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Math Help - Vector Calculus, sketching regions in R3

  1. #1
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    Vector Calculus, sketching regions in R3

    sketch the following region in R3

    V={r:y>=0,0<=x^2+y^2<=1,0<=z<=2y}

    ok i have that xy plane is a quarter circle, on the zy plane is the line z=2y.. im trying to picture but i just cant, im thinking its like a quarter cone, but the apex is not at (0,0,0) please can anyone try and draw it for me?
    Last edited by fredrick08; May 7th 2009 at 04:33 AM.
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  2. #2
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    ok from that, i think i can see that x is from 0 to root(1-y^2), y is from 0 to 1 and z is from 0 to 2y.. so they are bounds of my region.
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  3. #3
    Senior Member Spec's Avatar
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    I'm not sure if you're aware of it, but the post explaining the problem is completely unreadable.
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  4. #4
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    thankyou i will try and fix it
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  5. #5
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    ok, have been thinking now an can plz someone confirm, or help with this..

    on xy plane x^2+y^2=1 is semicircle with radius 1. but is constriced to a quarter circle because x [0,1] and y [0,1], on the yz plane, the line z=2y... putting them together gives an upside down quarter cone.... i think, but not sure because wouldnt that imply that z=2x as well?
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  6. #6
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    Quote Originally Posted by fredrick08 View Post
    ok, have been thinking now an can plz someone confirm, or help with this..

    on xy plane x^2+y^2=1 is semicircle with radius 1. but is constriced to a quarter circle because x [0,1] and y [0,1], on the yz plane, the line z=2y... putting them together gives an upside down quarter cone.... i think, but not sure because wouldnt that imply that z=2x as well?
    I can see nowhere that you said that x was restricted to [0,1]. Knowing that [tex]x^2+ y^2= 1[/itex] and that [itex]y\ge 0[/itex] only tells you that [itex]-1\le x\le 1[/itex]. Also z= 2y is not a line it is a z plane. And, no, z= 2y does not mean z= 2x because x in not, in general, equal to y. The solid has a half circle base with the plane z= 2y a top, extending from the x-axis up to the point (0, 1, 2).

    This is not, however, a bounded region because you have not specified a "bottom". Are you also requiring, say, z\ge 0?
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  7. #7
    MHF Contributor Calculus26's Avatar
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    It's a wedge See attachment
    Attached Files Attached Files
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  8. #8
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    ok now i see, what you mean, sorry i orignally had the right answer.. but keep on doubting myself... because i couldnt picture the graph properly.. thanks heaps, very much appreciated = )
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