Calculus of variations: find $y(a/2)$ if $y(x)$ maximizes the volume of rotation A curve $y(x)$ of length $2a$ is drawn between the points (0,0) and (a,0) in such a way that the solid obtained by rotating the curve about the $x$-axis has the largest possible volume.  Find $y\left(\frac{a}{2}\right)$.
 A: When rotating about $x-axis$ maximum volume is equal to maximum area in 2D. Therefore the question can be given as max area given the length. The area to be maximized can be defined as

$$A[y]=\int_{x_1}^{x_2}y\ dx$$
  subject to
  $$2a=\int_{x_1}^{x_2}\sqrt{1+y'\ ^2} dx$$

We can use Lagrange multiplier such as
$$H=y+\lambda \sqrt{1+y'\ ^2}$$
and
$$\frac{\partial H}{\partial y}-\frac{d}{dx}\bigg(\frac{\partial H}{\partial y'}\bigg)=0$$
$$1-\frac{d}{dx}\bigg(\frac{\lambda y'}{\sqrt{1+y'\ ^2}}\bigg)=0$$
$$\Rightarrow \frac{x-\alpha}{\lambda}=\frac{y'}{\sqrt{1+y'\ ^2}}$$
If $\bar x=x-\alpha$ 
$$\frac{\bar x}\lambda=\frac{y'}{\sqrt{1+y'\ ^2}}\Rightarrow y'=\frac{dy}{dx}=\frac{\bar x}{\sqrt{\lambda^2-\bar x^2}}$$
which can be solved for $y(x)$ such that
$$y(x)=-\sqrt{\lambda^2-\bar x^2}+\beta=\beta-\sqrt{\lambda^2-\big(x-\alpha\big)^2}$$
The equation must satisfy following conditions
$$(0,0)\Rightarrow 0=\beta-\sqrt{\lambda^2-\big(-\alpha\big)^2}$$
$$(a,0)\Rightarrow 0=\beta-\sqrt{\lambda^2-\big(a-\alpha\big)^2}$$
$$2a=\int_{0}^{a}\sqrt{1+y'\ ^2} dx\qquad y'=\frac{x-\alpha}{\sqrt{\lambda^2-\big(x-\alpha\big)^2}}$$
which you can solve to find $\alpha$, $\beta$ and $\lambda$. From $(0,0)$ condition it follows that 
$$\beta=\sqrt{\lambda^2-\big(\alpha\big)^2}$$
By replacing it into $(a,0)$ condition it follows that $\alpha=a/2$. We can rewrite the equation as a circle equation
$$\bigg(y-\sqrt{\lambda^2-\big(a/2\big)^2}\bigg)^2+\bigg(x-a/2\bigg)^2=\lambda^2$$
---------------EDIT-------------------
Although it produces a nice result; the equation is wrong. It assumes that every $dA$ produces the same amount of $dV$. It would have been right when the volume was produced by extrusion; but in rotational case $dA$ away from axis produces bigger $dV$.
