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Area of a plane that lies within a cylinder

  1. Apr 22, 2014 #1

    BiGyElLoWhAt

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    Say I have a plane, and it intersects with a [edited]cylinder*. What kind of method should I use to go about solving this?

    I've tried setting up a ##\int \int dA## situation, but wasn't sure that was applicable because it's in 3-space (also my plane is in terms of x y and z).

    I know it's an ellipse, I can solve for the points of intersection, any nudges that anyone cares to share? I'm drawing a blank (other than trying to solve for the axis lengths and looking the area of an ellipse) but that would be cheating ;-]

    (I'm not looking for a how to, just an idea to get me thinking so I can figure it out without looking up how to do it)
     
    Last edited: Apr 22, 2014
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  3. Apr 22, 2014 #2

    LCKurtz

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    Your title says cylinder and your question says cone. But they are similar. For the cylinder case just use cylindrical coordinates ##r,\theta## over the circle. The cone is trickier. Set the z's equal to get the xy equation of the cylindrical ellipse passing through their intersection. Put it in standard form and you may find the xy ellipse is not centered at the origin. You may find something in the form$$
    \frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2}=1$$In that case you can use a similar parameterization:$$
    x = x_0+ar\cos\theta,~y=y_0 + br\sin\theta$$with the correct ##dS## for the parameterization. Unfortunately, it could be even worse with the ellipse rotated.
     
  4. Apr 22, 2014 #3

    BiGyElLoWhAt

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    It is a cylinder, the cone is a typo, I'll edit it out.
    But... I'm not sure what you mean by using cylindrical coordinates. My cylinder equation (in cylindrical) would be ##r^2 = 4##, but I don't see how that helps me.

    So it's really not that complex of an equation.

    Also, 1 the z's are irrelevant I believe, as I could move the plane up, or down any amount and still have the same area.

    2 I have no z in my cylindrical expression.

    I'll throw the problem out there, maybe it'll help:
    ##x+2y+3z=1## is my plane, &
    ##x^2+y^2 = 4## is my cylinder
     
  5. Apr 22, 2014 #4

    LCKurtz

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    The area of the plane you want lies above the interior of the circle. You can set it all up in xy coordinates if you want. Solve the plane for z in terms of x and y and use the formula for dS. Then set it up to integrate over the circle. At that point you can change to cylindrical (polar in xy) coordinates to make the integration over the circle easier. Alternatively you can parameterize the plane with ##r,\theta## in the first place if you know how to calculate dS that way.
     
  6. Apr 22, 2014 #5

    BiGyElLoWhAt

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    So is this basically green's theorem?
     
  7. Apr 22, 2014 #6

    LCKurtz

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    No. Nothing to do with Green's theorem.
     
  8. Apr 22, 2014 #7

    BiGyElLoWhAt

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    Alright then I don't think I'm getting what you're saying. It looks like you're telling me to set up a line integral with dS.
    What does this particular dS represent?
     
  9. Apr 22, 2014 #8

    LCKurtz

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    You were asking about a surface area. Isn't that what you are studying? Your book should have a formula for the differential element of surface area ##dS##. If you have a function ##f(x,y)## on a region ##R## in the xy plane then$$
    dS =\sqrt{1 + f_x^2 + f_y^2}dydx$$and you calculate the area of the surface by$$
    Area =\iint_R 1~dS = \iint_R\sqrt{1 + f_x^2 + f_y^2}dydx$$Surely this is explained in your text. It has nothing to do with line integrals.
     
  10. Apr 22, 2014 #9

    BiGyElLoWhAt

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    That's not what I'm getting at. First of all its not a surface area, as I have no solid that I'm looking at. Second of all, if you look at the equation, the area I'm trying to find is definitely not in the xy plane.

    What I need, and what I think I'm going to need to refresh up on, is projections on to a plane that doesn't lie on an axis.
    I.e. x +2y + 3z=1

    I have a circle, project that circle onto the given plane, and find the area within the ellipse that is made. I could look it up and have someone "give" me the answer, I will guarantee the formula is in my book. I don't want the answer given to me, if I did I would've looked it up yesterday. I want to overcome the challenge of figuring it out by myself. The reason I posted this is because I feel I am overlooking something. The question is: what?

    Also from the formula you gave (forgive me, I'm on my phone and can't itex well)
    ##\sqrt{dx^2dy^2 + \frac{dx^2dy^2dz^2}{dx^2} + \frac{dx^2dy^2dz^2}{dy^2}}## is that my area? If it is I'm not seeing it. I'm not saying it's wrong, I just don't see it.
     
    Last edited: Apr 22, 2014
  11. Apr 22, 2014 #10

    LCKurtz

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    First you say it is not an area you want, then you say it is. And I didn't say anything about a solid or that the plane was in the xy plane.

    I suppose you could write dS that way but why would you want to? Just solve your plane for z in terms of x and y and use the formula I gave you. Or look in your book.
     
  12. Apr 23, 2014 #11

    BiGyElLoWhAt

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    Ahh, I see what you were saying. You were talking about the circle lying in the xy plane. Sorry, I was tired last night.
     
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