Vector Calc: Find the volume [using triple integrals]

In summary, the conversation discusses finding the volume using triple integrals for a region in the first octant beneath the plane 2x+3y+2z=6. The solution involves finding two vectors parallel to the plane and using them to compute a normal vector. The process also includes setting limits and describing the region in the xy plane for the integrals. The final answer is 3, but there was a mistake in the calculation of the normal vector.
  • #1
DOH-WLcat
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.
 
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  • #2
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.
 
  • #3
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 .
 
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  • #4
SammyS said:
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
Screwdriver said:
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.
 
  • #6
DOH-WLcat said:
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:
  • #7
[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
 

1. What is vector calculus?

Vector calculus is a branch of mathematics that deals with the study of vector fields and their derivatives in multiple dimensions. It is used to analyze and describe physical phenomena such as fluid flow, electromagnetism, and motion in three-dimensional space.

2. What is a triple integral?

A triple integral is a mathematical tool used to find the volume of a three-dimensional object. It involves integrating a function over a three-dimensional region in space, and the result is a single value representing the volume of the object.

3. How is a triple integral calculated?

A triple integral is calculated by first determining the limits of integration for each variable (x, y, and z) and then setting up the integral with the function to be integrated. The integral is then evaluated using techniques such as substitution or integration by parts.

4. When is triple integration used?

Triple integration is used in various fields of science and engineering, such as physics, chemistry, and computer graphics. It is used to find the volume of three-dimensional objects, calculate mass and center of mass, and solve problems involving fluid flow and electric fields.

5. What are some practical applications of triple integration?

Triple integration has many practical applications, such as calculating the volume of irregularly shaped objects, determining the mass and density of an object, finding the force of gravity between two masses, and analyzing the flow of fluids in pipes and channels. It is also used in computer graphics to create three-dimensional images and animations.

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