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.

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- Mar 11th 2010, 02:45 PMgreencheesecaSurface 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, 03:33 PMScott H
The function defines a parabolic sheet of constant altitude extending over the -axis. The cross-sections of the portion of the surface that lies over are the same for all values of .

Hint: We may find the surface area by first calculating the length of the curve defined by from to . - Mar 11th 2010, 04:53 PMgreencheesecaI'm lost
- Mar 12th 2010, 07:12 AMJeroentje
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

Hope this will help you! - Mar 14th 2010, 10:38 PMgreencheeseca
shouldn't there be 2 integrals to solve this problem?