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9.3: Double Integrals in Polar Coordinates

This page is a draft and is under active development. 

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Double integrals are sometimes much easier to evaluate if we change rectangular coordinates to polar coordinates. However, before we describe how to make this change, we need to establish the concept of a double integral in a polar rectangular region.

Polar Rectangular Regions of Integration

When we defined the double integral for a continuous function in rectangular coordinates—say, g over a region R in the xy-plane—we divided R into subrectangles with sides parallel to the coordinate axes. These sides have either constant x-values and/or constant y-values. In polar coordinates, the shape we work with is a polar rectangle, whose sides have constant r-values and/or constant θ-values. This means we can describe a polar rectangle as in Figure 9.3.1a, with R={(r,θ)|arb,αθβ}.

alt
Figure 9.3.1: (a) A polar rectangle R (b) divided into subrectangles Rij (c) Close-up of a subrectangle.

In this section, we are looking to integrate over polar rectangles. Consider a function f(r,θ) over a polar rectangle R. We divide the interval [a,b] into m subintervals [ri1,ri] of length Δr=(ba)/m and divide the interval [α,β] into n subintervals [θi1,θi] of width Δθ=(βα)/n. This means that the circles r=ri and rays θ=θi for 1im and 1jn divide the polar rectangle R into smaller polar subrectangles Rij (Figure 9.3.1b).

As before, we need to find the area ΔA of the polar subrectangle Rij and the “polar” volume of the thin box above Rij. Recall that, in a circle of radius r the length s of an arc subtended by a central angle of θ radians is s=rθ. Notice that the polar rectangle Rij looks a lot like a trapezoid with parallel sides ri1Δθ and riΔθ and with a width Δr. Hence the area of the polar subrectangle Rij is

ΔA=12Δr(ri1Δθ+r1δθ).

Simplifying and letting rij=12(ri1+ri), we have ΔA=rijΔrΔθ.

Therefore, the polar volume of the thin box above Rij (Figure 9.3.2) is

f(rij,θij)bΔA=f(rij,θij)rijΔrΔθ.

alt
Figure 9.3.2: Finding the volume of the thin box above polar rectangle Rij.

Using the same idea for all the subrectangles and summing the volumes of the rectangular boxes, we obtain a double Riemann sum as

mi=1nj=1f(rij,θij)rijΔrΔθ.

As we have seen before, we obtain a better approximation to the polar volume of the solid above the region R when we let m and n become larger. Hence, we define the polar volume as the limit of the double Riemann sum,

V=limm,nmi=1nj=1f(rij,θij)rijΔrΔθ.

This becomes the expression for the double integral.

Definition: The double integral in polar coordinates

The double integral of the function f(r,θ) over the polar rectangular region R in the rθ-plane is defined as

Again, just as in section on Double Integrals over Rectangular Regions, the double integral over a polar rectangular region can be expressed as an iterated integral in polar coordinates. Hence,

\iint_R f(r, \theta)\,dA = \iint_R f(r, \theta) \,r \, dr \, d\theta = \int_{\theta=\alpha}^{\theta=\beta} \int_{r=a}^{r=b} f(r,\theta) \,r \, dr \, d\theta.

Notice that the expression for dA is replaced by r \, dr \, d\theta when working in polar coordinates. Another way to look at the polar double integral is to change the double integral in rectangular coordinates by substitution. When the function f is given in terms of x and y using x = r \, \cos \, \theta, \, y = r \, \sin \, \theta, and dA = r \, dr \, d\theta changes it to

\iint_R f(x,y) \,dA = \iint_R f(r \, \cos \, \theta, \, r \, \sin \, \theta ) \,r \, dr \, d\theta.

Note that all the properties listed in section on Double Integrals over Rectangular Regions for the double integral in rectangular coordinates hold true for the double integral in polar coordinates as well, so we can use them without hesitation.

Example \PageIndex{1A}: Sketching a Polar Rectangular Region

Sketch the polar rectangular region

R = \{(r, \theta)\,|\,1 \leq r \leq 3, 0 \leq \theta \leq \pi \}. \nonumber

Solution

As we can see from Figure \PageIndex{3}, r = 1 and r = 3 are circles of radius 1 and 3 and 0 \leq \theta \leq \pi covers the entire top half of the plane. Hence the region R looks like a semicircular band.

imageedit_1_9607827905.png
Figure \PageIndex{3}: The polar region R lies between two semicircles.

Now that we have sketched a polar rectangular region, let us demonstrate how to evaluate a double integral over this region by using polar coordinates.

Example \PageIndex{1B}: Evaluating a Double Integral over a Polar Rectangular Region

Evaluate the integral \displaystyle \iint_R 3x \, dA over the region R = \{(r, \theta)\,|\,1 \leq r \leq 2, \, 0 \leq \theta \leq \pi \}.

Solution

First we sketch a figure similar to Figure \PageIndex{3}, but with outer radius r=2


9.3: Double Integrals in Polar Coordinates is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by LibreTexts.

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