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how would i set this up?
by getting the right picture :) still trying to hash it out tho
z=0 is a x-y plane x=z is a plane passing through y-axis and equal distance from both x and z axis y=z goes the similar above
|dw:1332612327328:dw| and x=y
how'd you get the 1/2 for y?
the intersection in the xy plane of z=x+y and x=y is what i think i see there
okay so 2y=z so y=1/2?
okay I'm good so far!
there is a plane that needs to be defined as our "cap" in order to integrate these limits under
i believe it is created by the points: (0,1,0) , (1,0,1), and (1/2,1/2,0)
(0 ,1 ,0) , (1 ,0, 1), and (1/2,1/2,0) -.5-.5-0 -.5-.5-0 -.5-.5-0 ----------------------------------- <-.5,.5,0> <.5,-,5,1> <0,0,0> <1,-1,0> cross <1,-1,1> would get us a normal vector to the plane
this seems wayyy too complicated
it's supposed to be a medium difficulty problem....
ohhhhhhhh it's x+y=2 not z
x 1 1 x= -1 y -1 -1 -y= -1+1 z 0 1 z= 1 <-1,0,1> -(x-1)+(z-1) = 0 x+z = 0 :)
well, it was good practice anyhoos :)
my y spose to = 1 in the cross
okay, so now y is from 0 to 2-x, and z is 0 to x
but how would i do x?
if you can, type up the good information up in a new question so I can get a fresh look at it :)
bounded by plane z=x, y=x, x+y=2, z=0
i figured out all the bounds now, but what do i integrate?
by determining the plane that caps it all; say h(x,y,z) then\[\int_z\int_y\int_x h(x,y,z)\ dx.dy.dz\]
the xyz need not be in that order, but as long as your intervals match up to the appropriate d your good
is it x?
z=x, y=x, x+y=2, z=0|dw:1332613613393:dw|
|dw:1332613924696:dw| the points (0,0,0), (2,0,2), (0,2,0) are in the cap plane if I see it right
h(x,y,x) = -x+z
as long as your bonds are good, that should be easy enough to peel out