---
title: "Volume of astroid about x-axis"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:volume-astroid-about-x-axis"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/volume-astroid-about-x-axis
site: BCT Engineering
---

# Volume of astroid about x-axis

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2073 Shrawan
- Video: [Volume of the solid generated by revolving astroid 2022PU Mathmatics-I BE Civil first sem new course](https://www.youtube.com/watch?v=4tAENnpPYvo) — Civil Engineering Solution Channel
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/volume-astroid-about-x-axis
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Find the volume generated by revolution of astroid x^(2/3) + y^(2/3) = a^(2/3) about x-axis.

## Short answer

Parametrize with x=a cos³t, y=a sin³t. The volume becomes 6πa³∫₀^{π/2} sin⁷t cos²t dt, giving 32πa³/105 cubic units.

## Answer

Astroid: x^(2/3) + y^(2/3) = a^(2/3). The solid is symmetrical about the y-axis, so V = 2π ∫₀^a y² dx.

Use x = a cos³t, y = a sin³t, dx = −3a cos²t sin t dt. As x goes 0 → a, t goes π/2 → 0.

```text
V = 2π ∫_(π/2)^0 a² sin⁶t (−3a cos²t sin t) dt
  = 6π a³ ∫₀^(π/2) sin⁷t cos²t dt
```
Beta function: ∫₀^(π/2) sin⁷t cos²t dt = Γ(4) Γ(3/2) / (2Γ(11/2)).

Γ(4) = 6; Γ(3/2) = √π/2; Γ(11/2) = (9/2)(7/2)(5/2)(3/2)(1/2)√π = 945√π/32.

= 6 · (√π/2) / (2 · 945√π/32) = 3 · 32/(2 · 945) = 48/945 = 16/315.

V = 6πa³ · 16/315 = 96πa³/315

**V = 32πa³/105 cubic units.**

## Check yourself

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

1. Volume about the x-axis for the astroid is
   A. π integral y dx from 0 to a
   B. 2π integral y^2 dx from 0 to a
   C. 2π integral x y dx
   D. 4π integral y^2 dx
   Answer: B. Symmetric about y-axis, so double the half. (Key point: Formula)

2. The integral reduces to
   A. 6π a^3 integral sin^6 t cos^2 t dt
   B. 3π a^3 integral sin^7 t cos^2 t dt
   C. 6π a^3 integral sin^7 t cos^2 t dt
   D. 6π a^3 integral sin^2 t cos^7 t dt
   Answer: C. y^2 dx = a^2 sin^6 t · 3a cos^2 t sin t dt. (Key point: Reduction)

3. The integral of sin^7 t cos^2 t equals
   A. 8/315
   B. 32/105
   C. 16/105
   D. 16/315
   Answer: D. Γ(4)Γ(3/2)/(2Γ(11/2)) = 16/315. (Key point: Beta value)

4. The volume is
   A. 32πa^3/105
   B. 16πa^3/105
   C. 32πa^3/315
   D. 96πa^3/105
   Answer: A. 6π a^3 · 16/315 = 32πa^3/105. (Key point: Result)

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