Surface Area — Question 1

Question 1

Curve: $y=x$ on $1\le x\le3$.
Axis of rotation: the $x$-axis.
Task: Find the lateral surface area, then verify the result with the frustum formula.

Question 1 – Solution

Step 1: Differentiate and find $ds$. $$y'=1,\qquad ds=\sqrt{1+(y')^2}\,dx=\sqrt2\,dx.$$ Step 2: Apply the surface-area formula. $$\begin{align*} S&=2\pi\int_1^3y\,ds\\ &=2\pi\int_1^3x\sqrt2\,dx\\ &=2\pi\sqrt2\left[\frac{x^2}{2}\right]_1^3\\ &=\pi\sqrt2(9-1)=8\pi\sqrt2. \end{align*}$$ Step 3: Verify with the frustum formula. The radii are $r_1=1$ and $r_2=3$. The slant height is $$s=\sqrt{(3-1)^2+(3-1)^2}=2\sqrt2.$$ Therefore, $$S=\pi(r_1+r_2)s=\pi(1+3)(2\sqrt2)=8\pi\sqrt2.$$ $$\boxed{S=8\pi\sqrt2}$$