Double Integrals with Polar Coordinates

Introduction

There are cases where it is more convenient to use polar coordinates to evaluate a double integral. Recall that polar and rectangular coordinates are related by the following identities: x=rcos⁡θ     and     y=rsin⁡θr2=x2+y2     and     θ=tan⁡−1yx\begin{aligned} x=r \cos \theta \ \ \ \ &\textrm{ and }\ \ \ \ y=r \sin \theta\ r^2=x^2+y^2 \ \ \ \ &\textrm{ and }\ \ \ \ \theta = \tan^{-1} \dfrac{y}{x} \end{aligned}

Therefore, the function f(x,y)f(x,y) can be written f(rcos⁡θ,rsin⁡θ)f(r \cos θ,r \sin θ): ∬Rf(x,y) dA=∬Rf(rcos⁡θ,rsin⁡θ) ?\iint_R f(x,y)\ dA = \iint_R f(r \cos \theta, r \sin \theta)\ ?

The question is, what is dAdA?

To answer this, we'll draw a small polar "rectangle" as shown below and shrink it down until the difference between the radii is a differential drdr, and the difference between the angles is dθdθ.

Polar slice

To find the area of this slice, note that the area of the large wedge is the area of the large circle times the proportion of the circle that it covers: A2=πr22(θ2−θ12π)A_2 = \pi r_2^2 \left(\dfrac{\theta_2 - \theta_1}{2\pi}\right)

The area of the small wedge is similar: A1=πr12(θ2−θ12π)A_1 = \pi r_1^2 \left(\dfrac{\theta_2 - \theta_1}{2\pi}\right)

Therefore, the area of the shaded rectangle is Aslice=π(r22−r12)(θ2−θ12π)=(r2+r1)(r2−r1)(θ2−θ12)A_{slice} = \pi (r_2^2 - r_1^2) \left(\dfrac{\theta_2 - \theta_1}{2\pi}\right) = (r_2+r_1)(r_2-r_1) \left(\dfrac{\theta_2 - \theta_1}{2}\right)

If Δθ=θ2−θ1Δθ=θ_2−θ_1 and Δr=r2−r1Δr=r_2−r_1, then Aslice=r2+r12 Δr Δθ.A_{slice} = \dfrac{r_2+r_1}{2}\ \Delta r\ \Delta \theta.

As the slice gets smaller, going toward differentials, r2→r1r_2→r_1, so the differential area is dA=r dr dθ.dA = r\ dr\ d\theta.

Therefore, the rectangular double integral can be rewritten in polar coordinates as follows: (don't forget the rr before dr dθdr\ dθ) ∬Rf(x,y) dA=∬Rf(rcos⁡θ,rsin⁡θ) r dr dθ\iint_R f(x,y)\ dA = \iint_R f(r \cos \theta, r \sin \theta)\ r\ dr\ d\theta

What you'll see as we do examples is that we tend to use polar coordinates when we see x2+y2x^2+y^2 somewhere in the double integrals, since we can replace that with r2r^2.

Examples

Find the volume of the solid bounded by the paraboloid z=9−x2−y2z=9-x^2-y^2 and the xyxy plane.

Solution

We notice that switching to polar coordinates will simplify the integral: ∬R9−x2−y2 dA=∬R(9−r2) r dr dθ\iint_R 9-x^2-y^2\ dA = \iint_R (9-r^2)\ r\ dr\ d\theta

To find the limits of integration (ranges of rr and θθ), note that when the paraboloid intersects the xyxy plane (where z=0z=0), x2+y2=9x^2+y^2=9, which describes a circle centered at the origin with radius 33. Therefore, rr ranges from 00 (the center of this circle) to 33 (the edge of the circle).

0≤r≤3     and     0≤θ≤2π0 \leq r \leq 3\ \ \ \ \textrm{ and }\ \ \ \ 0 \leq \theta \leq 2\pi

The double integral, then, is ∫02π∫039r−r3 dr dθ=∫02π[92r2−14r4]03 dθ=∫02π814 dθ=814θ∣02π=81π2\begin{aligned} \int_0^{2\pi} \int_0^3 9r-r^3\ dr\ d\theta &= \int_0^{2\pi} \bigg[\dfrac{9}{2}r^2 - \dfrac{1}{4}r^4 \bigg]_0^3\ d\theta\ &= \int_0^{2\pi} \dfrac{81}{4}\ d\theta = \dfrac{81}{4}\theta \bigg|_0^{2\pi} = \ans{\dfrac{81\pi}{2}} \end{aligned}

Find the volume of the solid under the paraboloid z=x2+y2z=x^2+y^2, above the xyxy plane, and inside the cylinder x2+y2=2xx^2+y^2=2x.

Solution

The function x2+y2x^2+y^2 will be replaced with r2r^2, so all that remains is to find the limits represented by the cylinder x2+y2=2xx^2+y^2=2x: x2+y2=2xr2=2rcos⁡θr(r−2cos⁡θ)=0r=0,2cos⁡θ\begin{aligned} x^2+y^2 &= 2x\ r^2 &= 2 r \cos \theta\ r(r-2\cos\theta) &= 0\ r &= 0, 2 \cos \theta \end{aligned}

What about limits on θθ? Note that x2+y2=2xx^2+y^2=2x can be rewritten (x−1)2+y2=1(x-1)^2+y^2=1, which represents a circle of radius 11 centered at (1,0)(1,0), so −π2≤θ≤π2-\dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2}

Therefore, the double integral is ∬Rx2+y2 dA=∬Rr2⋅r dr dθ=∫−π/2π/2∫02cos⁡θr3 dr dθ=∫−π/2π/24cos⁡4θ dθ=…=3π2\begin{aligned} \iint_R x^2+y^2\ dA &= \iint_R r^2 \cdot r\ dr\ d\theta\ &= \int_{-\pi/2}^{\pi/2} \int_0^{2\cos\theta} r^3\ dr\ d\theta = \int_{-\pi/2}^{\pi/2} 4\cos^4 \theta\ d\theta = \ldots = \ans{\dfrac{3\pi}{2}} \end{aligned}

Find the volume of the solid bounded by the paraboloids z=x2+y2z=x^2+y^2 and z=8−x2−y2z=8-x^2-y^2.

Solution

To find the limits, find where these surfaces intersect: x2+y2=8−x2−y22(x2+y2)=8x2+y2=4\begin{aligned} x^2+y^2 &= 8-x^2-y^2\ 2(x^2+y^2) &= 8\ x^2+y^2 &= 4 \end{aligned}

which is a circle of radius 22 centered at the origin, so 0≤r≤2     and     0≤θ≤2π0 \leq r \leq 2 \ \ \ \ \textrm{ and }\ \ \ \ 0 \leq \theta \leq 2\pi

Therefore, the double integral to find this volume is ∬R(8−x2−y2)−(x2+y2) dA=∬[(8−r2)−r2] r dr dθ=∫02π∫028r−2r3 dr dθ=∫02π[4r2−12r4]02 dθ=∫02π8 dθ=16π\begin{aligned} \iint_R &(8-x^2-y^2)-(x^2+y^2)\ dA\ &= \iint [(8-r^2)-r^2]\ r\ dr\ d\theta\ &= \int_0^{2\pi} \int_0^2 8r-2r^3\ dr\ d\theta\ &= \int_0^{2\pi} \bigg[ 4r^2-\dfrac{1}{2}r^4 \bigg]_0^2\ d\theta\ &= \int_0^{2\pi} 8\ d\theta = \ans{16\pi} \end{aligned}

Find the volume of the solid bounded by the paraboloid z=4−x2−y2z=4−x^2−y^2 above the region that is outside the circle r=2r=2 and inside the circle r=4cos⁡θr=4 \cos θ (these two circles are shown below).

Two circles, one centered at the origin and the other centered at (2,0)

Solution

The limits on rr are pretty straightforward: 2≤r≤4cos⁡θ2 \leq r \leq 4 \cos \theta but what about the limits on θθ? In other words, what is θθ when the two circles intersect? 4cos⁡θ=2⟶cos⁡θ=12⟶θ=±π34 \cos \theta = 2 \longrightarrow \cos \theta = \dfrac{1}{2} \longrightarrow \theta = \pm \dfrac{\pi}{3}

Therefore, the volume of this solid is ∫−π/3π/3∫24cos⁡θ(4−r2) r dr dθ=∫−π/3π/3[2r2−14r4]24cos⁡θ dθ=∫−π/3π/332cos⁡2θ−64cos⁡4θ−4 dθ=−63−8π\begin{aligned} \int_{-\pi/3}^{\pi/3} &\int_2^{4\cos \theta} (4-r^2)\ r\ dr\ d\theta\ &= \int_{-\pi/3}^{\pi/3} \bigg[ 2r^2-\dfrac{1}{4}r^4 \bigg]2^{4\cos\theta}\ d\theta\ &= \int{-\pi/3}^{\pi/3} 32 \cos^2 \theta - 64\cos^4 \theta - 4\ d\theta = \ans{-6\sqrt{3}-8\pi} \end{aligned}

  1. Evaluate ∬Rx2+y2 dA\displaystyle\iint_R x^2+y^2\ dA, where R={(r,θ) : 0≤r≤4,0≤θ≤2π}.R={(r,\theta)\ :\ 0 \leq r \leq 4, 0 \leq \theta \leq 2\pi}.

    2. ∫02π∫04r3 dr dθ=128π\displaystyle\int_0^{2\pi} \displaystyle\int_0^4 r^3\ dr\ d\theta = 128\pi

  2. Evaluate ∬R2xy dA\displaystyle\iint_R 2xy\ dA, where R={(x,y) : x2+y2≤9,y≥0}.R={(x,y)\ :\ x^2+y^2 \leq 9, y \geq 0}.

    3. ∫0π∫032r3cos⁡θ sin⁡θ dr dθ=0\displaystyle\int_0^{\pi} \displaystyle\int_0^3 2r^3 \cos\theta\ \sin\theta\ dr\ d\theta = 0

  3. Evaluate ∬R116−x2−y2 dA\displaystyle\iint_R \dfrac{1}{\sqrt{16-x^2-y^2}}\ dA, where R={(x,y) : x2+y2≤4,x≥0,y≥0}.R={(x,y)\ :\ x^2+y^2 \leq 4,x \geq 0, y \geq 0}.

    4. ∫0π/2∫02r16−r2 dr dθ=(2−3)π\displaystyle\int_0^{\pi/2} \displaystyle\int_0^2 \dfrac{r}{\sqrt{16-r^2}}\ dr\ d\theta = (2-\sqrt{3})\pi

  4. Evaluate the integral ∫−11∫−1−x21−x2(x2+y2)3/2 dy dx.\displaystyle\int_{-1}^1 \displaystyle\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} (x^2+y^2)^{3/2}\ dy\ dx.

    5. ∫02π∫01r4 dr dθ=2π5\displaystyle\int_0^{2\pi} \displaystyle\int_0^1 r^4\ dr\ d\theta = \dfrac{2\pi}{5}

  5. Evaluate the integral ∫0π/4∫0sin⁡θr dr dθ.\displaystyle\int_{0}^{\pi/4} \displaystyle\int_{0}^{\sin\theta} r\ dr\ d\theta.

    6. 14θ−18cos⁡(2θ)∣0π/4=π16+18\dfrac{1}{4}\theta - \dfrac{1}{8}\cos(2\theta) \bigg|_0^{\pi/4} = \dfrac{\pi}{16} + \dfrac{1}{8}

  6. Evaluate ∬Rx2+y2 dA\displaystyle\iint_R \sqrt{x^2+y^2}\ dA, where R={(x,y) : 1≤x2+y2≤4}.R={(x,y)\ :\ 1 \leq x^2+y^2 \leq 4}.

    7. ∫02π∫12r2 dr dθ=14π3\displaystyle\int_0^{2\pi} \displaystyle\int_1^2 r^2\ dr\ d\theta = \dfrac{14\pi}{3}