---
title: "Volume under z=x²+y²+2 inside cylinder"
subject: "Mathematics"
chapter: 2
key: "semester_2/mathematics:volume-under-z-x2-y2"
source: https://www.bctengineer.com/q/semester_2/mathematics/volume-under-z-x2-y2
site: BCT Engineering
---

# Volume under z=x²+y²+2 inside cylinder

Mathematics · Chapter 2: Multiple Integrals

- Asked in: 2068 Bhadra, 5 marks
- Video: [Volume bounded by Cylinder and Planes | Volume by Double Intgeral | Higher Engineering Mathematics](https://www.youtube.com/watch?v=KihpcvdN7aE) — Spectrum of Mathematics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/volume-under-z-x2-y2
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-2

## Question

Find by double integration, the volume bounded by the plane z = 0, surface z = x²+y²+2 and the cylinder x²+y² = 4.

## Short answer

Use polar coordinates: V = ∫₀²π ∫₀² (r²+2) r dr dθ = 2π [r⁴/4 + r²]₀² = 16π cubic units.

## Answer

**Setup.** The solid lies between the plane z = 0 and the surface z = x² + y² + 2, inside the cylinder x² + y² = 4. Its base in the xy-plane is the disc x² + y² ≤ 4.

V = ∬_R z dx dy = ∬_R (x² + y² + 2) dx dy

#### Polar coordinates
x = r cos θ, y = r sin θ, dx dy = r dr dθ, with 0 ≤ r ≤ 2 and 0 ≤ θ ≤ 2π.

```text
V = ∫₀^{2π} ∫₀² (r² + 2) r dr dθ
  = ∫₀^{2π} dθ · ∫₀² (r³ + 2r) dr
  = 2π [ r⁴/4 + r² ]₀²
  = 2π (4 + 4)
  = 16π
```

**Answer: V = 16π cubic units.**

## Check yourself

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

1. How is the volume set up as a double integral?
   A. V = double integral of (x^2 + y^2 + 2) dx dy over the disc x^2 + y^2 <= 2
   B. V = double integral of (x^2 + y^2 + 2) dx dy over the disc x^2 + y^2 <= 4
   C. V = double integral of (x^2 + y^2) dx dy over the disc x^2 + y^2 <= 4
   D. V = double integral of 4 dx dy over the disc x^2 + y^2 <= 4
   Answer: B. The height is z = x^2 + y^2 + 2 above the base disc cut out by the cylinder, radius 2. (Key point: Setting up the integrand)

2. Which polar limits and area element are used?
   A. 0 <= r <= 2, 0 <= theta <= 2 pi, dx dy = r dr d theta
   B. 0 <= r <= 4, 0 <= theta <= 2 pi, dx dy = r dr d theta
   C. 0 <= r <= 2, 0 <= theta <= pi, dx dy = r dr d theta
   D. 0 <= r <= 2, 0 <= theta <= 2 pi, dx dy = dr d theta
   Answer: A. The disc has radius 2 and is covered by a full turn, and the Jacobian is r. (Key point: Polar transformation)

3. What is the integral of (r^3 + 2r) dr from 0 to 2?
   A. 4
   B. 12
   C. 8
   D. 16
   Answer: C. [r^4/4 + r^2] from 0 to 2 = 4 + 4 = 8. (Key point: Radial integral)

4. What is the volume?
   A. 8 pi
   B. 16 pi
   C. 4 pi
   D. 32 pi
   Answer: B. V = 2 pi x 8 = 16 pi cubic units. (Key point: Final volume)

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