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

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Homework Statement



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

Homework Equations





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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  • #2
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Homework Statement



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

Homework Equations





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.
 
  • #3
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Coodgee, if you want an easier integration domain, try a change of coordinates.
 
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LCKurtz
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To add to Mark44's comments: The xy domain is the unit square. The problem is that you have two different upper surfaces. Note that x+y+z=2 passes through (1,0,1),(1,1,0), and (0,1,1) which are 3 corners of the cube. Draw that triangle since it the only portion of that plane that is relevant. If you look at the shadow of that triangle on the xy plane you will see a triangular domain, which is half the unit square, for which the upper surface is the slanted plane. On the other half of the unit square, the top of the cube itself is the upper surface. You have to set up two integrals.

[Edit, added later] If you want to do it as a single triple integral, consider integrating in the x direction first. Then the "front" surface is the plane x+y+z=2 and the domain in the yz plane is a triangle.
 
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