---
title: "Triple integral (2x+y) over cylinder region"
subject: "Mathematics"
chapter: 2
key: "semester_2/mathematics:triple-integral-2x-y-over"
source: https://www.bctengineer.com/q/semester_2/mathematics/triple-integral-2x-y-over
site: BCT Engineering
---

# Triple integral (2x+y) over cylinder region

Mathematics · Chapter 2: Multiple Integrals

- Asked in: 2075 Baisakh, 5 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/triple-integral-2x-y-over
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-2

## Question

Evaluate ∭_R (2x+y) dx dy dz where R is the closed region bounded by the cylinder z = 4 − x² and the planes x = 0, y = 0, y = 2, z = 0.

## Short answer

Solid under z=4−x², above z=0, x≥0, 0≤y≤2. Limits: 0≤x≤2, 0≤y≤2, 0≤z≤4−x². Integrate (2x+y) over z,y,x: result 80/3.

## Answer

**Region.** The solid under the cylinder z = 4 − x² (a parabolic cylinder, parallel to the y-axis), above z = 0, between y = 0 and y = 2, and with x ≥ 0. On z = 0, 4 − x² = 0 gives x = 2. So the limits are

0 ≤ x ≤ 2, 0 ≤ y ≤ 2, 0 ≤ z ≤ 4 − x².

```text
I = ∫₀² ∫₀² ∫₀^{4−x²} (2x + y) dz dy dx
```

#### z-integration
∫₀^{4−x²} (2x + y) dz = (2x + y)(4 − x²).

#### y-integration
```text
∫₀² (2x + y) dy = 4x + 2
→ (4x + 2)(4 − x²) = 16x − 4x³ + 8 − 2x²
```

#### x-integration
```text
I = ∫₀² (16x − 4x³ − 2x² + 8) dx
  = [ 8x² − x⁴ − (2/3)x³ + 8x ]₀²
  = 32 − 16 − 16/3 + 16
  = 32 − 16/3
  = 80/3
```

**Answer: 80/3.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. x runs from 0 to:
   A. √2
   B. 1
   C. 4
   D. 2
   Answer: D. 4−x² = 0. (Key point: Limits)

2. Integrating z gives:
   A. 2x+y
   B. (2x+y)x²
   C. (2x+y)(4−x²)
   D. 4−x²
   Answer: C. Integrand independent of z. (Key point: After z)

3. Value:
   A. 16/3
   B. 32
   C. 80/3
   D. 64/3
   Answer: C. 32 − 16/3. (Key point: Result)

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