Surface Area — Question 6

Question 6

Curve: $y=x^2$ on $0\le x\le1$.
Axes of rotation: first the $x$-axis, then the $y$-axis.
Task: Write both surface-area integrals and determine which is larger without evaluating either integral.

Question 6 – Solution

Step 1: Find the common arc-length factor. $$y'=2x,\qquad ds=\sqrt{1+4x^2}\,dx.$$ Step 2: Rotate about the $x$-axis. The radius is $y=x^2$: $$S_x=2\pi\int_0^1x^2\sqrt{1+4x^2}\,dx.$$ Step 3: Rotate about the $y$-axis. The radius is $x$: $$S_y=2\pi\int_0^1x\sqrt{1+4x^2}\,dx.$$ Step 4: Compare the integrands. For $0<x<1$, $$x>x^2.$$ Multiplying by the positive common factor $2\pi\sqrt{1+4x^2}$ preserves the inequality. Therefore, $$\boxed{S_y>S_x}.$$