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About LinkBacksFind 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.
The functiondefines 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 byfrom
to
.
Originally Posted by Scott H
The function
Hint: We may find the surface area by first calculating the length of the curve defined by
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..
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!
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