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Area of one cylinder inside another intersecting cylinder

  1. May 20, 2010 #1
    1. The problem statement, all variables and given/known data :

    Find surface area of part of cylinder [itex]x^2 + z^2 = 1[/itex] that is inside the cylinder [itex]x^2 + y^2 = 2ay[/itex] and also in the positive octant ( [itex]x \geq 0, y \geq 0, z \geq 0 [/itex] ). Assume a > 0.

    2. Relevant equations

    [tex]x^2 + z^2 = 1[/tex]
    [tex]x^2 + y^2 = 2ay[/tex]
    ( [itex]x \geq 0, y \geq 0, z \geq 0 [/itex] )
    a > 0

    3. The attempt at a solution

    Cylinder expression [itex]x^2 + z^2 = 1[/itex] can be written as [itex] z = \sqrt{ ( a^2 - x^2 ) }[/itex].This surface can be projected on X-Y plane by using following parametrization -

    x = x
    y = y
    z = f( x, y ) = [itex]\sqrt{ a^2 - x^2 }[/itex]

    Hence the surface area of cylinder [itex]x^2 + z^2 = 1[/itex] for above parametrization is given by integral -

    Area integral = [itex]\iint{\sqrt{ ( 1 + f'_{x})^2 + ( f'_{y})^2 }} dx dy[/itex]

    Now, [itex] \sqrt{ 1 + (f'_{x})^2 + ( f'_{y})^2 }[/itex] = [itex] \sqrt{ 1 + \frac{x^2}{a^2 - x^2} + 0 }[/itex] = [itex] \frac{a}{ \sqrt{ a^2 - x^2 }}[/itex]


    Area integral = [itex] \iint{ \frac{a}{ \sqrt{ a^2 - x^2 }}} dx dy[/itex]

    The limits of this integral will be decided by second insecting cylinder [itex]x^2 + y^2 = 2ay[/itex]

    Working out for finding limits -

    [itex] y^2 + (-2a) y + x^2 = 0 [/itex] treating second order equation in ' y ' we can write -

    [itex] y = a \pm \sqrt { a^2 - x^2 } [/itex] selecting positive root for given conditions for positive octant.

    Hence limits are – for x -> 0 to a and for y -> 0 to [itex] a + \sqrt { a^2 - x^2 }[/itex]

    The integral becomes


    Area integral = [itex] \int_{x = 0}^{ x = a } \int_{ y = 0 }^{ y = a + \sqrt { a^2 - x^2 } }\frac{a}{ \sqrt{ a^2 - x^2 }} dy dx[/itex]

    = [itex] \int_{0}^{a} \frac{ a ( a + \sqrt { a^2 - x^2 } )}{ \sqrt { a^2 - x^2 }} dx[/itex]

    = [itex] a^2 \int_{0}^{a} \frac{ dx }{ \sqrt { a^2 - x^2 } } + a \int_{0}^{a}} dx [/itex]
    = [itex] a^2 \left[ \arcsin \left ( \frac{x}{a} \right) \right]_{0}^{a} + a^2[/itex]

    Required area = [itex] a^2 \left[ \frac{\pi}{2} +1 \right][/itex]


    The answer given in the book is [itex]2a^2[/itex] . Am I going wrong somewhere ? Please help.
     
  2. jcsd
  3. May 20, 2010 #2

    lanedance

    User Avatar
    Homework Helper

    shouldn't
    [tex]z = f( x, y ) = \sqrt{ 1 - x^2 }[/tex]?

    also you need to assume a < 1, which you're doing, but should be aware of

    otherwise i think your mothod is ok ;)
     
  4. May 21, 2010 #3
    @lanedance - Thanks for your reply.

    You have pointed out correct typing mistake due to some 'copy paste'. Sorry for that.

    Corrected problem is as follows -


    1. The problem statement, all variables and given/known data :

    Find surface area of part of cylinder [itex]x^2 + z^2 = a^2[/itex] that is inside the cylinder [itex]x^2 + y^2 = 2ay[/itex] and also in the positive octant ( [itex]x \geq 0, y \geq 0, z \geq 0 [/itex] ). Assume a > 0.

    2. Relevant equations

    [tex]x^2 + z^2 = a^2[/tex]
    [tex]x^2 + y^2 = 2ay[/tex]
    ( [itex]x \geq 0, y \geq 0, z \geq 0 [/itex] )
    a > 0

    3. The attempt at a solution

    Cylinder expression [itex]x^2 + z^2 = a^2[/itex] can be written as [itex] z = \sqrt{ ( a^2 - x^2 ) }[/itex].This surface can be projected on X-Y plane by using following parametrization -

    x = x
    y = y
    z = f( x, y ) = [itex]\sqrt{ a^2 - x^2 }[/itex]

    Hence the surface area of cylinder [itex]x^2 + z^2 = a^2[/itex] for above parametrization is given by integral -

    Area integral = [itex]\iint{\sqrt{ ( 1 + f'_{x})^2 + ( f'_{y})^2 }} dx dy[/itex]

    Now, [itex] \sqrt{ 1 + (f'_{x})^2 + ( f'_{y})^2 }[/itex] = [itex] \sqrt{ 1 + \frac{x^2}{a^2 - x^2} + 0 }[/itex] = [itex] \frac{a}{ \sqrt{ a^2 - x^2 }}[/itex]


    Area integral = [itex] \iint{ \frac{a}{ \sqrt{ a^2 - x^2 }}} dx dy[/itex]

    The limits of this integral will be decided by second insecting cylinder [itex]x^2 + y^2 = 2ay[/itex]

    Working out for finding limits -

    [itex] y^2 + (-2a) y + x^2 = 0 [/itex] treating second order equation in ' y ' we can write -

    [itex] y = a \pm \sqrt { a^2 - x^2 } [/itex] selecting positive root for given conditions for positive octant.

    Hence limits are – for x -> 0 to a and for y -> 0 to [itex] a + \sqrt { a^2 - x^2 }[/itex]

    The integral becomes


    Area integral = [itex] \int_{x = 0}^{ x = a } \int_{ y = 0 }^{ y = a + \sqrt { a^2 - x^2 } }\frac{a}{ \sqrt{ a^2 - x^2 }} dy dx[/itex]

    = [itex] \int_{0}^{a} \frac{ a ( a + \sqrt { a^2 - x^2 } )}{ \sqrt { a^2 - x^2 }} dx[/itex]

    = [itex] a^2 \int_{0}^{a} \frac{ dx }{ \sqrt { a^2 - x^2 } } + a \int_{0}^{a} dx [/itex]
    = [itex] a^2 \left[ \arcsin \left ( \frac{x}{a} \right) \right]_{0}^{a} + a^2[/itex]

    Required area = [itex] a^2 \left[ \frac{\pi}{2} +1 \right][/itex]


    The answer given in the book is [itex]2a^2[/itex] . Am I going wrong somewhere ? Please help.
     
  5. May 21, 2010 #4

    lanedance

    User Avatar
    Homework Helper

    ok so i had another look & i think your lower limit for the y integration step is incorrect, it should also be dependent on x.

    it may help to change the integration order, do x then y.
     
  6. May 21, 2010 #5

    lanedance

    User Avatar
    Homework Helper

    to help see it, the vertical cylinder given the bounds is defined by the circle
    [tex] x^2 + (y-a)^2 = a^2[/tex]
    so radius a, centred on (0,a)
     
  7. May 25, 2010 #6
    @lanedance

    You are right. After analyzing the problem again, posting the corrected area integral -


    Area integral = [tex] \int_{x = 0}^{ x = a } \int_{ y = a - \sqrt { a^2 - x^2 } }^{ y = a + \sqrt { a^2 - x^2 } }\frac{a}{ \sqrt{ a^2 - x^2 }} dy dx [/tex]


    = [tex] \int_{0}^{a} \frac{ a ( a + \sqrt { a^2 - x^2 } - a + \sqrt { a^2 - x^2 }) }{ \sqrt { a^2 - x^2 }} dx [/tex]


    = [tex]2a \int_{0}^{a} dx [/tex]

    Required area = [tex] 2a^2[/tex]

    Which is same as required !

    Above problem solved in cylindrical parametrization is posted here -

    http://www.cramster.com/answers-may...-the-surface-area-of-that-part-of_854407.aspx

    In addition to above two methods I have also tried above problem using Green's method given here -
    http://mathworld.wolfram.com/SteinmetzSolid.html

    ( answer by Green's Theorem method is posted in my next post )


    I just want confirm that, everything is solved correctly, and steps are appropriate for above three methods.
     
    Last edited by a moderator: Apr 25, 2017
  8. May 25, 2010 #7
    Method III - Using Green's theorem

    Given equation [itex] x^2 + z^2 = a^2 [/itex] can be parametrized as follows -

    x = x
    z = [itex] \sqrt{ a^2 - x^2 }[/itex]

    According to result obtained from Green's theorem we can write -

    Area = [itex] \int{ y ds }[/itex]


    where ds = [tex] \sqrt{ 1 + \left(\frac{dz}{dx} \right )^2 } dx[/tex]


    ds = [tex] \frac{a}{\sqrt{ a^2 - x^2 } } dx [/tex]


    Now, y can be calculated from cylinder equation [itex] x^2 + y^2 = 2ay[/itex] as

    [itex] y = a \pm \sqrt{ a^2 - x^2 }[/itex]

    Here we have mapped required surface area on x-y plane. Mapped area region is divided into two regions for which y takes two different values calculated. Hence, required area is given by -

    Area = [tex]\left( \int_{x = 0}^{ x = a }\frac{a( a + \sqrt{ a^2 - x^2 } ) dx } {\sqrt{ a^2 - x^2 }}\right) - \left( \int_{x = 0}^{ x = a }\frac{a( a - \sqrt{ a^2 - x^2 } ) dx } {\sqrt{ a^2 - x^2 }}\right)[/tex]


    Negative sign of second integral indicates opposite orientation of second part of surface.

    Area = [itex] 2a \int_{0}^{a} dx [/itex]

    Area = [itex] 2a^2[/itex]


    Corrections and suggestions will be highly appreciated.
     
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