Math Insight

Parametrized surface area example


Find the surface area of the cone $\dls$ \begin{align*} \dlsp(r,\theta) = (r \cos \theta, r \sin\theta, r) \end{align*} for $0 \le \theta \le 2\pi$ and $0 \le r \le 1$.

Solution: The generic formula for surface area is \begin{align*} A=\iint_\dlr \left\| \pdiff{\dlsp}{\spfv}(\spfv,\spsv) \times \pdiff{\dlsp}{\spsv}(\spfv,\spsv) \right\| d\spfv\,d\spsv. \end{align*} In this case, the variables are $r$ and $\theta$ rather than $\spfv$ and $\spsv$. Hence, we need to integrate \begin{align*} \left\|\pdiff{\dlsp}{r}(r,\theta) \times \pdiff{\dlsp}{\theta}(r,\theta)\right\| \end{align*} over the region $\dlr$ defined by $0 \le \theta \le 2\pi$ and $0 \le r \le 1$.

We calculate the cross product as follows. \begin{align*} \pdiff{\dlsp}{r} &= (\cos\theta, \sin\theta ,1)\\ \pdiff{\dlsp}{\theta} &= (-r\sin\theta, r \cos\theta, 0)\\ \pdiff{\dlsp}{r} \times \pdiff{\dlsp}{\theta} &= \left| \begin{array}{ccc} \vc{i} & \vc{j} & \vc{k}\\ \cos\theta & \sin\theta & 1\\ -r \sin\theta & r \cos\theta & 0 \end{array} \right|\\ &= \vc{i}(-r\cos\theta) -\vc{j} (r\sin\theta)\\ &\quad + \vc{k} r(\cos^2\theta+\sin^2\theta)\\ &=-r\cos\theta \vc{i} - r\sin\theta \vc{j} + r \vc{k}\\ \left\|\pdiff{\dlsp}{r} \times \pdiff{\dlsp}{\theta}\right\| &=\sqrt{r^2\cos^2\theta + r^2\sin^2\theta + r^2}\\ &=\sqrt{2r^2} = r\sqrt{2} \end{align*}

The area of cone $\dls$ is \begin{align*} A(\dls) &=\int_0^1\int_0^{2\pi} \left\|\pdiff{\dlsp}{r} \times \pdiff{\dlsp}{\theta}\right\| d\theta\, dr\\ &=\int_0^1\int_0^{2\pi} r\sqrt{2} d\theta\, dr\\ &=\int_0^1 2\pi r \sqrt{2} dr\\ &= \left.\left.\pi r^2 \sqrt{2} \right|_0^1\right. = \pi\sqrt{2}. \end{align*}