First, let's see what the graph of the plane looks like. When y=0 and z=0, x=a, SO It is the plane touching (0,0,c) (a,0,0) and (0,b,0).
Now, x/a+y/b=1: This will be a plane also, you have a line on the x-y plane that extends upward on the z-axis. You have the points (a,0,z) and (0,b,z) These points intersect with the plane when z=0, because it gives you (a,0,0) and (0,b,0) from above. This is where you are cutting off the picture: the z=0 plane.
Well, since the graph would extend off to x= negative infinite, and the volume would be infinite, I'll assume that you mean it's bounded by the x, y, and z plane:
SO
that makes it a lot easier.
Take (0,0,c) (a,0,0) and (0,b,0). All you're doing is finding the volume underneath the plane between where it intersects with the z=0 plane and x=0 and y=0 planes. The line tells you to stop before z becomes negative. So:
Draw the area you're integrating over: x=0, y=0, x/a+y/b=1
y=b-bx/a
I(0,a) I(0,b-bx/a) z dy dx= I(0,a) I(0,b-bx/a) c(1-x/a-y/b) dy dx
And you can integrate from that equation : )
2007-12-02 15:23:50
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answer #1
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answered by Anonymous
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This question seems to have some problems.
1) First, planes don't have volume. I assume you mean to find the volume enclosed by the given planes.
2) There is no enclosed volume. Notice that only the first plane has any constraint on z at all. So there would be two infinite spaces, one above the first plane and the other below.
Also, it takes a minimum of four distinct planes to enclose a volume. However, you only have three distinct planes. The second and third planes are coincident.
x = a – ya/b
x + ya/b = a
x/a + y/b = 1 (This is the same as the second plane.)
Please check to see if you have presented the problem correctly here.
2007-12-03 17:16:25
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answer #2
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answered by Northstar 7
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