Volume in the first octant bounded by the coordinate planes and x + 2y + z = 4.

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Homework Help Overview

The discussion revolves around finding the volume in the first octant bounded by the coordinate planes and the plane defined by the equation x + 2y + z = 4. Participants are exploring the setup of the problem and the interpretation of the volume element in the context of triple integrals.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the limits of integration and the definition of the infinitesimal volume element. There is a focus on whether the volume in the first octant should be represented as 1/8 of the total volume and the implications of this interpretation. Some participants suggest re-evaluating the setup of the integrals and the order of integration.

Discussion Status

The conversation is ongoing, with participants providing insights and questioning assumptions about the volume element and the interpretation of the first octant. Some guidance has been offered regarding the correct form of the volume element and the nesting of integrals, but no consensus has been reached on the interpretation of the volume in the first octant.

Contextual Notes

There is a mention of previous chapters discussing the volume in the first octant as being 1/8 of the total volume, which is being questioned in the current context. Participants are also clarifying the definitions of the coordinate planes and their relevance to the problem.

agnimusayoti
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Homework Statement
Find the volume in the first octant bounded by the coordinate planes and x + 2y + z = 4.
Relevant Equations
Multiple integrals.
First, I try to make a sketch and from that I take limit of integration from:
1. ##z_1 = 0## to ##z_2 = 4 - x -2y##
2. ##x_1 = 0## to## x_2 = 4- 2y ##
3. ##y_1 = 0## to ##y_2 = 2##

Then, I define infinitesimal volume element in the first octant as ##dV = 1/8 dz dz dy##.
Therefore,
$$V=1/8 \int_{y_1=0}^{y_2 = 2} \int_{x_1=0}^{x_2=4-2y} \int_{z_1=0}^{z_2 = 4 - x -2y} dz dy dx$$
$$V=1/8 \int_{y_1=0}^{y_2 = 2} \int_{x_1=0}^{x_2=4-2y} \int_{z_1=0}^{z_2 = 4 - x -2y} dz dy dx$$
$$V=1/8 \int_{y_1=0}^{y_2 = 2} \int_{x_1=0}^{x_2=4-2y} (4 - x -2y) dy dx$$
$$V=1/8 \int_{y_1=0}^{y_2 = 2} (8 - 8y +6y^2) dx$$
$$V=1/8 (16) = 2$$.

But, final answer should be 16/3. So, I think I made a mistake when interpreting "the volume in the first octant" as 1/8 dV. Why the denominator is 3? Does the volume in the first octant is 1/3 of total volume? Thanks.
 
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agnimusayoti said:
when interpreting .. 1/8 dV
Your intuition is correct: infinitesimal volume element is simply ##dV = dx\,dy\,dz##. The integral bounds take care of what you tried to insert in ##dV##.

Still some work to check : 8 x 2 ##\ne## 16/3 !"Tip": make a habit of properly nesting the integrals, so $$ \int_{z_1}^{z_2} \Biggl ( \int_{y_1}^{y_2} \left ( \int_{x_1}^{x_2} ... \ dx\right ) \,dy \Biggr ) \,dz $$
 
Last edited:
agnimusayoti said:
Homework Statement:: Find the volume in the first octant bounded by the coordinate planes and x + 2y + z = 4.
Relevant Equations:: Multiple integrals.

First, I try to make a sketch and from that I take limit of integration from:
1. ##z_1 = 0## to ##z_2 = 4 - x -2y##
2. ##x_1 = 0## to## x_2 = 4- 2y ##
3. ##y_1 = 0## to ##y_2 = 2##

So you are going to integrate in the ##z## direction first, the ##x## direction second, and the ##y## direction last. Ok, that means ##dV = dzdxdy## in that order.

Then, I define infinitesimal volume element in the first octant as ##dV = 1/8 dz dz dy##.

Why the ##\frac 1 8##? That shouldn't be there. And it should be ##dV=dzdxdy## (a typo?)

Therefore,
$$V=1/8 \int_{y_1=0}^{y_2 = 2} \int_{x_1=0}^{x_2=4-2y} \int_{z_1=0}^{z_2 = 4 - x -2y} dz dy dx$$
Should be$$
V= \int_{y_1=0}^{y_2 = 2} \int_{x_1=0}^{x_2=4-2y} \int_{z_1=0}^{z_2 = 4 - x -2y} dz dx dy$$

Work it carefully from here. You will get ##\frac {16} 3##.
 
Yes, in my work, I'm integrate from dz, dx, and dy. However, maybe I need more time to get used with Latex Code (for me it's difficult) to write down my work.

Nah, ``1/8 is from details " volume in first octant". In previous chapter, volume in first octant means 1/8 from total volume. So, I think that dV in first octant is 1/8 dV which dV is element volume (total).

So, what does volume in first octant mean?
 
agnimusayoti said:
So, what does volume in first octant mean?
That the bounds of the tertraeder region are the given plane plus the three coordinate planes ##x=0, y=0, z=0##.
 
Yes, I made a mistake in last steps. So the result is 16/3. So, in this case, the total volume bounded by the region is in first octant?

Three coordinates plane ##x=0## is yz plane? And how about the second octant? Please explain it to me. Thankss
 
agnimusayoti said:
plane x=0 is yz plane
Correct.
agnimusayoti said:
And how about the second octant?
Good question. I don't know if there is a particular ordering for these eight octants. But for sure the first octant is the one with all x,y,z ##\ge## 0.
 
agnimusayoti said:
Yes, in my work, I'm integrate from dz, dx, and dy. However, maybe I need more time to get used with Latex Code (for me it's difficult) to write down my work.

Nah, ``1/8 is from details " volume in first octant". In previous chapter, volume in first octant means 1/8 from total volume. So, I think that dV in first octant is 1/8 dV which dV is element volume (total).

So, what does volume in first octant mean?
The slanted plane leans up against the coordinate planes in the first octant. It is the only octant where the plane bounds a finite volume with the coordinate planes. So the 1/8 makes no sense. There is no larger volume that this is a portion of.
 
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