Surface Area — Question 8

Question 8

Curve: $y=\sqrt{R^2-x^2}$ on $a\le x\le b$, where $R>0$ and $-R\le a<b\le R$.
Axis of rotation: the $x$-axis.
Task: For fixed $R$, show that the surface area depends only on the zone height $h=b-a$.

Question 8 – Solution

Step 1: Differentiate. $$y'=-\frac{x}{\sqrt{R^2-x^2}}.$$ Step 2: Simplify the arc-length factor. $$\begin{align*} \sqrt{1+(y')^2} &=\sqrt{1+\frac{x^2}{R^2-x^2}}\\ &=\sqrt{\frac{R^2}{R^2-x^2}} =\frac{R}{\sqrt{R^2-x^2}}. \end{align*}$$ Step 3: Calculate the surface area. $$\begin{align*} S&=2\pi\int_a^by\sqrt{1+(y')^2}\,dx\\ &=2\pi\int_a^b\sqrt{R^2-x^2} \frac{R}{\sqrt{R^2-x^2}}\,dx\\ &=2\pi R\int_a^bdx\\ &=2\pi R(b-a)=2\pi Rh. \end{align*}$$ $$\boxed{S=2\pi R(b-a)=2\pi Rh}$$