Multivariable calculus and vector analysis

Surface area example

Example

Find the surface area of the cone $S$

$$\begin{array}{lll}\hfill \mathbf{\Phi }\left(r,\theta \right)=\left(rcos\theta ,rsin\theta ,r\right)& & \hfill \end{array}$$

for $0\le \theta \le 2\pi$ and $0\le r\le 1$.

Solution: Recall that the generic formula for surface area is

$$\begin{array}{lll}\hfill A={\iint }_D∥\frac{\partial \mathbf{\Phi }}{\partial u}\left(u,v\right)\times \frac{\partial \mathbf{\Phi }}{\partial v}\left(u,v\right)∥dudv.& & \hfill \end{array}$$

In this case, the variables are $r$ and $\theta$ rather than $u$ and $v$. Hence, we need to integrate

$$\begin{array}{lll}\hfill ∥\frac{\partial \mathbf{\Phi }}{\partial r}\left(r,\theta \right)\times \frac{\partial \mathbf{\Phi }}{\partial \theta }\left(r,\theta \right)∥& & \hfill \end{array}$$

over the region $D$ defined by $0\le \theta \le 2\pi$ and $0\le r\le 1$.

We calculate the cross product as follows.

$$\begin{array}{llll}\hfill \frac{\partial \mathbf{\Phi }}{\partial r}& =\left(cos\theta ,sin\theta ,1\right)& \hfill & \\ \hfill \frac{\partial \mathbf{\Phi }}{\partial \theta }& =\left(-rsin\theta ,rcos\theta ,0\right)& \hfill & \\ \hfill \frac{\partial \mathbf{\Phi }}{\partial r}\times \frac{\partial \mathbf{\Phi }}{\partial \theta }& =\left|\begin{array}{ccc}\hfill \mathbf{i}\hfill & \hfill \mathbf{j}\hfill & \hfill \mathbf{k}\hfill \\ \hfill cos\theta \hfill & \hfill sin\theta \hfill & \hfill 1\hfill \\ \hfill -rsin\theta \hfill & \hfill rcos\theta \hfill & \hfill 0\hfill \end{array}\right|& \hfill & \\ \hfill & =\mathbf{i}\left(-rcos\theta \right)-\mathbf{j}\left(rsin\theta \right)& \hfill & \\ \hfill & +\mathbf{k}r\left({cos}^2\theta +{sin}^2\theta \right)& \hfill & \\ \hfill & =-rcos\theta \mathbf{i}-rsin\theta \mathbf{j}+r\mathbf{k}& \hfill & \\ \hfill ∥\frac{\partial \mathbf{\Phi }}{\partial r}\times \frac{\partial \mathbf{\Phi }}{\partial \theta }∥& =\sqrt{r^2{cos}^2\theta +r^2{sin}^2\theta +r^2}& \hfill & \\ \hfill & =\sqrt{2r^2}=r\sqrt{2}& \hfill & \end{array}$$

The area of cone $S$ is

$$\begin{array}{llll}\hfill A\left(S\right)& ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{2\pi }∥\frac{\partial \mathbf{\Phi }}{\partial r}\times \frac{\partial \mathbf{\Phi }}{\partial \theta }∥d\theta dr& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^1{\int \; <!--nolimits\; -->}_0^{2\pi }r\sqrt{2}d\theta dr& \hfill & \\ \hfill & ={\int \; <!--nolimits\; -->}_0^{1}2\pi r\sqrt{2}dr& \hfill & \\ \hfill & ={\left.\pi r^2\sqrt{2}\right|}_0^1=\pi \sqrt{2}.& \hfill & \end{array}$$

Contents by topic

1. Geometry in higher dimensions

2. Differential Calculus

3. Integral calculus

4. Vector calculus

4a. Vector field basics

4b. Parametrized curves

4c. Parametrized surfaces

• Parametrized surfaces
• Parametrized surface examples
• Surface area of parametrized surfaces
• Surface area calculation
• Normal vector of parametrized surfaces
• Orienting surfaces
• A surface that is not orientable
• Surface area example

4d. Path integrals and line integrals

4e. Surface integrals

5. The fundamental theorems of vector calculus

6. Review material