---
title: "Polar coordinates: ∫∫(x²+y²) dy dx"
subject: "Mathematics"
chapter: 2
key: "semester_2/mathematics:polar-coordinates-x2-y2-dy"
source: https://www.bctengineer.com/q/semester_2/mathematics/polar-coordinates-x2-y2-dy
site: BCT Engineering
---

# Polar coordinates: ∫∫(x²+y²) dy dx

Mathematics · Chapter 2: Multiple Integrals

- Asked in: 2071 Bhadra, 5 marks; 2067 Chaitra, 5 marks (asked 2 times)
- Video: [MA25C01-Applied Calculus | MA3151 |  Change of variables from Cartesian to polar co-ordinates](https://www.youtube.com/watch?v=Jdy-FT4YsCs) — Mathematics Kala
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/polar-coordinates-x2-y2-dy
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-2

## Question

Evaluate the following integral by changing to polar coordinates: ∫₀¹ ∫ₓ^√(2x−x²) (x²+y²) dy dx.

## Short answer

Region between y=x (θ=π/4) and circle r=2cosθ in first quadrant. Polar limits: θ from π/4 to π/2, r from 0 to 2cosθ. ∫∫ r³ dr dθ = 4∫ cos⁴θ dθ = 3π/8 − 1.

## Answer

I = ∫₀¹ ∫ₓ^{√(2x−x²)} (x² + y²) dy dx

#### Region
- Lower limit y = x is a straight line through the origin (θ = π/4).
- Upper limit y = √(2x − x²) gives y² = 2x − x², i.e. (x − 1)² + y² = 1: the upper half of the circle with centre (1, 0), radius 1. In polar form: r² = 2r cos θ, so r = 2 cos θ.
- The line and circle meet at (0,0) and (1,1).
- For 0 ≤ x ≤ 1 the region lies above y = x and inside the circle.

```text
y
1 |      *(1,1)
  |    /' .
  |  /  region  .
  |/ _ _ _ _ _ _ . x
 0                 2
 (line y=x, circle r = 2cosθ)
```

So θ runs from π/4 (the line) to π/2 (the y-axis, where the circle meets the origin), and r runs from 0 to 2 cos θ.

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

```text
I = ∫_{π/4}^{π/2} ∫₀^{2cosθ} r³ dr dθ
  = ∫_{π/4}^{π/2} (2cosθ)⁴/4 dθ
  = 4 ∫_{π/4}^{π/2} cos⁴θ dθ
```

Use cos⁴θ = (3 + 4 cos 2θ + cos 4θ)/8:

```text
I = (1/2) [ 3θ + 2 sin 2θ + (1/4) sin 4θ ]_{π/4}^{π/2}
  = (1/2) [ (3π/2 + 0 + 0) − (3π/4 + 2 + 0) ]
  = (1/2)(3π/4 − 2)
  = 3π/8 − 1
```

**Answer: I = 3π/8 − 1 ≈ 0.178.**

## Check yourself

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

1. y = √(2x−x²) in polar form is:
   A. r = 2
   B. r = 2 sin θ
   C. r = 2 cos θ
   D. r = cos θ
   Answer: C. r² = 2r cosθ. (Key point: Circle)

2. θ runs from:
   A. π/4 to π
   B. 0 to π/4
   C. 0 to π/2
   D. π/4 to π/2
   Answer: D. Line y = x to y-axis. (Key point: θ limits)

3. The integral reduces to:
   A. ∫cos²θ dθ
   B. 4∫cos²θ dθ
   C. 4∫cos⁴θ dθ
   D. 2∫cos⁴θ dθ
   Answer: C. (2cosθ)⁴/4. (Key point: Integral)

4. The value is:
   A. π/4 − 1
   B. 3π/4 − 2
   C. 3π/8
   D. 3π/8 − 1
   Answer: D. (1/2)(3π/4−2). (Key point: Result)

## Close variants set in other exams

- [Polar coordinates: y²√(x²+y²)](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Apolar-coordinates-y2sqrt-x2-y2) — Mathematics, chapter 2, asked 7×

## Related in Mathematics

- [Polar coordinates: e^−(x²+y²) integral](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Apolar-coordinates-e-x2-y2) — Mathematics, chapter 2, asked 5×
- [Polar coordinates: √(x²+y²) integral](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Apolar-coordinates-sqrt-x2-y2) — Mathematics, chapter 2, asked 2×
- [Evaluate ∫∫ e^(x²) dx dy](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aevaluate-e-x2-dx-dy) — Mathematics, chapter 2
- [Evaluate ∫∫ e^(y/x) dy dx](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aevaluate-e-y-x-dy) — Mathematics, chapter 2
- [Transform to polar and evaluate ∫∫(x²+y²), 2ax−x²](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Atransform-polar-evaluate-x2-y2) — Mathematics, chapter 2
- [Triple integral (2x+y) over cylinder region](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Atriple-integral-2x-y-over) — Mathematics, chapter 2

## Related in other subjects

- [Solve dy/dx = (x+2y−3)/(2x+y−3)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-homogeneous-linear-x-2y-3) — Mathematics I, chapter 4
- [Solve (1+y²)+(x−e^(tan⁻¹y))dy/dx=0](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-1-plus-y2-x-minus-e-atan-y) — Mathematics I, chapter 4, asked 6×
- [Solve dy/dx = y/x + tan(y/x)](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-dy-dx-y-over-x-tan) — Mathematics I, chapter 4, asked 6×
- [Solve dy/dx − 2y tan x = y² tan x](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Ade-bernoulli-2y-tan-x) — Mathematics I, chapter 4, asked 6×