Surface Area — Question 5

Question 5

Curve: $y=x$ on $0\le x\le a$, where $a>0$.
Axis of rotation: the $x$-axis.
Task: Find $a$ so that the generated surface area is $9\pi\sqrt2$.

Question 5 – Solution

Step 1: Find the arc-length factor. $$y'=1,\qquad\sqrt{1+(y')^2}=\sqrt2.$$ Step 2: Write and evaluate the area. $$\begin{align*} S&=2\pi\int_0^ay\sqrt{1+(y')^2}\,dx\\ &=2\pi\int_0^ax\sqrt2\,dx\\ &=2\pi\sqrt2\left[\frac{x^2}{2}\right]_0^a =\pi\sqrt2a^2. \end{align*}$$ Step 3: Set the area equal to $9\pi\sqrt2$ and solve. $$\begin{align*} \pi\sqrt2a^2&=9\pi\sqrt2,\\ a^2&=9,\\ a&=3 \end{align*}$$ because $a>0$. $$\boxed{a=3}$$