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Mth 277
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question_12_4_9
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Problem number 9 from section 12.4 of vector calc. Find the surface area of: the portion of the surface z= x^2 over the square region with vertices (0,0,0), (0,4,0), (4,0,0), (4,4,0).
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Since your region is a square in the xy plane and your surface is in the xz plane, how is it that you have any surface area at all over that region? The book says the answer is : 8*sqrt(65) + ln(8 + sqrt(65) ). Even if you did try to do the surface area you would get (double integral) sqrt( 4x^2 + 1) dx dy. where x ranges from 0 to 4 and y ranges from 0 to 4. But is that integrable? Am I doing something stupid again?
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The surface z = x^2 does include the curve z = x^2 in the x-z plane. So for example the surface contains the point (2, 0, 4), where x = 2 and x = 4.
However the surface also contains, for example, the point (2, 1, 4). As you can verify, the coordinates of this point satisfy z = x^2.
In fact the coordinates of any point (2, y, 4) satisfy the condition that z = x^2.
So the surface z = x^2 contains not just the point (2, 0, 4) but the entire line (2, y, 4) for all possible values of y.
This is just the line through (2, 0, 4) parallel to the y axis.
Now (2, 0, 4) is just one point on the curve z = x^2 in the x z plane. The same thing would be true for any point on this curve, which means that the surface x = z^2 can be constructed by constructing the curve z = x^2 in the x-z plane, then constructing through every point of this curve the straight line parallel to the y axis.
So the surface is a sort of parabolic trough symmetric about the yz plane.
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A normal vector is z_x `i + z_y `j + `k.
z_x = 2 x
z_y = 0
so the normal vector is
`N = 2 x `i + 0 `j + `k
and a unit normal is
`n = (2 x `i + `k) / sqrt( 4 x^2 + 1).
Thus for evey point (x, y, x^2) on the surface we have
cos(gamma) = `n dot `k = 1 / sqrt( 4 x^2 + 1)
so that
sec(gamma) = sqrt(4 x^2 + 1).
The surface area of the increment above your typical `dx by `dy increment of the xy plane is just sec(gamma) * `dx * `dy, where sec(gamma) is evaluated at the sample point (x*, y*). Thus
sec(gamma) = sqrt( 4 x* ^ 2 + 1).
You sum the sqrt(4 x*^2 + 1) `dx `dy increments, which approach the integral of sqrt(4 x^2 + 1) over the rectangular region of the xy plane.
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