Hi. As the problem is stated, the volume is infinite because there is no lower bound onOriginally Posted by bjorn85
The constraints on can be written
But for then for the constraint is
hi, i got some problems with solving a volume calculation, it looks like this:
Find the volume of the space that is created between the planes x=0, y=0; x+y+z=3 and x+2y=2
by the looks of it it seems fairly simple but i can't find a way to solve it, my main problem is to define the boundaries needed for the integral. anyone with some ideas?
Let the master try.
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1) , Take 3 non-colinear point I thing the bests are
And place them on the x-y-z graph and connect them. So you have a triangle. (In red)
2) , since this does not contain you are going to do a regular two dimensional graph and them project is (move it staight) up and down to get the full plane. Take two points, say, (In blue).
3)Note the region, on the bottom, it was supplied by the diagram previously.
4)Now you can see what you are integrating.
Thus, you have
But,
the lower curve,
And,
the upper curve,
Now switch to the region .
It is a type I region thus,
Where, is upper curve which is when in the curve thus, thus,
And is lower cuve which is, which is equivalent to,
Thus, the volume integral thus far is,
Now, we move to find this is the easiest not the region starts from and ends on
Thus, we have (just like before),
After the first integration,
Thus,
Thus,
Thus,
Open and combine,
I think I did this correctly. Not sure why answeres do not match. Maybe, I understood this problem slightly diffrenctly.
If you change the upper bound on to 4, the iterated integral evaluates to 8/3 as it should. The diagram of (3) assumed which implied the upper bound for was 3. This was part of the problem first posted but not of the corrected problem.Originally Posted by ThePerfectHacker
Originally Posted by bjorn85