use a triple integral to find volume of the tetrahedron enclosed by the coordinate planes and the plane
2x+y+z=22
i did ʃʃʃdzdydx
0<=x<=2
0<=y<=4-2x
0<=z<=22-2x-y
i got the answer to be 72 and thats wrong. i tried to go back to see where i messed up but nothing caught my eye
2007-12-05 15:24:30 · 4 answers · asked by ♡♥EM♡♥ 4 in Science & Mathematics ➔ Mathematics
your limits are incorrect... try these...
0 ≤ x ≤ 11
0 ≤ y ≤ 22-2x
0 ≤ z ≤ 22-2x-2y
they should give the correct value....
§
2007-12-05 15:36:02 · answer #1 · answered by Alam Ko Iyan 7 · 0⤊ 0⤋
Think of the plane 2x+y+z=22 as forming the slanted "roof" of the tetrahedron. Then the region of the xy-plane bounded by the positive x-axis, the positive y-axis, and the intersection of the "roof" plane with the xy-plane is the "floor". The equation of the xy-plane is z=0. So in the xy-plane, this intersection has equation 2x+y+0 = 22.
So y goes from 0 to 22 - 2x, and x goes from 0 to 11.
2007-12-05 15:44:03 · answer #2 · answered by Ron W 7 · 0⤊ 0⤋
try:
the bounds for y are 0 and 22-2x
the bounds for x are 0 and 11
keep z the same
2007-12-05 15:32:59 · answer #3 · answered by S K 2 · 0⤊ 0⤋
Dissect a monkey. The answer is located in his stomach.
2007-12-05 15:28:21 · answer #4 · answered by Anonymous · 0⤊ 1⤋