Parametric Surfaces — Question 10

Question 10 – Solution

Strategy. First parametrize a radius-$1$ circle centered $3$ units from the axis, then rotate it about the $z$-axis.

Step 1: Parametrize $$\boxed{\mathbf r(u,v)=\langle(3+\cos v)\cos u,(3+\cos v)\sin u,\sin v\rangle},$$ with $0\le u\le 2\pi$ and $0\le v\le 2\pi$.

Step 2: Compute a normal Let $A=3+\cos v$. Then $$\mathbf r_u=\langle-A\sin u,A\cos u,0\rangle,$$ $$\mathbf r_v=\langle-\sin v\cos u,-\sin v\sin u,\cos v\rangle.$$ Their cross product is $$\boxed{\mathbf r_u\times\mathbf r_v =A\langle\cos u\cos v,\sin u\cos v,\sin v\rangle},$$ which points from the tube’s center circle toward the surface, hence outward.

Step 3: Regularity and seams The unit vector in brackets has magnitude $1$, while $A=3+\cos v\ge 2$. Thus the cross product never vanishes. The edges $u=0,2\pi$ are identified, as are $v=0,2\pi$.

Verification The distance from the $z$-axis is $3+\cos v$ and the height is $\sin v$, so $(\sqrt{x^2+y^2}-3)^2+z^2=1$, the intended torus.