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2 cylinders intersect, area of resulting parametric surface

  1. Jun 18, 2015 #1
    1. The problem statement, all variables and given/known data

    I want to know if I got the answer correct and if my reasoning is sound. The text answers and solutions manual only gives answers/solutions for odd numbered problems.

    Here is the problem: Tko1xFh.png

    And a direct link to the imgur page: http://i.imgur.com/Tko1xFh.png

    2. Relevant equations

    [tex]A(S) = \int \int_D \left|\vec{r}_u\times\vec{r}_v\right|dA[/tex]

    3. The attempt at a solution

    Instead of trying to come up with a parametric vector equation for the entire surface, I started subdividing the surface as much as possible using symmetry. I got it down to one sixteenth of the whole surface. Basically, just consider the surface in the first octant, and then divide that in two again along the line y = x.

    By doing that I get the following:

    [tex]A(S) = 16 \times \int\int_D\left|\vec{r}_x\times\vec{r}_y\right|dA[/tex]

    where [itex]\vec{r} = \left< x,y,\sqrt{1-x^2}\right>[/itex] and [itex]D = \{(x,y)|0\leq x \leq 1, 0 \leq y \leq x\}[/itex]

    so

    [itex]\vec{r}_x = \langle 1,0,\frac{-x}{\sqrt{1-x^2}} \rangle[/itex] and [itex]\vec{r}_y = \langle 0,1,0 \rangle[/itex]

    [tex]\vec{r}_x \times \vec{r}_y = \langle \frac{x}{\sqrt{1-x^2}},0,1 \rangle[/tex]

    [tex]\left| \vec{r}_x \times \vec{r}_y \right| = \sqrt{1+\frac{x^2}{1-x^2}} = \frac{1}{1-x^2}[/tex]

    Now

    [tex]A(S) = 16 \times \int_0^1\int_0^x \frac{1}{1-x^2} dydx[/tex]

    [tex]A(S) = 16 \times \int_0^1 \frac{x}{1-x^2} dx = -16 \times\left[\sqrt{1-x^2}\right]_0^1 = 16[/tex]
     
  2. jcsd
  3. Jun 18, 2015 #2

    LCKurtz

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    Shouldn't that be$$
    \frac{1}{\sqrt{1-x^2}}$$
     
  4. Jun 19, 2015 #3
    Typo :) In my original work that typo isn't there.
     
  5. Jun 19, 2015 #4

    haruspex

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    I haven't tried to follow your working (yet), but a mental calculation using a simpler method gives me 16 also.
     
  6. Jun 19, 2015 #5

    haruspex

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    My method:
    Convert the xz plane to polar, measuring theta from lowest point.
    Consider face on the positive x side.
    Between ##\theta## and ##\theta+d\theta##, there is a rectangle measuring ##r.d\theta## by ##2r\sin(\theta)##. Integrate 0 to ##\pi## and multiply by 4.
     
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