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Math Help - Surface area of cone lying inside a cylinder

  1. #1
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    Cool Surface area of cone lying inside a cylinder

    Find the surface area of that part of the cone z^2 = x^2 + y^2 which lies inside the cylinder x^2 + y^2 = 2x

    OK SO: z is a circle (x − 1)^2 + y^2 = 1. We use polar coordinates for x
    and y
    x =  pcos(t) , y = psin(t)  .
    Then, since we are interested in a surface living on the cone, its equation yields z^2 = p^2.
    The sought surface is made of two parts, one for positive z i.e. z = p and one for negative z i.e. z = −p. We need to determine the bounds for p and t. This is given by the fact that we are interested in the interior of the cylinder

    x^2 + y^2 − 2x ≤ 0. This gives
    2 − 2pcos(t) ≤ 0
    Combined with the fact that p ≥ 0 in polar coordinates we get
    0 ≤ p ≤ 2 cos (t)
    which also shows that 0 ≤ pcos(t) so t E [−pi/2,pi/2 ]. Collecting everything the parametrization for the top part is

    −>
    F 1 : [0, 2 cos (t)] [−pi/2,pi/2] → R^3

    (p,(t)) → pcos(t)i+psin(t)j+pk

    and for the bottom part

    −>
    F 1 : [0, 2 cos (t)] [−pi/2,pi/2] → R^3

    (p,(t)) → pcos(t)i+psin(t)j-pk


    I JUST DONT UNDERSTAND HOW THE PARAMETRIZATION OF THE TOP AND BOTTOM PART HAVE BEEN WORKED OUT!
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  2. #2
    Senior Member AllanCuz's Avatar
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    Quote Originally Posted by zizou1089 View Post
    Find the surface area of that part of the cone z^2 = x^2 + y^2 which lies inside the cylinder x^2 + y^2 = 2x

    OK SO: z is a circle (x − 1)^2 + y^2 = 1. We use polar coordinates for x
    and y
    x = pcos(t) , y = psin(t) .
    Then, since we are interested in a surface living on the cone, its equation yields z^2 = p^2.
    The sought surface is made of two parts, one for positive z i.e. z = p and one for negative z i.e. z = −p. We need to determine the bounds for p and t. This is given by the fact that we are interested in the interior of the cylinder

    x^2 + y^2 − 2x ≤ 0. This gives
    2 − 2pcos(t) ≤ 0
    Combined with the fact that p ≥ 0 in polar coordinates we get
    0 ≤ p ≤ 2 cos (t)
    which also shows that 0 ≤ pcos(t) so t E [−pi/2,pi/2 ]. Collecting everything the parametrization for the top part is

    −>
    F 1 : [0, 2 cos (t)] [−pi/2,pi/2] → R^3

    (p,(t)) → pcos(t)i+psin(t)j+pk

    and for the bottom part

    −>
    F 1 : [0, 2 cos (t)] [−pi/2,pi/2] → R^3

    (p,(t)) → pcos(t)i+psin(t)j-pk

    I JUST DONT UNDERSTAND HOW THE PARAMETRIZATION OF THE TOP AND BOTTOM PART HAVE BEEN WORKED OUT!
    I really wouldn't attempt this question like this, it doesn't make it any easiar.

    But you know what z=\sqrt {x^2 + y^2 } looks like right? It's a cone with its vertex at the origin and its base facing upwards. Of course this face would extend forever, but in this case is bounded by a cylinder.

    Parametrization is defined as

    r= x \hat i + y \hat j + z \hat k

    Well, we are letting

    x=pcos(t) and y=pcos(t) so  z^2 = p^2cos^2t + p^2sin^2t = p^2 so  z = p

    Putting this into r

     r = pcos(t) \hat i + psin(t) \hat j + p \hat k

    Which is what you have. I don't see where the problem is? You understood how to do all the above, it's simply a matter of putting in into r.

    This question is actually not helped by a parametrization (at least not in this form). I mean at the end of the day it's going to be the exact same, but why bother going through the extra step they're making you walk here?

    What we're actually looking for is

     \iint dS where

     dS = \sqrt { 1 + ( \frac{x}{ \sqrt { x^2 + y^2 }} )^2 + ( \frac {y}{ \sqrt{x^2 + y^2} } )^2 } dxdy

    Thus,

    dS = \sqrt { 1 + \frac{x^2 + y^2} {x^2 + y^2} } dxdy = \sqrt {2} dxdy

    So what we need is

    \iint \sqrt{2} dxdy

    XY domain is bounded by  x^2 + y^2 = 2x so let us transform to polar.

    p^2 = 2pcos( \theta )

    Thus, we know

     0 \le p \le 2cos ( \theta )

    And our theta is bounded from

     0 \le \theta \le \frac { \theta } {2}

    Our integral is then

     2 \int_0^{ \frac { \theta }{2} } d \theta \int_0^{ 2cos ( \theta ) } p\sqrt{2} dp

    Which is an easy integral!
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