Calculus of Variations-question on rotating curve of max volume

Question

My calc of variations is still rusty. I'm assuming implementation of arclength revolution formula is necessary, but how to find y(1/2a)?

Answer 1

We can decrease the problem to find the largest area in 2D because it will give the largest volume in rotating the curve. The are 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 $\bar x=\frac{x-\alpha}{\lambda}$ $$\bar x=\frac{y'}{\sqrt{1+y'\ ^2}}\Rightarrow y'=\frac{dy}{dx}=\frac{\bar x}{\sqrt{1-\bar x^2}}$$ which can be solved for $y(x)$ such that $$y(x)=-\sqrt{1-\bar x^2}+\beta=\beta-\sqrt{1-\bigg(\frac{x-\alpha}{\lambda}\bigg)^2}$$ The equation must satisf boundary conditions $$(0,0)\Rightarrow 0=\beta-\sqrt{1-\bigg(\frac{-\alpha}{\lambda}\bigg)^2}$$ $$(a,0)\Rightarrow 0=\beta-\sqrt{1-\bigg(\frac{a-\alpha}{\lambda}\bigg)^2}$$ $$2a=\int_{0}^{a}\sqrt{1+y'\ ^2} dx\qquad y'=\frac{x-\alpha}{\lambda^2 \sqrt{1-\frac{\big(\alpha-x\big)^2}{\lambda^2}}}$$