Given $V= \pi \int_{1}^{c}-y \sqrt{1-y^2}\,\mathrm{d}y$, Find $f^\prime\left(x\right)$

Question

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.

Answer 1

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 $x<0$ $$f'(x) = \frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = -\frac{\sqrt{2}\cos(t)\sqrt{\sin(2t)}}{\cos(2t)}$$

Answer 2

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 $$f'(x)=-\frac{V'(x)}{\pi f(x)\sqrt{1-f^2(x)}}.$$