# Vector Calculus, sketching regions in R3

by fredrick08
Tags: calculus, regions, sketching, vector
 P: 376 solved
 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.
 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.
P: 376

## Vector Calculus, sketching regions in R3

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 Quote by fredrick08 1. The problem statement, all variables and given/known data sketch the following region in R3 V={$$\widehat{}r$$:y$$\geq$$0,0$$\leq$$x2+y2$$\leq$$1,0$$\leq$$z$$\leq$$2y} You have $y\ge 0$, $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$$\leq$$x2$$\leq$$1... and 0$$\leq$$z$$\leq$$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 by fredrick08 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$, $0\le x^2+ y^2\le 1$, $z\le 2y$

first draw an xy-plane. $y\ge 0$ means your graph is in the upper half plane. the 0 in $0\le x^2+ y^2\le 1$ doesn't really tell us anything because $x^2+ y^2$ can't be negative anyway. But $\x^2+ y^2= 1$ 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.
 P: 376 ok sorry, i have an anxiety problem.. thankyou very much, i will try and draw it and post it up.
 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 = )
 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
 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)
 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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