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

In summary: 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?
  • #1
agnimusayoti
240
23
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.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
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:
  • #3
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##.
 
  • #4
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?
 
  • #5
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##.
 
  • #6
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
 
  • #7
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.
 
  • #8
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.
 
  • Like
Likes SammyS

1. What is the formula for finding the volume in the first octant bounded by the coordinate planes and x + 2y + z = 4?

The formula for finding the volume in this scenario is simply the area of the base (x + 2y + z = 4) multiplied by the height (which is the distance from the origin to the plane).

2. How do I determine the boundaries for the first octant in this scenario?

The first octant is defined as the region where all three coordinates (x, y, and z) are positive. In this case, the boundaries would be x = 0, y = 0, and z = 0.

3. Can the volume in the first octant bounded by the coordinate planes and x + 2y + z = 4 be negative?

No, the volume cannot be negative as it represents a physical space and cannot have a negative value.

4. How can I visualize the volume in the first octant bounded by the coordinate planes and x + 2y + z = 4?

You can visualize the volume by graphing the equation x + 2y + z = 4 and identifying the points where it intersects with the coordinate planes (x = 0, y = 0, and z = 0). These points will form a triangle in the first octant, with the origin as one of the vertices. The height of this triangle will determine the volume in the first octant.

5. Are there any real-life applications of finding the volume in the first octant bounded by the coordinate planes and x + 2y + z = 4?

Yes, this scenario can be applied in physics and engineering to calculate the volume of a solid object or container that is bounded by three planes. It can also be used in geometry to find the volume of a triangular prism.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
455
  • Calculus and Beyond Homework Help
Replies
15
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
549
  • Calculus and Beyond Homework Help
Replies
6
Views
861
  • Calculus and Beyond Homework Help
Replies
5
Views
298
  • Calculus and Beyond Homework Help
Replies
2
Views
518
  • Calculus and Beyond Homework Help
Replies
14
Views
670
  • Calculus and Beyond Homework Help
Replies
19
Views
970
  • Calculus and Beyond Homework Help
Replies
2
Views
397
  • Calculus and Beyond Homework Help
Replies
3
Views
286
Back
Top