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Triple Integrals

Date: 12/07/97 at 14:31:28
From: Alicia Sussman
Subject: Triple integrals

Hi again. Thank you for answering my last question, it was extremely
helpful. I have another one:

Evaluate /  /  /
        |  |  |
        |  |  |  y dx dy dz, where T is the solid in the first octant
        /  /  /
          T      bounded by y=1, y=x, z=x+1 and the coordinate planes.

I think that the constraints would be  1<y<x, and x+1<z<y+1 but I 
can't figure out for x. Am I on the right track? I know the answer to 
the problem is 11/24. Please let me know. 

Thanks again!

Date: 12/08/97 at 06:51:12
From: Doctor Pete
Subject: Re: Triple integrals


To determine the limits, try drawing a picture first. Say you are 
looking down on the xy-plane, so that the first octant is in the 
upper-right corner (just like in the two-dimensional case).  Then the 
region of integration looks like a triangle, with vertices (0,0), 
(0,1), and (1,1). You can't see the constraint z = x+1 from this 
angle, unfortunately. But what we do see from this view is that 
0 < x < 1, and x < y < 1. 

Now, to understand how I got this from seeing just a triangle, let's 
suppose we pick a point, say p, along the x-axis, between x = 0 and 
x = 1. Draw a vertical line parallel to the y-axis, which is the line 
x = p. Then it will intersect the edges of the triangle at two points, 
namely (p,1) and (p,p). Therefore, the y-coordinates of these two 
points are your limits on y; hence 0 < x < 1 and x < y < 1.

As for z, we see that it is bounded below by z = 0 (the xy-plane), and 
above by another plane, z = x+1. So this limit is simply 0 < z < x+1.  
If you picked any point (p,q,0) in the little triangle we talked 
about, then the line parallel to the z-axis from that point is going 
to intersect the xy-plane at z = 0, and the plane z = x+1 at p+1.  
Therefore, your integral is

     /x=1 /y=1 /z=x+1
     |    |    |      y dz dy dx.
     /x=0 /y=x /z=0

Notice I have rearranged the integrals, because both the integrals 
over y and z are dependent on the value of x. (Does it matter if I put 
y first or z first?  Why?)

When you try to determine limits of a triple integral, it is often 
very helpful to draw pictures of what the region looks like when you 
see it from different angles. Since not many people can visualize 3-D 
objects in their heads, it's much easier to look at 2-D projections on 
paper. Notice that I also drew lines intersecting the region T to see 
why the limits were the way they were.

-Doctor Pete,  The Math Forum
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Associated Topics:
High School Calculus

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