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Vector Calc: Find the volume [using triple integrals]

  • Thread starter DOH-WLcat
  • Start date
5
0
1. Find the volume, using triple integrals, of the region in the first octant beneath the plane 2x+3y+2z = 6



2. http://tutorial.math.lamar.edu/Classes/CalcIII/TripleIntegrals.aspx



SOLUTION:
1. Assume X and Y are 0. Solve for Z: 2(0)+3(0)+2z=6 => z=3 (0,0,3)
2. Assume X and Z are 0. Solve for Y: 2(0)+3(y)+2(0)=6 => y=2 (0,2,0)
3. Assume Y and Z are 0. Solve for X: 2(x)+3(0)+2(0)=6 => x=3 (3,0,0)
4. The plane in question passes through three points: (0,0,3), (0,2,0), (3,0,0). Find two vectors parallel to this plane.
a. V = (3,0,0) - (0,2,0) = (3,-2,0)
b. W = (3,0,0) - (0,0,3) = (3,0,-3)
5. Compute a Normal vector N of this plane with the two vectors obtained in the previous two steps.
a. N = v*w = (3,-2,0) [(3,0,-3)]
b. i.=6, j.= - 9 and k.=6
c. (6x-9y+6z) = ((x-3)i+yj+zk) => 6x-9y+6z=18 => Solve for Z: z= -x+3y/2+3
d. Solve for y => (using the slope equation: (0-2)/(3-0)=> -2x/3+2 =y
e. Solve for X => We already know that it is 3.
f. Plug in x, y, and z information into the formula listed above.
e. My answer is 9, but the [final exam review for this problem is 3]

Please help. I am taking my final this week to finish my summer school before school starts next week.

Thank you all for your help.

DOH-WLcat.
 
129
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The method outlined in Paul's math notes is nothing like that (unless this whole normal vector thing is some genius obscure way of doing these.) The first thing you should do is isolate [itex]z[/itex] to get:

[tex]z=(1/2)(6 - 2x - 3y)[/tex]

Since we only want the volume in the first octant, the lower limit is the [itex]xy[/itex] plane, which is when [itex]z=0[/itex]. Thus the first integral should have limits with [itex]0 \leq z \leq (1/2)(6 - 2x - 3y)[/itex]. From there, set [itex]z=0[/itex] and describe the region in the [itex]xy[/itex] plane for the other two integrals.
 

SammyS

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Look at Example #2 in the http://tutorial.math.lamar.edu/Classes/CalcIII/TripleIntegrals.aspx" [Broken] you provided.

By the way: The vector, 2i + 3j + 2k, is normal to the plane, 2x+3y+2z = 6 .
 
Last edited by a moderator:
5
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Look at Example #2 in the http://tutorial.math.lamar.edu/Classes/CalcIII/TripleIntegrals.aspx" [Broken] you provided.

By the way: The vector, 2i + 3j + 2k, is normal to the plane, 2x+3y+2z = 6 .
I got 9 [same as my old answer]
 
Last edited by a moderator:
5
0
The method outlined in Paul's math notes is nothing like that (unless this whole normal vector thing is some genius obscure way of doing these.) The first thing you should do is isolate [itex]z[/itex] to get:

[tex]z=(1/2)(6 - 2x - 3y)[/tex]

Since we only want the volume in the first octant, the lower limit is the [itex]xy[/itex] plane, which is when [itex]z=0[/itex]. Thus the first integral should have limits with [itex]0 \leq z \leq (1/2)(6 - 2x - 3y)[/itex]. From there, set [itex]z=0[/itex] and describe the region in the [itex]xy[/itex] plane for the other two integrals.
This was my first step. I tried to upload a problem that is similar to this problem, but my post got deleted.
 
129
0
This was my first step. I tried to upload a problem that is similar to this problem, but my post got deleted.
I don't make the rules or anything around here; I'm just some guy. But now you should try sketching the region in the [itex]xy[/itex] plane. What does it look like? How can you describe it in terms of [itex]x[/itex] and [itex]y[/itex]?
 
Last edited:
5
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[SOLVED] Vector Calc: Find the volume [using triple integrals]

I got it right. It was 3. I made a mistake at N = v*w = (3,-2,0) [(3,0,-3)]. The answer for this would be i.=6, j.= 9 and k.=6. Not i.=6, j.= - 9 and k.=6. I got screwed up by a negative sign. :(
BTW, my method was right. Felt so dumb :P
 

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