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Next: Mass of a thin Up: Surface integrals and Surface Previous: Surface integral.

Special formulas for Surface Area.

The surface $\mathcal{S}$ is given by an equation of the form z=f(x,y), where $(x,y) \in R$; then

\begin{displaymath}\text{Area of } \mathcal{S} = \iint_{R} \sqrt{f_x^2+f_y^2+1} \; dx \; dy
\end{displaymath}

Example 8.29   The area of the surface cut from the bottom of the paraboloid $\mathcal{P}:\; z= x^2+y^2$ by the plane whose equation is z=1 is:
\begin{align*}A & =\iint{x^2+y^2 \leq 1} \sqrt{(2x)^2+(2y)^2+1}\; dx \; dy \\
\...
... \frac {1}{12} \right)\; d \theta
=\frac {(5 \sqrt{5} -1) \pi}{6}.
\end{align*}

***

The surface $\mathcal{S}$ is given by an equation of the form x=f(y,z), where $(y,z) \in R$; then

\begin{displaymath}\text{Area of } \mathcal{S} = \iint_{R} \sqrt{f_y^2+f_z^2+1} \; dy \; dz
\end{displaymath}

Example 8.30   The area of the surface cut from the bottom of the paraboloid $\mathcal{P}:\; x= 1 - z^2 - y^2$ by the yz-plane is:
\begin{align*}A & =\iint_{y^2+z^2 \leq 1} \sqrt{(-2y)^2+(-2z)^2+1} \; dy \; dz \...
...; dz \\
\quad &= \frac {(5 \sqrt{5} -1) \pi}{6} (\text{as above}).
\end{align*}

***

The surface $\mathcal{S}$ is given by an equation of the form y=f(x,z), where $(x,z) \in R$; then

\begin{displaymath}\text{Area of } \mathcal{S} = \iint_{R} \sqrt{f_x^2+f_z^2+1} \; dx \; dz
\end{displaymath}

Example 8.31   The area of the surface cut in the first octant from the cylinder whose equation is y=z2 by the planes whose respective equationsare x=1 and y=1 is:
\begin{align*}A & =\int_0^1 \int_0^1 \sqrt{0^2+(2z)^2+1} \; dx \; dz \\
\quad &...
...^2+1} \right]_0^1 =\frac 14 \ln (\sqrt{5}+2) + \frac {\sqrt{5}}{2}.
\end{align*}

  
Figure 8:
\begin{figure}
\mbox{\epsfig{file=AreaCylinder.eps,height=5cm} }
\end{figure}


next up previous contents
Next: Mass of a thin Up: Surface integrals and Surface Previous: Surface integral.
Noah Dana-Picard
2001-05-30