Evaluate this triple integral for calculating volume

[Granger]

I'm tring to find this volume using a triple integral in the form $dy$ $dx$ $dz$

However I think I'm evaluating the wrong integral because the result is 1 when the volume should be 1/6... Can someone help me find out what I'm doing wrong?

The set is

$$V=\{(x,y,z)\in \mathbb{R^3}: x+y+2z \leq 1; x+y-2z \leq 1 ; x \geq 0; y \geq 0\}$$

and my integral is:

$$\int_{-1/2}^{0} (\int_{0}^{1+2z} (\int_{1+2z-x}^{1+2z} 1 dy + \int_{1+2z}^{1-2z-x} 1 dy) dx + \int_{1+2z}^{1-2z} (\int_{0}^{1-2z-x} 1 dy) dx) dz + \int_{0}^{1/2} (\int_{0}^{1-2z} (\int_{1-2z-x}^{1-2z} 1 dy + \int_{1-2z}^{1+2z-x} 1 dy) dx + \int_{1-2z}^{1+2z} (\int_{0}^{1+2z-x} 1 dy) dx) dz$$NOTE: The goal of the question is to use this integral in this specific order of integration. Please do not suggest other orders

 

[jvicens]

of integration or methods of solving the triple integral.

Hello,

First of all, it is great that you are using a triple integral to find the volume of this set. However, it seems like there may be a mistake in your integral. Let's break it down step by step to see where the error might be.

First, let's look at the innermost integral:

$$\int_{1+2z-x}^{1+2z} 1 dy + \int_{1+2z}^{1-2z-x} 1 dy$$

Notice that the upper limit of the first integral is $1+2z$, while the lower limit of the second integral is $1+2z$. This means that these two integrals are actually evaluating the same thing, and you are essentially adding the same value twice. This could be why your result is coming out to be 1 instead of 1/6.

To fix this, you could change the lower limit of the second integral to $1+2z-x$, so that it is evaluating a different part of the function.

Next, let's look at the middle integral:

$$\int_{0}^{1+2z} (\int_{1+2z-x}^{1+2z} 1 dy) dx$$

Here, the limits of the inner integral are the same as the limits of the outer integral. This means that you are essentially integrating the same value over the same interval, which again could be why your result is coming out to be 1 instead of 1/6.

To fix this, you could change the upper limit of the inner integral to $1+2z$ so that it is evaluating a different part of the function.

Finally, let's look at the outermost integral:

$$\int_{-1/2}^{0} (\int_{0}^{1+2z} (\int_{1+2z-x}^{1+2z} 1 dy) dx) dz$$

Here, the limits of the middle integral are the same as the limits of the outer integral. This means that you are again integrating the same value over the same interval, which could be why your result is coming out to be 1 instead of 1/6.

To fix this, you could change the upper limit of the middle integral to $0$ so that it is evaluating a different part of the function.

Overall