---
title: "Volume integral over a region F=2xzi−xj+y²k"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:volume-integral-over-a-region-f"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/volume-integral-over-a-region-f
site: BCT Engineering
---

# Volume integral over a region F=2xzi−xj+y²k

Mathematics III · Chapter 2: Line, Surface and Volume Integrals

- Asked in: 2069 Chaitra, 5 marks
- Video: [Problems on Volume Integral bonded by the surface / vector integral calculus](https://www.youtube.com/watch?v=0eBuLVWrCqI) — TEAM Education
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/volume-integral-over-a-region-f
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Evaluate the volume integral ∭_V F dv, where V is the region bounded by the surfaces x = 0, y = 0, y = 6, z = x², z = 4 and F = 2xz i − x j + y² k.

## Short answer

Integrate componentwise over 0≤x≤2, 0≤y≤6, x²≤z≤4. ∭ 2xz dv = 128, ∭ x dv = 24, ∭ y² dv = 384. Result: 128 i − 24 j + 384 k.

## Answer

The integral of a vector function over a volume is taken component by component: ∭_V F dv = i ∭ 2xz dv − j ∭ x dv + k ∭ y² dv.

#### Limits
The region is bounded by x = 0, y = 0, y = 6, z = x² and z = 4. The parabolic cylinder z = x² meets z = 4 at x = 2. So

0 ≤ x ≤ 2, 0 ≤ y ≤ 6, x² ≤ z ≤ 4.

#### Components
```text
∫_{x²}⁴ 2xz dz = x (16 − x⁴)
∫₀² x(16 − x⁴) dx = [8x² − x⁶/6]₀² = 32 − 64/6 = 64/3
∭ 2xz dv = 6 × 64/3 = 128

∭ x dv = 6 ∫₀² x (4 − x²) dx = 6 [2x² − x⁴/4]₀² = 6 (8 − 4) = 24

∭ y² dv = (∫₀⁶ y² dy) ∫₀² (4 − x²) dx = 72 × (8 − 8/3) = 72 × 16/3 = 384
```

**Answer: ∭_V F dv = 128 i − 24 j + 384 k.**

## Check yourself

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

1. Vector volume integral computed?
   A. Component by component
   B. By dotting
   C. By curl
   D. By divergence
   Answer: A. Each component separately. (Key point: Method)

2. Limits?
   A. 0<=x<=4
   B. 0<=x<=2, 0<=y<=6, x^2<=z<=4
   C. 0<=z<=x^2
   D. 0<=y<=2
   Answer: B. z=x^2 meets z=4 at x=2. (Key point: Limits)

3. Integral of 2xz?
   A. 64/3
   B. 24
   C. 128
   D. 384
   Answer: C. 6 x 64/3. (Key point: x component integral)

4. Answer?
   A. 128i + 24j + 384k
   B. 24i - 128j
   C. 128i - 24j
   D. 128i - 24j + 384k
   Answer: D. j has minus. (Key point: Result)

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