---
title: "Transform to polar and evaluate ∫∫(x²+y²), 2ax−x²"
subject: "Mathematics"
chapter: 2
key: "semester_2/mathematics:transform-polar-evaluate-x2-y2"
source: https://www.bctengineer.com/q/semester_2/mathematics/transform-polar-evaluate-x2-y2
site: BCT Engineering
---

# Transform to polar and evaluate ∫∫(x²+y²), 2ax−x²

Mathematics · Chapter 2: Multiple Integrals

- Asked in: 2067 Mangsir, 5 marks
- Video: [Change into Polar Co-ordinate ll Multiple Integral ll Double Integral In Hindi](https://www.youtube.com/watch?v=jWVxo3embLk) — MATHS AKHADA
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/transform-polar-evaluate-x2-y2
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-2

## Question

Transform to polar coordinates and complete the integral ∫₀^(2a) ∫₀^√(2ax−x²) (x²+y²) dy dx.

## Short answer

Region is upper half of (x−a)²+y²=a². In polars, r ≤ 2a cosθ, 0 ≤ θ ≤ π/2. Integrate r²·r dr dθ → 4a⁴ ∫₀^{π/2} cos⁴θ dθ = 3π a⁴/4.

## Answer

I = ∫₀^{2a} ∫₀^{√(2ax−x²)} (x² + y²) dy dx

#### Region
y ranges from 0 to √(2ax − x²), i.e. the upper half of the circle y² = 2ax − x², or (x − a)² + y² = a². This circle has centre (a, 0), radius a, and passes through the origin. x ranges from 0 to 2a. So the region is the upper half-disc.

In polar form, x² + y² = 2ax becomes r² = 2a r cos θ, i.e. r = 2a cos θ. The half-disc above the x-axis corresponds to 0 ≤ θ ≤ π/2 and 0 ≤ r ≤ 2a cos θ.

```text
y
a |    . - .
  |  /       \
  | /         \
  |/___________\__ x
  O     a      2a
```

#### Integral
x² + y² = r², dy dx = r dr dθ:

```text
I = ∫₀^{π/2} ∫₀^{2a cosθ} r² · r dr dθ
  = ∫₀^{π/2} (2a cosθ)⁴/4 dθ
  = 4a⁴ ∫₀^{π/2} cos⁴θ dθ
```

By Wallis' formula, ∫₀^{π/2} cos⁴θ dθ = (3/4)(1/2)(π/2) = 3π/16.

```text
I = 4a⁴ · 3π/16 = 3π a⁴/4
```

**Answer: 3π a⁴/4.**

## Check yourself

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

1. What region does y from 0 to sqrt(2ax - x^2), x from 0 to 2a describe?
   A. The full disc with centre (a, 0) and radius a
   B. The upper half of the disc with centre (0, a) and radius a
   C. The upper half of the disc with centre (a, 0) and radius a
   D. The upper half of the disc with centre (0, 0) and radius 2a
   Answer: C. y^2 = 2ax - x^2 gives (x - a)^2 + y^2 = a^2, and y >= 0 selects the upper half. (Key point: Identifying the region)

2. What are the polar limits for this region?
   A. 0 <= theta <= pi/2 and 0 <= r <= 2a cos(theta)
   B. 0 <= theta <= pi and 0 <= r <= 2a cos(theta)
   C. 0 <= theta <= pi/2 and 0 <= r <= 2a
   D. 0 <= theta <= pi/2 and 0 <= r <= 2a sin(theta)
   Answer: A. The circle becomes r = 2a cos(theta), and the upper half-disc has theta from 0 to pi/2. (Key point: Polar limits)

3. After transformation, what does the r-integral give?
   A. (2a cos theta)^3 / 3
   B. (2a cos theta)^4 / 4 = 4a^4 cos^4 theta
   C. (2a cos theta)^2 / 2
   D. (2a cos theta)^4
   Answer: B. The integrand r^2 times the Jacobian r is r^3, whose integral is r^4/4. (Key point: Integrand after transformation)

4. Using Wallis' formula, what is the value of the integral?
   A. 3 pi a^4 / 4
   B. pi a^4 / 4
   C. 3 pi a^4 / 16
   D. 3 pi a^4 / 2
   Answer: A. The integral of cos^4 is 3 pi/16, and 4a^4 x 3 pi/16 = 3 pi a^4/4. (Key point: Final value)

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