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Math Help - Shapes of surfaces in R^3

  1. #1
    iva
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    Shapes of surfaces in R^3

    I have to describe the shapes given by some formulas, would this be the way to do it ,I'll do it with one of my problems:

    (the x2 and y2 below are meant to be x squared and y squared)

    f(x,y) = 2 √ (x2 + y2)

    So to get a projection on the Y-Z axis I put x=0 then z=2y
    Same for the X-Z axis, if y=0 then z=2x So this forms 2 straight lines where the z value is always double the x or y value so all of this together gives a cone along the z axis.

    Then we must find the contour lines, so if we put z=1 then this is a circle with a radius of 1/4
    Similarly if z=2 this is a circle with radius 1 and so on.

    So this definitely gives a cone starting at the origin and formed along the z axis.

    Is that right?

    Thanks!
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    iva
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    Agh, another thing is that I have to describe the intersection of this equation with this other equation which is a hemisphere (h(x,y) = √ (1-x2 - y2). When I put these two equal to each other f = h i get the equation of

    x2 + y2 = 1/5 which is a circle. I can more or less imagine this, ie the cone intersecting the hemisphere at the radius z = 1/5 , where the borders of the cone and hemisphere intersect? Is that right? But then my other question is how can the intersection be 2 dimensional and not 3?

    Many thanks again
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  3. #3
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    I found all of that admirably clear. (Whether it meets the particular expectations for your assignment I can't be sure, but would bet.) As for your last question, you wouldn't object to the intersection of two lines being a point, would you?
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  4. #4
    iva
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    Good point ..BUT, ok, before working out the intersection, i visualised it, and thought that it would be a cone shape ( like an ice cream cone with scoop (dome) on top) but then when i worked out the equation it worked out to just a circle, i mean, well the intersection will have to include part of the cone that sits inside the hemisphere! Its 3D right???? but the equation tells me it just intersects on the outline of the hemisphere in a circle? What about the inside??? OR doesn't it count because its only the outline of the sphere that is actually part of the graph? THAT could make sense right?
    Thanks!
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  5. #5
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    Quote Originally Posted by iva View Post
    OR doesn't it count because its only the outline of the sphere that is actually part of the graph? THAT could make sense right?
    Thanks!
    Precisely - the volume inside the sphere would be represented by a corresponding inequality for that equation.
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