Surface Area with Parametric Equations — Question 5

Question 5 – Solution

Solution

1. The interval $0\le t\le\pi$ traces the upper semicircle, while $\pi\le t\le2\pi$ traces the lower semicircle.

2. When either semicircle is rotated about the $x$-axis, it generates the entire unit sphere. A point $(x,y)$ and its reflection $(x,-y)$ produce the same circle of revolution because both are the same distance $|y|$ from the axis.

3. If the full interval is used with the correct radius $|y|$, then $ds=dt$ and $$2\pi\int_0^{2\pi}|\sin t|\,dt=8\pi,$$ which is twice the unit sphere’s area. The formula counts the repeated sweep.

4. If one incorrectly uses signed $y=\sin t$ instead, the upper and lower contributions cancel and give zero, which is also not physical surface area.

5. Use one generating semicircle only, for example $$S=2\pi\int_0^\pi\sin t\,dt=\boxed{4\pi}.$$ Thus the full-circle interval is unsuitable because it double-covers the surface.