# Vector Calculus, sketching regions in R3

• May 6th 2009, 11:31 PM
fredrick08
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?
• May 6th 2009, 11:32 PM
fredrick08
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.
• May 7th 2009, 02:58 AM
Spec
I'm not sure if you're aware of it, but the post explaining the problem is completely unreadable.
• May 7th 2009, 03:29 AM
fredrick08
thankyou i will try and fix it
• May 7th 2009, 04:55 AM
fredrick08
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?
• May 7th 2009, 05:15 AM
HallsofIvy
Quote:

Originally Posted by fredrick08
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, \$\displaystyle z\ge 0\$?
• May 7th 2009, 05:21 AM
Calculus26
It's a wedge See attachment
• May 7th 2009, 05:32 AM
fredrick08
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 = )