Surface Area

  • Thread starter Dafe
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  • #1
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Homework Statement


[tex]S={(x,y,z): z=x+y^2, 0\leq x\leq1 , x\leq y \leq 1 [/tex]



Homework Equations


A=a(S) = [tex]$
\int\int_R \sqrt{(1+(\frac{\partial f}{\partial x})^2 + (\frac{\partial f}{\partial y})^2\,dy\,dx[/tex]


The Attempt at a Solution


[tex]S={(x,y,z): z=x+y^2, 0\leq x\leq1 , x\leq y \leq 1 [/tex]

I suppose I can write this as:

[tex]S={(x,y,z): z=x+y^2, y\leq x\leq1 , 0\leq y \leq 1 [/tex]
And so i think:

A=a(S) = [tex]$
\int_0^1\int_y^1 \sqrt{2+4y^2}\,dy\,dx[/tex]

If I calculate this I don't get the answer that I should..
 
Last edited:

Answers and Replies

  • #2
Dick
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It would help a great deal to know what you got and how you got it.
 
  • #3
arildno
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SKETCH the region given!!

Do it in the following way:

1. Draw the rectangle strip [tex]0\leq{x}\leq{1}[/tex] in the xy-plane.
2. Since y<=1, draw the line y=1. You are to be below this line, within the strip from 1.

3. Now, you have x<=y. Draw the line x=y, you are to be above that line.

4. Thus, you may represent the region as follows:
[tex]0\leq{x}\leq{1}, x\leq{y}\leq{1}[/tex]
These limits on y were gained by looking at the vertical line segments the region consists of for all x-positions of these segments from 0 to 1.

Alternatively, we may consider the horizontal line segments the region consists of; this yields the equally valid representation:
[tex]0\leq{y}\leq{1}, 0\leq{x}\leq{y}[/tex]


These are the two simplest correct region representations, yours is not correct.
 

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