Sketch & Evaluate Region of Integration

[korr2221]

http://qaboard.cramster.com/Answer-Board/Image/cramster-equation-20081241616516336400421111250005486.gif

Homework Statement

1. Sketch the region of integration
2. Rewrite the integral in the order: dydxdz
3. Evaluate the integral (using either order)The attempt at a solution

Attached.

alternative link:

http://www.yourfilelink.com/get.php?fid=476767

 

[korr2221]

corrections made about my sketch

alternative:
http://www.yourfilelink.com/get.php?fid=476770

 

[HallsofIvy]

You seem to have draw the base as a rectangle when it should be a triangle but since you have not marked which axis is the x-axis, which the y-axis, and which the z-axis, I can't be sure.
If you intended the axis on which "2" is marked to be the x-axis, then you have drawn a "left handed" coordinate system when the standard is "right handed"- curling the fingers of your right hand from the x-axis to the y-axis, your thumb should point in the direction of the positive z-axis.

Since the "outermost" integral is from x= 0 to x= 2, mark vertical lines at x= 0 and x= 2 on the xy-plane.

Since the second integral is from y= 0 to y= x, draw the line y= x on the xy-plane. That gives a triangle with vertices at (0,0), (2,0), and (2,2).

Since "innermost" integral is from z= 0 to z= 2- x, that triangle is the base and the top of the region is the plane z= 2-x. That slopes downward from x=0, z= 2 on the left to x=2, z= 0 on the right.

 

[korr2221]

You seem to have draw the base as a rectangle when it should be a triangle but since you have not marked which axis is the x-axis, which the y-axis, and which the z-axis, I can't be sure.
If you intended the axis on which "2" is marked to be the x-axis, then you have drawn a "left handed" coordinate system when the standard is "right handed"- curling the fingers of your right hand from the x-axis to the y-axis, your thumb should point in the direction of the positive z-axis.

Since the "outermost" integral is from x= 0 to x= 2, mark vertical lines at x= 0 and x= 2 on the xy-plane.

Since the second integral is from y= 0 to y= x, draw the line y= x on the xy-plane. That gives a triangle with vertices at (0,0), (2,0), and (2,2).

Since "innermost" integral is from z= 0 to z= 2- x, that triangle is the base and the top of the region is the plane z= 2-x. That slopes downward from x=0, z= 2 on the left to x=2, z= 0 on the right.

uh, i believe you aren't looking at the corrections, the first one, in the first page is something else... sorry