Given $V= \pi \int_{1}^{c}-y \sqrt{1-y^2}\,\mathrm{d}y$, Find $f^\prime\left(x\right)$ Suppose that I have a function, $f\left(x\right)$ s.t. $f\left(x\right) > 0$ on $\left[0,a\right]$ and $f\left(0\right)=1$ and $f\left(a\right) = c$. The volume $V$ of the function is $$V= \pi \int_{1}^{c}-y \sqrt{1-y^2}\,\mathrm{d}y$$ (and $y=f\left(x\right)$). How do I obtain $f^\prime\left(x\right)$? On the surface, this looks like a question where I should integrate the volume. However, I have no way of finding out $f^\prime\left(x\right)$ using the fundamental theorem since the bound $c$ is unknown. Help would be appreciated.
 A: As you gave to less context I can only guess what you meant.
You start with the function $$V(a):=\pi\int_1^{f(a)}-y\sqrt{1-y^2}\mathrm dy.$$
As you mentioned that you want to apply the fundamental theorem this brings you to
$$V'(a)=\pi\left(\frac{\rm d}{\mathrm dx}\int_1^{x}-y\sqrt{1-y^2}\mathrm dy\right)_{x=f(a)}\cdot \left(\frac{\mathrm d f(a)}{\mathrm da}\right)=-\pi f(a)\sqrt{1-f^2(a)}\cdot f'(a). $$
Thus the derivation is given by
$$f'(x)=-\frac{V'(x)}{\pi f(x)\sqrt{1-f^2(x)}}.$$
A: Disclaimer: It could be possible that I am misunderstanding the question because of the lack of context. So maybe take this with a grain of salt?
There is definitely some issue with this question. Primarily because, the volume of revolution about the $y-$axis for any given curve $y = f(x)$ $(x \in [a,b])$ can be obtained by first writing $x$ in terms of $y$ (if the inverse image of $f$ splits into branches; we take the branch that corresponds to our problem); and then computing the integral
$$V = \pi\int_{f(a)}^{f(b)} x^2 dy$$
In the light of this, when we consider your case, by looking at your question, it is reasonable to assume $c$ is arbitrary. So, we consider the case $c>1$, and here we see that we have $$x^2 = -y\sqrt{1-y^2}$$
Clearly, if we are doing real calculus; we want $y<0$ for any meaningful results; however it is clearly stated in your question that $y = f(x) > 0$. In other words, a strictly positive function that satisfies the given volume integral does not exist (for $c>1$). Here's a desmos plot of the two branches that may perhaps give you more intuition on why this problem has something wrong with it.
This is more speculation: Here's maybe a more nuanced problem. Let us assume $0<c<1$; this means we can rewrite the volume integral as
$$V = \pi\int_c^1 y\sqrt{1-y^2} dy$$
and in this case,
$$x^2 = y\sqrt{1-y^2} $$
and everything might look well on the onset; but even in this very nice case; $y$ is not a function of $x$; so we have to resort to simply looking at different branches of $y$.
$$x^2 = y\sqrt{1-y^2} \implies x^4 = y^2(1-y^2)$$
Letting $y^2 = t$, we have
$$t^2 - t + x^4 = 0$$
we use the quadratic formula to see
$$y^2 = t = \frac{1 \pm \sqrt{1-4x^4}}{2}$$
So, since $y$ is always positive, we can only consider the positive square root
$$y = \sqrt{\frac{1 \pm \sqrt{1-4x^4}}{2}}$$
Now the problem becomes more clear; the two branches of solutions correspond to different parts of the closed curve; (you can view this plot to gather more intuition) and will necessarily have two different derivatives depending on which branch you look at! (in other words, $f'(x)$ is not well defined at a given $x$). So, really; we might have to parameterize the problem; but that gets out of hand real fast.
I will perhaps edit this answer later; once I am able find a proper parameterization.
Edit: A possible parameterization for this problem could be
when $x\geq 0$, $\left(\frac{1}{\sqrt{2}}\sqrt{\sin(2t)}, \sin(t)\right)$
and when $x<0$, $\left(-\frac{1}{\sqrt{2}}\sqrt{\sin(2t)}, \sin(t)\right)$
for $t \in [0,\pi]$.
So, when $x\geq 0$ $$f'(x) = \frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = \frac{\sqrt{2}\cos(t)\sqrt{\sin(2t)}}{\cos(2t)}$$
and when $x<0$
$$f'(x) = \frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = -\frac{\sqrt{2}\cos(t)\sqrt{\sin(2t)}}{\cos(2t)}$$
