Say I have a circle of radius 6 ft. How do I know that my hypotenuse is also 6ft?
It makes sense to me that from a point of origin $(0,0)$ on the circle, the x-axis and y-axis both stop at $6$ft.
So when finding the hypotenuse we say that the hypotenuse, $c^2=6^2+6^2$. So c cannot possibly be 6?
I am asking because I want to find x, the horizontal interval, where it is not a constant i.e. place my newfound c into the equation, $x^2 = c^2 - y^2$, where $c$ is defined as a constant and both $x$ and $y$ are not.
I gather that you are trying to understand Ross Millikan's answer to this question about the work done when oil is lifted out of a spherical tank.
A reason for your confusion is that you are drawing the wrong diagram. Consider the diagram below:
The circle with radius $r_c$ is a cross-section of a sphere with radius $r_s$ at a distance of $|z|$ from the center. By the Pythagorean Theorem, $$r_c^2 + |z|^2 = r_s^2$$ Since the square of a real number is the square of its absolute value, $$r_c^2 + z^2 = r_s^2$$ Solving for $r_c$ yields \begin{align*} r_c^2 & = r_s^2 - z^2\\ r_c & = \sqrt{r_s^2 - z^2} \end{align*} In that problem, the sphere has radius $6$, so the radius of the cross-section is $$r_c = \sqrt{36 - z^2}$$ Therefore, the area of the cross-section is $$A = \pi r_c^2 = \pi\left(\sqrt{36 - z^2}\right)^2 = \pi(36 - z^2)$$ Thus, the integral $$V = \int_{-6}^{0} \pi(36 - z^2)~\text{dz}$$ yields the volume of the oil in the bottom half of the sphere. Multiplying by the density $\rho$ of the oil yields the mass $$m = \int_{-6}^{0} \rho\pi(36 - z^2)dz$$ of the oil. The work done to lift the oil out of the sphere is $$W = \int_{-6}^{0} 9.8\rho(8 - z)\pi(36 - z^2)dz$$ where $9.8~\text{m}/\text{s}^2$ is the upward acceleration against gravity and $8 - z$ is the distance oil at a height $z$ is lifted to reach the top of the spout.