Vector Calculus, sketching regions in R3


by fredrick08
Tags: calculus, regions, sketching, vector
fredrick08
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#2
May7-09, 01:38 AM
P: 376
or if possible, be able to tell me, how to draw regions in maple? so i can at least picture it, and try to understand how to draw it.
fredrick08
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#3
May7-09, 02:03 AM
P: 376
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.

fredrick08
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#4
May7-09, 03:45 AM
P: 376

Vector Calculus, sketching regions in R3


please anyone?
HallsofIvy
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#5
May7-09, 05:38 AM
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Quote Quote by fredrick08 View Post
1. The problem statement, all variables and given/known data
sketch the following region in R3

V={[tex]\widehat{}r[/tex]:y[tex]\geq[/tex]0,0[tex]\leq[/tex]x2+y2[tex]\leq[/tex]1,0[tex]\leq[/tex]z[tex]\leq[/tex]2y}
You have [itex]y\ge 0[/itex], [itex]0\le x^2+ y^2\le

3. The attempt at a solution
ok I havn't done this a lot, and dont quite understand, i think that when y=0, 0[tex]\leq[/tex]x2[tex]\leq[/tex]1... and 0[tex]\leq[/tex]z[tex]\leq[/tex]2y... but i dont understand how to sketch these, in previous examples, we had z=f(x,y)... but this one is different... could someone please explain how to sketch this... coz im not sure.. x2+y2 is a dome shape that is greater then 0 but less the 1... but i just cant picture/sketch it.. apparently it is meant to look like a wedge shape, can anyone help me plz.

Thanks.
Quote Quote by fredrick08 View Post
please anyone?
Don't expect a response within a few hours. People aren't just sitting at their computers waiting for problems. I like to sleep at night!

You have [itex]y\ge 0[/itex], [itex]0\le x^2+ y^2\le 1[/itex], [itex]z\le 2y[/itex]

first draw an xy-plane. [itex]y\ge 0[/itex] means your graph is in the upper half plane. the 0 in [itex]0\le x^2+ y^2\le 1[/itex] doesn't really tell us anything because [itex]x^2+ y^2[/itex] can't be negative anyway. But [itex]\x^2+ y^2= 1[/itex] is a circle with center at (0,0) and radius 1. Since y must be positive, draw the upper semi-circle. Now draw a yz-plane beside your xy-graph. z= 2y is a line from (0,0) to (2, 1). If you now imagine that z-axis coming directly up from your first graph, with the y-axes aligned, you should see that plane forms a "top" on the parabolic cylinder. Yes, it is basically a "wedge" shape with the x= 0 plane forming one side, the plane top going down to cut the parabola. It is NOT a bounded region because there is no "bottom" unless you left out z= 0, say.
fredrick08
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#6
May7-09, 05:50 AM
P: 376
ok sorry, i have an anxiety problem.. thankyou very much, i will try and draw it and post it up.
fredrick08
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#7
May7-09, 05:55 AM
P: 376
ok, skip on the drawing part, lol i cant draw at all, so to confirm the shape is a .... quarter of a cone, with base radius 1, and height 2?

very well explained, by the way = )
fredrick08
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#8
May7-09, 06:14 AM
P: 376
or maybe not, because z=2y cuts through the origin and ends at (2,1,0) but a cone would have and apex at (0,0,0) instead of a cone, does it look like a pyramid with curved surface... oh i don't Im confused now
fredrick08
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#9
May7-09, 07:27 AM
P: 376
ok, been thinking, i think i got it now its an upside down quarter cone, with z edges of the line z=y... therefore the apex would be at (0,0,0)
fredrick08
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#10
May7-09, 07:32 AM
P: 376
no wait, coz if was a cone that would imply that z=2x aswelll.... omg i dont know, help....


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