Visualizing a Triple Integral Bounded by Planes

[arl146]

Homework Statement

the function is xyz²

V is bounded by y=1-x, z=0, and z=y.

The Attempt at a Solution

the limits are:

x is from -1 to 1 ?
y is from 0 to (1-x^2) ?
z is from 0 to y ?

the question asks for a picture ... how should that look? there are points on (1,0,0), (-1,0,0), (0,0,1) and (0,1,0) ?

 

[Quinzio]

You need one more bound.
This is likely x=c.
c is come constant

 

[arl146]

i don't get that, i mean like how that helps

 

[Quinzio]

What you are likely to get is a tetrahedron, which requires 4 faces.
The 4 faces are your 4 bounds, which are planes.
But you have wrote 3 bounds, what about the 4th?

 

[HallsofIvy]

Homework Statement

the function is xyz²

V is bounded by y=1-x, z=0, and z=y.

The Attempt at a Solution

the limits are:

x is from -1 to 1 ?
y is from 0 to (1-x^2) ?
z is from 0 to y ?

the question asks for a picture ... how should that look? there are points on (1,0,0), (-1,0,0), (0,0,1) and (0,1,0) ?

Draw a graph. x is from -1 to 1 so draw two vertical lines at x=-1 and x= 1. y if from 0 to 1- x^2 so draw the the line y= 0 and the parabola y= 1- x^2. The region to be integrated is inside that parabola above y= 0. Finally, the plane z= y crosses the y-axis up to (x, 1, 1) and your three dimensional region comes up to that. As Quinzio said, you need another bound on z or that region is not bounded. As it is, z could go up from that plane to infinity of down to negative infinity. Is there another limit, perhaps 0, on the z-integration?

 

[arl146]

Well I just gave what the problem gave. The upperimit for z is y?