Write a triple integral, including limits of integration, that gives the volume between 3x +2y + z = 3 and 5x + 3y + z = 3 and above x+y≤ 1, x ≥ 0, y ≥ 0. volume = where a = C = e = b = d = and f = d d d

Calculus: Early Transcendentals
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Chapter1: Functions And Models
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**Triple Integral for Volume Calculation**

*Instructions*: Write a triple integral, including limits of integration, that gives the volume between the planes \(3x + 2y + z = 3\) and \(5x + 3y + z = 3\), and is above the region defined by \(x + y \leq 1\), \(x \geq 0\), \(y \geq 0\).

### Volume Calculation

\[
\text{volume} = \int_{a}^{b} \int_{c}^{d} \int_{e}^{f} \, dz \, dy \, dx
\]

### Limits of Integration

- \(a =\)  
- \(b =\)  
- \(c =\)  
- \(d =\)  
- \(e =\)  
- \(f =\) 

These limits define the region where the triple integral will be evaluated to find the volume between the given planes. The conditions \(x + y \leq 1\), \(x \geq 0\), and \(y \geq 0\) describe a triangular region in the xy-plane which bounds the area for integration. The functions \(3x + 2y + z = 3\) and \(5x + 3y + z = 3\) represent the planes confining the volume in the z-direction.
Transcribed Image Text:**Triple Integral for Volume Calculation** *Instructions*: Write a triple integral, including limits of integration, that gives the volume between the planes \(3x + 2y + z = 3\) and \(5x + 3y + z = 3\), and is above the region defined by \(x + y \leq 1\), \(x \geq 0\), \(y \geq 0\). ### Volume Calculation \[ \text{volume} = \int_{a}^{b} \int_{c}^{d} \int_{e}^{f} \, dz \, dy \, dx \] ### Limits of Integration - \(a =\) - \(b =\) - \(c =\) - \(d =\) - \(e =\) - \(f =\) These limits define the region where the triple integral will be evaluated to find the volume between the given planes. The conditions \(x + y \leq 1\), \(x \geq 0\), and \(y \geq 0\) describe a triangular region in the xy-plane which bounds the area for integration. The functions \(3x + 2y + z = 3\) and \(5x + 3y + z = 3\) represent the planes confining the volume in the z-direction.
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