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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.

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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