The original function **cannot be recovered** from the result of a definite integral unless it is known that $f(x)$ is a constant.
## N-dimensional integrals
Much like how $dx$ represents an infinitely small line, $dx\cdot dy$ represents an infinitely small rectangle. This means that the surface area of an object can be expressed as:
$$dS=dx\cdot dy$$
Therefore, the area of a function can be expressed as:
$$S=\int^x_0\int^y_0 dy\ dx$$
where $y$ is usually equal to $f(x)$, changing on each iteration.
!!! example
The area of a circle can be expressed as $y=\pm\sqrt{r^2-x^2}$. This can be reduced to $y=2\sqrt{r^2-x^2}$ because of the symmetry of the equation.
$$
\begin{align*}
A&=\int^r_0\int^{\sqrt{r^2-x^2}}_0 dy\ dx \\
&=\int^r_0\sqrt{r^2-x^2}\ dx
\end{align*}
$$
!!! warning
Similar to parentheses, the correct integral squiggly must be paired with the correct differential element.
Although differential elements can be blindly used inside and outside an object (e.g., area), the rules break down as the **boundary** of an object is approached (e.g., perimeter). Applying these rules to determine an object's perimeter will result in the incorrect deduction that $\pi=4$.
The **mass distribution** of an object varies depending on its surface density $\rho_s$. In objects with uniformly distributed mass, the surface density is equal to the total mass over the total area.
$$dm=\rho_s\ dS$$
The formula for the **moment of inertia** of an object is as follows, where $r_\perp$ is the distance from the axis of rotation:
If the axis of rotation is perpendicular to the plane of the object, $r_\perp=r$. If the axis is parallel, $r_\perp$ is the shortest distance to the axis. Setting an axis along the axis of rotation is easier.
- The **polarity** of a particle is whether it is positive or negative.
The law of **conservation of charge** states that electrons and charges cannot be created nor destroyed, such that the **net charge in a closed system stays the same**.
The law of **charge quantisation** states that charge is discrete — electrons have the lowest possible quantity.
Please see [SL Physics 1#Charge](/sph3u7/#charge) for more information.
Because Coulomb's law is an experimental law, it does not quite cover all of the nuances of electrostatics. Notably:
- $Q_1$ and $Q_2$ must be point charges, making distributed charges inefficient to calculate, and
- the formula breaks down once charges begin to move (e.g., if a charge moves a lightyear away from another, Coulomb's law says the force changes instantly. In reality, it takes a year before the other charge observes a difference.)
An **electric dipole** is composed of two equal but opposite charges $Q$ separated by a distance $d$. The dipole moment is the product of the two, $Qd$.
The charge experienced by a positive test charge along the dipole line can be reduced to as the ratio between the two charges decreases to the point that they are basically zero:
Compared to Coulomb's law, $Q_1$ creates an electric field around itself — each point in space is assigned a vector that depends on the distance away from the charge. $Q_2$ *interacts* with the field. According to Maxwell, as a charge moves, it emits a wave that carries information to other charges.
If charge is distributed over a three-dimensional object, integration similar to moment of inertia can be used. Where $dQ$ is an infinitely small point charge at point $P$, $d\vec E$ is the electric field at that point, and $r$ is the vector representing the distance from any arbitrary point:
$$d\vec E = \frac{kdQ}{r^2}\hat r$$
!!! warning
As the arbitrary point moves, both the direction and the magnitude of the distance from the desired point $P$ change (both $\hat r$ and $r$).
Generally, if a decomposing the vector into Cartesian forms $d\vec E_x$, $d\vec E_y$, and $d\vec E_z$ is helpful even if it is easily calculated in polar form because of the significantly easier ability to detect symmetry in the shape. Symmetry about the axis allows deductions such as $\int d\vec E_y=0$, which makes calculations easier.
In a **one-dimensional** charge distribution (a line), the charge density is used in a similar way as moment of inertia's surface density:
$$dQ=\rho_\ell d\ell$$
**Two-dimensional** charge distributions are more or less the same, but polar or Cartesian forms of the surface area work depending on the shape.
$$dQ=\rho_s dS$$
!!! example
A rod of uniform charge density and length $L$ has a charge density of $p_\ell=\frac{Q}{L}$.
- A **closed surface** is any closed three-dimensional object.
- **Electric flux** represents the number of electric field lines going through a surface.
At an arbitrary surface, the **normal** to the plane is its vector form:
$$\vec{dS}=\vec n\cdot dS$$
The **electric flux density** $\vec D$ is an alternate representation of electric field strength. In a vacuum:
$$\vec D = \epsilon_0\vec E$$
**Electric flux** is the electric flux density multiplied by the surface area at every point of an object.
$$\phi_e=\epsilon_0\int_s\vec E\bullet\vec{dS}$$
The flux from charges outside a closed surface will **always be zero at the surface**. A point charge in the centre of a closed space has a flux equal to its charge. Regardless of the charge distribution or shape, the **total flux** through a closed surface is equal to the **total charge within** the closed surface.
$$\oint \vec D\bullet\vec{dS}=Q_\text{enclosed}$$
This implies $\phi_e>0$ is a net positive charge enclosed.
!!! warning
Gauss's law only applies when $\vec E$ is from all charges in the system
To apply this strategy, the following conditions must hold:
- $Q$ must not vary with the length of the cylinder or $\phi$
- The charge must be distributed over either a cylindrical surface or the volume of the cylinder.
- In the real world, $r$ must be significantly smaller than $L$ as an approximation.
- The strategy is more accurate for points closer to the centre of the wire.
Please see [Maxwell's integral equations#Gauss's law](https://en.wikiversity.org/wiki/MyOpenMath/Solutions/Maxwell%27s_integral_equations) for more information.
**Outside** the radius $R$ of the cylinder of the Gaussian surface, the enclosed charge is, where $L$ is the length of the cylinder:
$$Q_{enc}=\pi R^2\rho_0L$
such that the field at any radius $r>R$ is equal to:
**Inside** the radius $R$ of the cylinder, the enclosed charge depends on $r$. For a uniform charge density:
$$Q_{enc}=\pi r^2\rho_0L$$
such that the field at any radius $r<R$isequalto:
$$\vec E(r)=\frac{\rho_0}{2\epsilon_0}r\hat r$$
The direction of $\vec E$ should always be equal to that of $\vec r$. Generally, where $lim$ is $r$ if $r$ is *inside* the cylinder or $R$ otherwise, $\rho_v$ is the function for charge density based on radius, and $r_1$ is hell if I know:
To apply this strategy, the following conditions must hold:
- $Q$ must not vary with the lengths of the plane
- The charge must be distributed over a plane or slab
- In the real world, the thickness $z$ must be significantly smaller than the lengths as an approximation
Where $\rho_v$ is an **even** surface density function and $lim$ is from $0$ to $z$ if the desired field is outside of the charge, or $0$ to field height $h$ if it is inside the charge:
$$\epsilon_0 E=\int_{lim}\rho_v\ dh_1$$
Any two points have equal electric fields regardless of distance due to the construction of a uniform electric field.
Where $\rho_v$ is not an even surface density function, $d$ is the thickness of the slab, and $E$ is the electric field **outside** the slab:
$$2\epsilon_0E = \int^d_0\rho_v(A)dh_1$$
Where $E$ is the electric field **inside** the slab at some height $z$:
At a point $P$, the electrostatic potential $V_p$ or voltage is the work done per unit positive test charge from infinity to bring it to point $P$ by an external agent.
$$
V_p=\lim_{q\to 0^+}\frac{W_i}{q} \\
W_I=\int^p_\infty\vec F_I\bullet \vec{dl}
$$
Because the desired force acts opposite to the force from the electric field, as long as $\vec E$ is known at each point:
Where $\vec dl$ is the path of the test charge from infinity to the point, and $\vec dr$ is the direct path from the origin through the point to the charge, because $dr=-dl$:
