Section 6.6 Parametric Surface Area

6.6 Parametric Surface Area

In this section, we are interested in finding the surface area when we are given the parametric representation of the surface. It is naturally we know that the formula has to come from Riemann sum. 

Definition: 

A parameterized surface is given by a description of the form $\overrightarrow{r(u,v)}=<x(u,v),y(u,v),z(u,v)>$ where $(u,v)$ ranges throughout the parameters domain $D$ then the surface area of $S$ is 

$$A(S)=\int {\int }_D||\overrightarrow{r_u}\times \overrightarrow{r_v}||dA.$$

Example 1: Find the surface area of $\overrightarrow{r(u,v)}=<u-v,1+2u,1-v>$, $1\le u\le 2$, $0\le v\le 1$. 

Exercise 1: Find the surface area of $\overrightarrow{r(u,v)}=<u+v,3+v,2+u>$, $0\le u\le 1$, $2\le v\le 3$. 

Example 2: Find the surface area of $\overrightarrow{r(u,v)}=<v,u\text{cos}(v),u\text{sin}(v)>$, $0\le u\le 2$, $0\le v\le \pi$. 

Exercise 2: Find the surface area of $\overrightarrow{r(u,v)}=<u\text{sin}(v),u\text{cos}(v),v>$, $0\le u\le 1$, $0\le v\le \frac{\pi }{2}$. 

For a surface $z=f(x,y)$, we can find the surface area using the following

$$A(S)=\int {\int }_D\sqrt{1+f_x^2+f_y^2}dA.$$

Example 3: Find the surface area of $z=4x-5y$, $-1\le x\le 1$, $0\le y\le 1$. 

Exercise 3: Find the surface area of $z=-x+y$, $0\le x\le 1$, $-1\le y\le 2$. 

Example 4: Find the surface area of $z=4-2x^2+y$ that lies above the triangle $(0,0)$, $(1,0)$ and $(1,1)$. 

Exercise 4: Find the surface area of $z=16+x+y^2$, that lies above the triangle $(0,0)$, $(-1,1)$ and $(0,1)$. 

Example 5: Find the surface area of the sphere $x^2+y^2+z^2=25$ that lies inside the cylinder $x^2+y^2=4$.

Group work:

1. Find the surface area of the sphere $x^2+y^2+z^2=16$ that lies inside the cylinder $x^2+z^2=1$.

2. Find the surface area of $z=(x^{\frac{3}{2}}+y^{\frac{3}{2}})$, $0\le x\le 1$, $0\le y\le 1$. 

3. Find the surface area of the plane $z=x+4$ that lies inside the cylinder $x^2+y^2=4.$

4. Find the equation of the tangent plane at $(\frac{1}{2},\frac{\sqrt{3}}{2},\frac{\pi }{3})$ of the surface $\overrightarrow{r(u,v)}=<u\text{cos}(v),u\text{sin}(v),v>$.