Surface area example

Example

Find the surface area of the cone S

$\displaystyle \Phi$(r,$ \theta$) = (r cos$ \theta$, r sin$ \theta$, r)    

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

Solution: Recall that the generic formula for surface area is

A = $\displaystyle \iint_{D}^{}$$\displaystyle \left\Vert\vphantom{ \frac{\partial \mathbf{\Phi}}{\partial u}(u,v) \times \frac{\partial \mathbf{\Phi}}{\partial v}(u,v) }\right.$$\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial u}}}$(u, v) x $\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial v}}}$(u, v)$\displaystyle \left.\vphantom{ \frac{\partial \mathbf{\Phi}}{\partial u}(u,v) \times \frac{\partial \mathbf{\Phi}}{\partial v}(u,v) }\right\Vert$du dv.    

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

$\displaystyle \left\Vert\vphantom{\frac{\partial \mathbf{\Phi}}{\partial r}(r,{... ...s \frac{\partial \mathbf{\Phi}}{\partial \theta}(r,{\textstyle \theta})}\right.$$\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial r}}}$(r,$ \theta$) x $\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial \theta}}}$(r,$ \theta$)$\displaystyle \left.\vphantom{\frac{\partial \mathbf{\Phi}}{\partial r}(r,{\tex... ...rac{\partial \mathbf{\Phi}}{\partial \theta}(r,{\textstyle \theta})}\right\Vert$    

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.

$\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial r}}}$ = (cos$ \theta$, sin$ \theta$, 1)    
$\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial \theta}}}$ = (- r sin$ \theta$, r cos$ \theta$, 0)    
$\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial r}}}$ x $\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial {\textstyle \theta}}}}$ = $\displaystyle \left\vert\vphantom{ \begin{array}{ccc} \mathbf{i} & \mathbf{j} &... ... -r \sin{\textstyle \theta}& r \cos{\textstyle \theta}& 0 \end{array} }\right.$$\displaystyle \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \cos{\t... ...heta}& 1 -r \sin{\textstyle \theta}& r \cos{\textstyle \theta}& 0 \end{array}$$\displaystyle \left.\vphantom{ \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \ma... ... \sin{\textstyle \theta}& r \cos{\textstyle \theta}& 0 \end{array} }\right\vert$    
  = i(- r cos$ \theta$) - j(r sin$ \theta$)    
       + kr(cos2$ \theta$ + sin2$ \theta$)    
  = - r cos$ \theta$i - r sin$ \theta$j + rk    
$\displaystyle \left\Vert\vphantom{\frac{\partial \mathbf{\Phi}}{\partial r} \times \frac{\partial \mathbf{\Phi}}{\partial {\textstyle \theta}}}\right.$$\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial r}}}$ x $\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial {\textstyle \theta}}}}$$\displaystyle \left.\vphantom{\frac{\partial \mathbf{\Phi}}{\partial r} \times \frac{\partial \mathbf{\Phi}}{\partial {\textstyle \theta}}}\right\Vert$ = $\displaystyle \sqrt{{r^2\cos^2{\textstyle \theta}+ r^2\sin^2{\textstyle \theta}+ r^2}}$    
  = $\displaystyle \sqrt{{2r^2}}$ = r$\displaystyle \sqrt{{2}}$    

The area of cone S is

A(S) = $\displaystyle \int_{0}^{1}$$\displaystyle \int_{0}^{{2\pi}}$$\displaystyle \left\Vert\vphantom{\frac{\partial \mathbf{\Phi}}{\partial r} \times \frac{\partial \mathbf{\Phi}}{\partial {\textstyle \theta}}}\right.$$\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial r}}}$ x $\displaystyle {\frac{{\partial \mathbf{\Phi}}}{{\partial {\textstyle \theta}}}}$$\displaystyle \left.\vphantom{\frac{\partial \mathbf{\Phi}}{\partial r} \times \frac{\partial \mathbf{\Phi}}{\partial {\textstyle \theta}}}\right\Vert$d$ \theta$ dr    
  = $\displaystyle \int_{0}^{1}$$\displaystyle \int_{0}^{{2\pi}}$r$\displaystyle \sqrt{{2}}$d$ \theta$ dr    
  = $\displaystyle \int_{0}^{1}$2$ \pi$r$\displaystyle \sqrt{{2}}$dr    
  = $ \pi$r2$\displaystyle \sqrt{{2}}$$\displaystyle \Big\vert _{0}^{1}$ = $\displaystyle \pi$$\displaystyle \sqrt{{2}}$.    


Duane Nykamp
nykamp@math.umn.edu
2005-11-10