surface area of torus of revolution

Solution 1:

It follows from your calculation that we have a map $F:[0,2\pi]\times[0,2\pi]\rightarrow\mathbb{R}^3$, where $$F(\theta,\psi)=((a + b \cos \psi)\cos \theta, b \sin \psi, (a + b \cos \psi)\sin \theta)$$ and the image under $F$ is the torus. Then we have $$F_\theta=(-(a + b \cos \psi)\sin \theta, 0, (a + b \cos \psi)\cos \theta),$$ $$F_\psi=(-b \sin \psi\cos \theta, b \cos \psi, -b \sin \psi\sin \theta).$$ Hence, $$F_\theta\times F_\psi=(-b (a + b \cos \psi)\cos \theta\cos \psi,-b(a + b \cos \psi)\sin \psi, -b(a + b \cos \psi)\sin \theta\cos\psi)$$ which implies that $$\|F_\theta\times F_\psi\|=b (a + b \cos \psi).$$ It's a standard fact from Calculus that the surface area is given by $$\int_0^{2\pi}\int_0^{2\pi}\|F_\theta\times F_\psi\|d\theta d\psi.$$ (see here: Parametric surface) Therefore the surface area of the torus is equal to $$\int_0^{2\pi}\int_0^{2\pi}b (a + b \cos \psi)d\theta d\psi=4ab\pi^2.$$