# Surface area problem

• Mar 11th 2010, 01:45 PM
greencheeseca
Surface area problem
Find the area of the surface.

z = f(x,y) = 9 - x^2 over the region R: Square with vertices (0,0), (3,0), (0,3), (3,3).

can anyone show me how to solve this problem? I seriously have no clue how to start.
• Mar 11th 2010, 02:33 PM
Scott H
The function $\displaystyle f(x,y)=9-x^2$ defines a parabolic sheet of constant altitude extending over the $\displaystyle y$-axis. The cross-sections of the portion of the surface that lies over $\displaystyle [0,3]\times[0,3]$ are the same for all values of $\displaystyle y$.

Hint: We may find the surface area by first calculating the length of the curve defined by $\displaystyle f(x)=9-x^2$ from $\displaystyle x=0$ to $\displaystyle x=3$.
• Mar 11th 2010, 03:53 PM
greencheeseca
I'm lost
Quote:

Originally Posted by Scott H
The function $\displaystyle f(x,y)=9-x^2$ defines a parabolic sheet of constant altitude extending over the $\displaystyle y$-axis. The cross-sections of the portion of the surface that lies over $\displaystyle [0,3]\times[0,3]$ are the same for all values of $\displaystyle y$.

Hint: We may find the surface area by first calculating the length of the curve defined by $\displaystyle f(x)=9-x^2$ from $\displaystyle x=0$ to $\displaystyle x=3$.

I can't get it. I just started surface integrals this week and I'm totally lost. I need to see some resolved problems so I can back engineer what's going on..
• Mar 12th 2010, 06:12 AM
Jeroentje
The surface you are looking can be devided in a part where the graph has not entered the square (x=0 to sqrt(6)) and a part where the graph has entered the square (x=sqrt(6) to 3).

The surface of the first part is easy. sqrt(6)x3

The surface of the second part can be written as an integral.

-x^2+9 integrated from sqrt(6) to 3 =
27-27/3-9*sqrt(6)-(sqrt(6)^3)/3=0.85

which leads to an surface of 8.20