I was wondering how to find the "new" equation if say you had the function
$f(x) = x^3$
bounded by the line
$x=0$
I need this to answer a, calculus question. To be more specific, the question is as follows:
Find the volume of the solid whose base is the region bounded between the curve y = x^3 and the y-axis from y = 0 to y = 1 and whose cross-sections taken perpendicular to the y-axis are squares.
Since the cross-sections are perpendicular to the $y$-axis, the cross-sections move in the $y$-direction, so the integral must be with respect to $y$. The volume of any solid with a known cross-section perpendicular to the $y$-axis is therefore given by:
$$\int_{c}^{d} A(y) \, \mathrm dy$$
where $c \leq y \leq d$ (these are your bounds).
If each cross-section is a square, then the area of each square will the the length of its side, $s$, squared:
$$A(s) = s^2$$
You have to find a way to write this in terms of $y$, so you can get $A(y)$.
First, notice that for any square cross-section, the base of the square corresponds to a horizontal line extending from the $y$-axis (where $x = 0$) to a point $(x, f(x))$ on the curve $f(x) = y = x^3$. Hence, the length of the line is simply $x$. This means $\color{blue}{s = x}$. If you draw a sketch, this should be clear. (Here's a graph in case you prefer a visual instead of a wordy explanation.)
Since you want $s$ to be in terms of $y$ and not $x$, note that $y = x^3$, so $x = \sqrt[3]{y}$. Hence, $\color{blue}{s = x = \sqrt[3]{y}}$. Now that $s$ is in terms of $y$, you can find $A(y)$:
$$A = s^2 = \left(\sqrt[3]{y}\right)^2 = y^{2/3} \iff \color{blue}{A(y) = y^{2/3}}$$
Since the lower and upper bounds are $y = 0$ and $y = 1$, respectively, those will be your bounds of integration:
$$V = \int_{c}^{d} A(y) \, \mathrm dy \longrightarrow V = \int_{0}^{1} y^{2/3} \, \mathrm dy$$
Evaluating this integral gives the exact volume of the solid.