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 $\pi=4$.
Therefore, further approximations can be made by making a length $\dl=\sqrt{(dx)^2+(dy)^2}$ to represent the perimeter.
!!! example
This reduces to $dl=\sqrt{\left(\frac{dy}{dx}\right)^2+1}$.
### Polar coordinates
Please see [MATH 115: Linear Algebra#Polar form](/1a/math115/#polar-form) for more information.
In polar form, the difference in each "rectangle" side length is slightly different.
| Vector | Length difference |
| --- | --- |
| $\hat r$ | $dr$ |
| $\hat\phi$ | $rd\phi$ |
Therefore, the change in surface area is equal to:
The axes in a Cartesian coordinate plane must be orthogonal so that increasing a value in one axis does not affect any other. The axes must also point in directions that follow the **right hand rule**.
