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Homework Help: Surface Area

  1. Mar 24, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]S={(x,y,z): z=x+y^2, 0\leq x\leq1 , x\leq y \leq 1 [/tex]



    2. Relevant 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]


    3. 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: Mar 24, 2007
  2. jcsd
  3. Mar 24, 2007 #2

    Dick

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    It would help a great deal to know what you got and how you got it.
     
  4. Mar 24, 2007 #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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