Double Integrals in Polar Coordinates

Double integrals in polar coordinates replace x and y with r and θ, using x = r cos θ and y = r sin θ. The area piece becomes dA = r dr dθ, with an extra factor of r. Use polar coordinates when the region is a disk, a ring or a sector, or when the integrand contains x² + y².

Before this lesson: Double Integrals over General Regions, Polar Coordinates and Area in Polar Coordinates

Why dA = r dr dθ

In x and y, you cut a region into small rectangles of area Δx Δy. In polar coordinates, you cut it with circles (r = constant) and rays (θ = constant). Each piece is a polar rectangle: a thin curved box.

Its depth is Δr. Its width is an arc of a circle of radius r, which has length r Δθ (see arc length). So its area is about r Δr Δθ. Pieces far from the origin are wider, and the factor r accounts for that.

Double integral in polar coordinates formula

\iint_D f(x, y)\,dA = \int_{\alpha}^{\beta} \int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta

1. Describe the region with r and θ. A disk of radius a centered at the origin is 0 ≤ r ≤ a and 0 ≤ θ ≤ 2π.
2. Replace x with r cos θ and y with r sin θ. Replace x² + y² with r².
3. Replace dA with r dr dθ. Do not forget the extra r.
4. Integrate in r first, then in θ.

Double integrals in polar coordinates examples

Example 1: A function of x² + y² on a disk

Evaluate \iint_D (x^2 + y^2)\,dA, where D is the disk x² + y² ≤ 4.

0 \le r \le 2, \quad 0 \le \theta \le 2\pi

\iint_D (x^2 + y^2)\,dA = \int_0^{2\pi} \int_0^2 r^2 \cdot r\,dr\,d\theta

\int_0^2 r^3\,dr = \Big[\dfrac{r^4}{4}\Big]_0^2

= 4

\int_0^{2\pi} 4\,d\theta = 8\pi

Answer: 8π ≈ 25.13. In x and y, this needs limits like ±√(4 − x²) and a much harder integral.

Example 2: A half ring

Evaluate \iint_D y\,dA, where D is the upper half of the ring between the circles x² + y² = 1 and x² + y² = 9.

1 \le r \le 3, \quad 0 \le \theta \le \pi

\iint_D y\,dA = \int_0^{\pi} \int_1^3 (r\sin\theta)\,r\,dr\,d\theta

\int_1^3 r^2 \sin\theta\,dr = \sin\theta\Big[\dfrac{r^3}{3}\Big]_1^3

= \dfrac{26}{3}\sin\theta

\int_0^{\pi} \dfrac{26}{3}\sin\theta\,d\theta = \dfrac{26}{3}\Big[-\cos\theta\Big]_0^{\pi}

= \dfrac{26}{3}(1 + 1)

= \dfrac{52}{3}

Answer: 52/3 ≈ 17.33.

Example 3: An integral you can’t do in x and y

Evaluate \iint_D e^{-(x^2 + y^2)}\,dA, where D is the disk x² + y² ≤ 9.

\iint_D e^{-(x^2 + y^2)}\,dA = \int_0^{2\pi} \int_0^3 e^{-r^2}\,r\,dr\,d\theta

\int_0^3 e^{-r^2}\,r\,dr = \Big[-\dfrac{1}{2}e^{-r^2}\Big]_0^3

= -\dfrac{1}{2}e^{-9} + \dfrac{1}{2}

= \dfrac{1 - e^{-9}}{2}

\int_0^{2\pi} \dfrac{1 - e^{-9}}{2}\,d\theta = \pi(1 - e^{-9})

\pi(1 - e^{-9}) \approx 3.1412

Answer: π(1 − e⁻⁹) ≈ 3.1412. The function e^(−x²) has no elementary antiderivative, so the extra r is what makes this work.

Example 4: Convert an iterated integral to polar

Evaluate \int_0^2 \int_0^{\sqrt{4 - x^2}} (x + y)\,dy\,dx by switching to polar coordinates.

0 \le x \le 2, \quad 0 \le y \le \sqrt{4 - x^2}

0 \le r \le 2, \quad 0 \le \theta \le \dfrac{\pi}{2}

\int_0^{\pi/2} \int_0^2 (r\cos\theta + r\sin\theta)\,r\,dr\,d\theta

\int_0^2 r^2(\cos\theta + \sin\theta)\,dr = \dfrac{8}{3}(\cos\theta + \sin\theta)

\int_0^{\pi/2} \dfrac{8}{3}(\cos\theta + \sin\theta)\,d\theta = \dfrac{8}{3}\Big[\sin\theta - \cos\theta\Big]_0^{\pi/2}

= \dfrac{8}{3}\Big((1 - 0) - (0 - 1)\Big)

= \dfrac{16}{3}

Answer: 16/3.

Example 5: Air inside a dome

A dome-shaped tent has the roof z = 9 − x² − y² over a circular floor, in feet. Find the volume of air inside.

9 - r^2 = 0 \Rightarrow r = 3

V = \int_0^{2\pi} \int_0^3 (9 - r^2)\,r\,dr\,d\theta

\int_0^3 (9r - r^3)\,dr = \Big[\dfrac{9r^2}{2} - \dfrac{r^4}{4}\Big]_0^3

= \dfrac{81}{2} - \dfrac{81}{4}

= \dfrac{81}{4}

\int_0^{2\pi} \dfrac{81}{4}\,d\theta = \dfrac{81\pi}{2}

\dfrac{81\pi}{2} \approx 127.2

Answer: about 127.2 ft³ of air.

Common mistakes

Related methods

Area of a polar region

With f = 1, the double integral gives the area: A = \int_{\alpha}^{\beta} \int_0^{r(\theta)} r\,dr\,d\theta = \int_{\alpha}^{\beta} \dfrac{1}{2}r(\theta)^2\,d\theta. That is the area in polar coordinates formula from Calculus 2.

Regions bounded by polar curves

The inner limits can depend on θ. For the disk inside r = 2 cos θ, use −π/2 ≤ θ ≤ π/2 and 0 ≤ r ≤ 2 cos θ, as in Practice 7.

When to stay in x and y

If the region is a rectangle or a triangle with straight sides, x and y are usually simpler. Polar coordinates pay off for circles and for integrands with x² + y².

Practice problems

1. Use a polar double integral to find the area of the disk x² + y² ≤ 25.
2. Evaluate \iint_D (x^2 + y^2)\,dA over the ring 1 ≤ x² + y² ≤ 4.
3. Evaluate \iint_D x\,dA over the right half of the disk x² + y² ≤ 9.
4. Evaluate \iint_D \sqrt{x^2 + y^2}\,dA over the disk x² + y² ≤ 9.
5. Evaluate \iint_D e^{x^2 + y^2}\,dA over the disk x² + y² ≤ 1.
6. Convert to polar and evaluate \int_{-1}^{1} \int_0^{\sqrt{1 - x^2}} (x^2 + y^2)^{3/2}\,dy\,dx.
7. Use a double integral to find the area inside the circle r = 2 cos θ.
8. Evaluate \iint_D xy\,dA over the quarter disk x² + y² ≤ 4 in the first quadrant.
9. Find the volume under z = 4 − x² − y² and above the xy-plane.
10. A sprinkler wets a disk of radius 6 m. The water depth is d = 0.03 − 0.005r meters at distance r from the sprinkler. How much water lands on the lawn?

A = \int_0^{2\pi} \int_0^5 r\,dr\,d\theta

\int_0^5 r\,dr = \Big[\dfrac{r^2}{2}\Big]_0^5

= \dfrac{25}{2}

\int_0^{2\pi} \dfrac{25}{2}\,d\theta = 25\pi

\int_0^{2\pi} \int_1^2 r^2 \cdot r\,dr\,d\theta

\int_1^2 r^3\,dr = \Big[\dfrac{r^4}{4}\Big]_1^2

= 4 - \dfrac{1}{4}

= \dfrac{15}{4}

\int_0^{2\pi} \dfrac{15}{4}\,d\theta = \dfrac{15\pi}{2}

-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}, \quad 0 \le r \le 3

\int_{-\pi/2}^{\pi/2} \int_0^3 (r\cos\theta)\,r\,dr\,d\theta

\int_0^3 r^2\,dr = 9

\int_{-\pi/2}^{\pi/2} 9\cos\theta\,d\theta = 9\Big[\sin\theta\Big]_{-\pi/2}^{\pi/2}

= 9(1 - (-1))

= 18

\int_0^{2\pi} \int_0^3 r \cdot r\,dr\,d\theta

\int_0^3 r^2\,dr = 9

\int_0^{2\pi} 9\,d\theta = 18\pi

\int_0^{2\pi} \int_0^1 e^{r^2}\,r\,dr\,d\theta

\int_0^1 e^{r^2}\,r\,dr = \Big[\dfrac{e^{r^2}}{2}\Big]_0^1

= \dfrac{e - 1}{2}

\int_0^{2\pi} \dfrac{e - 1}{2}\,d\theta = \pi(e - 1)

0 \le r \le 1, \quad 0 \le \theta \le \pi

\int_0^{\pi} \int_0^1 r^3 \cdot r\,dr\,d\theta

\int_0^1 r^4\,dr = \dfrac{1}{5}

\int_0^{\pi} \dfrac{1}{5}\,d\theta = \dfrac{\pi}{5}

-\dfrac{\pi}{2} \le \theta \le \dfrac{\pi}{2}, \quad 0 \le r \le 2\cos\theta

\int_0^{2\cos\theta} r\,dr = \Big[\dfrac{r^2}{2}\Big]_0^{2\cos\theta}

= 2\cos^2\theta

= 1 + \cos 2\theta

\int_{-\pi/2}^{\pi/2} (1 + \cos 2\theta)\,d\theta = \Big[\theta + \dfrac{\sin 2\theta}{2}\Big]_{-\pi/2}^{\pi/2}

= \dfrac{\pi}{2} - \Big(-\dfrac{\pi}{2}\Big)

= \pi

\int_0^{\pi/2} \int_0^2 (r\cos\theta)(r\sin\theta)\,r\,dr\,d\theta

\int_0^2 r^3\,dr = 4

\int_0^{\pi/2} 4\sin\theta\cos\theta\,d\theta = 4\Big[\dfrac{\sin^2\theta}{2}\Big]_0^{\pi/2}

= 4 \cdot \dfrac{1}{2}

= 2

4 - r^2 = 0 \Rightarrow r = 2

\int_0^2 (4 - r^2)\,r\,dr = \Big[2r^2 - \dfrac{r^4}{4}\Big]_0^2

= 8 - 4

= 4

\int_0^{2\pi} 4\,d\theta = 8\pi

V = \int_0^{2\pi} \int_0^6 (0.03 - 0.005r)\,r\,dr\,d\theta

\int_0^6 (0.03r - 0.005r^2)\,dr = \Big[0.015r^2 - \dfrac{0.005r^3}{3}\Big]_0^6

= 0.54 - 0.36

= 0.18

\int_0^{2\pi} 0.18\,d\theta = 0.36\pi

0.36\pi \approx 1.131

Double integrals in polar coordinates FAQ

Why is there an extra r in polar double integrals?

A small polar rectangle has depth Δr and curved width r Δθ, so its area is r Δr Δθ. Without the r, pieces far from the origin would count as too small.

When should I use polar coordinates for a double integral?

Use them when the region is a disk, a ring, a sector or a region bounded by a polar curve. They also help when the integrand contains x² + y², as in e^(−(x² + y²)).

How do you convert a double integral to polar coordinates?

Describe the region with r and θ, replace x with r cos θ and y with r sin θ, and replace dA with r dr dθ. Then integrate in r first and θ second.

What are the limits for a full disk in polar coordinates?

For the disk x² + y² ≤ a², use 0 ≤ r ≤ a and 0 ≤ θ ≤ 2π.

Summary

• Polar substitution: x = r cos θ, y = r sin θ, x² + y² = r².
• The area piece is dA = r dr dθ; never drop the extra r.
• Disk of radius a: 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π. Use smaller θ ranges for halves and quarters.
• Inner limits on r can depend on θ, as for r = 2 cos θ.
• Use polar coordinates for round regions and for integrands with x² + y².

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