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Find the area of the surface given by z = f(x, y) over the region R. (Hint: The...
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We have the surface
`f(x,y) = 64 + x^2 - y^2`
and the constraint `x^2+y^2 <=1` , `x` and `y` `in R`
This constraint can be rewritten as
`x^2 = 1 - y^2` and `y^2 <=1`
Substituting for `x^2` , we want the area of the bounded region
`f(y) = 65 - 2y^2`
where `0 <= y <=1 `.
Integrating, the bounded region is
`B = int_0^1 (65 - 2y^2) = (65y - 2/3y^3)|_0^1 = 65 -2/3 = 64 1/3`
The area is 64 + 1/3
Posted by mathsworkmusic on March 31, 2013 at 11:45 AM (Answer #1)
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