Question

need help re-expressing the bounds of a triple integral.

Q: consider the solid in the first octant bounded by the planes y=x, z=y and z=4 write the integral as an iterated triple integral with order of integration dxdydz, dzdxdy, and dydxdz

Answer 1

Can you include the picture? I've seen it before but it helps xP

Answer 2

yeah sure can hold on

Answer 3

Well, for dxdydz. x is going to range from 0 to y. Then y is gonna range from z to 4.
Then z is gonna range from 0 to 4. Does that make sense? (If I'm seeing this right)

Answer 4

dxdydz :i got
x: from 0 to y
y: from 0 to z
z: from 0 to 4

Answer 5

But that would give you the wrong region, look at the colored region, y doesn't start at zero always, it ranges. :/

Answer 6

As z because you have y=z

Answer 7

okay that makes sense now that im looking at the picture but how would it change if you were to do dzdxdy ?

Answer 8

y would range from 0 to 4 i think but what about the rest?

Answer 9

Okay. Well z would range from y to 4. Then x would range from from 0 to y then y would be 0 to 4.

Answer 10

Because you have z=y then the max is 4. X goes from x=y, it goes from zero since its bounded by y=0. Then y goes from zero to 4.

Answer 11

The easiest thing to do, is once you get the first bounds, let it go to zero then look at the "shadow" this will give you a triangle (for this problem) from there, it is easier (per se).

Answer 12

ohhhhh okay that helped alot. so for the last one dydxdz z will go from 0 to 4, x will be from 0 to y and y will go from z to 4?

Answer 13

As far as I can tell :)

Answer 14

sweet! thanks for the help this makes so much more sense now

Answer 15

You're welcome :) Any other questions just post them in the feed xP

Answer 16

thanks will do! :)