Triple integral problem - domain is that part of a cube btween 2 plane

Homework Statement

Evaluate the triple integral for the function $\int\int\int y dV$ over that part of the cube 0 $\leq$ x,y,z $\leq$ 1 lying above the plane y +z = 1 and below the plane x+y+z = 2

The Attempt at a Solution

This is the first attempt at a triple integral problem. The method I have been taught is to reduce the integral to a double integral problem by first integrating over the region perpendiculr to some base.

When I try to draw this thing, it seems like it's a half a cube, or a triangular prism, with a little bit cut off one corner. But I can't seem to find a base perpendicular to some dimmension that is bound by two simple functions. it seems that part of the object is bound by x =1, y =1, z = 1 and part by one of the other functions.

I am really lost..

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Mark44
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Homework Statement

Evaluate the triple integral for the function $\int\int\int y dV$ over that part of the cube 0 $\leq$ x,y,z $\leq$ 1 lying above the plane y +z = 1 and below the plane x+y+z = 2

The Attempt at a Solution

This is the first attempt at a triple integral problem. The method I have been taught is to reduce the integral to a double integral problem by first integrating over the region perpendiculr to some base.

When I try to draw this thing, it seems like it's a half a cube, or a triangular prism, with a little bit cut off one corner. But I can't seem to find a base perpendicular to some dimmension that is bound by two simple functions. it seems that part of the object is bound by x =1, y =1, z = 1 and part by one of the other functions.

I am really lost..
The two planes that serve as bounds for the region can be written as z = 1 - y and z = 2 - x - y. The difference, 2 - x - y - (1 - y), represents the vertical distance from the lower plane to the upper plane.

Coodgee, if you want an easier integration domain, try a change of coordinates.

Last edited:
LCKurtz