Triple IntegralsDate: 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 Hi, 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 Check out our web site! http://mathforum.org/dr.math/ |
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