What is the range for x in a triple integral for a wedge in the first octant?

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SUMMARY

The discussion focuses on determining the range for the variable x in a triple integral representing the volume of a wedge in the first octant, bounded by the cylinder defined by y² + z² ≤ 1 and the planes y = x, x = 0, and z = 0. The ranges for z and y are established as 0 ≤ z ≤ √(1 - y²) and 0 ≤ y ≤ 1, respectively. The challenge lies in defining the range for x, which could vary based on the integration setup, with suggestions that cylindrical coordinates may simplify the process.

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naspek
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Write a triple integral to represent the volume of the solid

The wedge in the first octant and from the cylinder y^2 + z^2 <= 1 by the planes
y=x, x=0, z=0

First..
i find the range for z..; 0 <=z<= sqrt(1- y^2)

then...
i find the range for y..; let z =0
0<=y<=1

next, if i let z = y = 0

how am i going to find the range for x?
 
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have you drawn a picture? it depends where you put it in the integration...

it could be 0 to y for example...

this may be easier in cylindrical co-ords, though i haven't tried it
 

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