = \oint_C y\,dx = \frac{1}{2} \oint_C x\,dy - y\,dx$$

Uses: later

Note: this is the 2-dimensional version of Stokes' theorem.

2.3Divergence theorem¶

$$\iint_S \vec F \cdot \vec n\,dS = \iiint_E \text{div }\vec F\,dV$$

where $E$ is the region bounded by the closed surface $S$ and $\vec n$ is the outward-pointing unit normal vector.

Also known as Gauss' theorem.

Uses: later

3Jacobians¶

3.1Change of variables¶

When transforming coordinates, don't forget the Jacobian. In this case, it's only a scaling factor, so take the absolute value. If you're given $x$ and $y$ and want to transform to $u$ and $v$, the "new variables" ($u$ and $v$) would be the denominator in the Jacobian, e.g.:

$$\frac{\partial (x, y)}{\partial (u, v)}$$

and since you can't evaluate that directly, take 1 over the reciprocal:

$$1 / \frac{\partial (u, v)}{\partial (x, y)}$$

The Jacobian is just the determinant of the following matrix (or its transpose, whichever way you prefer):

$$\begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix}$$

Multiply the integrand by the Jacobian, then express the result in terms of the new variables $u$ and $v$, then integrate with respect to $u$ and $v$ over the limits identified.

3.2Implicit functions¶

You also need the Jacobian for finding partial derivatives of implicit functions. In this case, you don't take the absolute value. Again, the "new variables" (i.e. the ones that are functions of the other variables) go on the bottom. Let's say we have the variables $x, y, u, v$ and want to express $u$ and $v$ in terms of $x$ and $y$ (so $u(x, y)$ and $v(x, y)$ are the functions, and $u$ and $v$ are the "new variables"). First you find the "regular" Jacobian, $J$:

$$J = \frac{\partial (F, G)}{\partial (u, v)} = \det\,\begin{pmatrix} F_u & F_v \\ G_u & G_v \end{pmatrix}$$

Finding the partial derivatives is really a matter of plugging in the values to a formula. Let's say we want to find the partial derivatives of $u$ and $v$ with respect to $x$ and $y$:

$$\frac{\partial u}{\partial x} = -\frac{1}{J} \frac{\partial (F, G)}{\partial (x, v)} \quad \frac{\partial u}{\partial y} = -\frac{1}{J} \frac{\partial (F, G)}{\partial (y, v)} \quad \frac{\partial v}{\partial x} = -\frac{1}{J} \frac{\partial (F, G)}{\partial (u, x)} \quad \frac{\partial v}{\partial y} = -\frac{1}{J} \frac{\partial (F, G)}{\partial (u, y)}$$

There's a pretty obvious pattern here - it's especially easy to see if you write out J as a fraction (then you can see what things cancel out and what remains afterwards, etc). In any case these types of problems are usually quite straightforward, so once you're comfortable with the formula they should be a breeze.

4Miscellaneous¶

4.1Flux¶

$$\iint_S\vec F \cdot \,d\vec S = \iint_S \vec F \cdot \vec n \,dS$$

Note that you don't have to divide by the magnitude of the normal vector as that is equal to the constant in $dS$ (the magnitude part thing) and they cancel each other out.

There are several ways of doing this, depending on the situation:

• It's a vector field in 3 dimensions, and the parametric equation you're given for the surface is a function of two variables, Take the dot product of $F(r(u,v)$ and $r_u \times r_v$ (i.e. the cross product of the partial derivatives of $r(u,v)$) and integrate that. It should be a double integral, with the limits being the limits provided for $u$ and $v$.
• Later

4.2Line integrals¶

... as in, what the fuck is a line integral and how do I do one??

Well, there are several methods for computing a line integral, depending on your situation:

• $\displaystyle \int_C \vec F \,d\vec r$, where $C$ is some boundary curve and $\vec F$ is a vector field - do it by parametrising the curve with $r(t)$, then take the dot-product of $F(r(t))$ and the derivative of $r(t)$ with respect to $t$, then integrate that over the limits for $t$. Link to HTSEFP section
• $\displaystyle \oint_C P(x,y)\,dy - Q(x,y)\,dx$ where $P$ and $Q$ are functions of $x$ and $y$ (can be 0) - do it by parametrising the curve with $r(t)$, then replacing $x$ and $y$ in $P$ and $Q$ with what come before $\hat i$ and $\hat j$ in $r(t)$, respectively. Then you just integrate over the limits for $t$ (given by the boundary $C$). Link to HTSEFP section
• Same as the above, except you're asked to use Green's theorem - use the formula $\displaystyle \oint P\,dx + Q\,dy = \iint_D \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\,dA$. Do the integral in Cartesian coordinates or polar or whatever if that makes more sense. Link to HTSEFP

4.3Coordinates of the centroid¶

The $x$ and $y$ coordinates (if it's a two-dimensional problem) of the centroid over a region $D$ are:

$$\bar x = \frac{\displaystyle \iint_D x\,ds}{\displaystyle \iint_D \,ds} \quad \bar y = \frac{\displaystyle \iint_D y\,ds}{\displaystyle \iint_D \,ds}$$

If it's in three dimensions, then make the above triple integrals with respect to $z$ as well. The $z$-coordinate is the obvious:

$$\bar z = \frac{\displaystyle \iiint_D z\,ds}{\displaystyle \iiint_D \,ds}$$