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3.4: Double Integration with Polar Coordinates

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We have used iterated integrals to evaluate double integrals, which give the signed volume under a surface, z=f(x,y), over a region R of the xy-plane. The integrand is simply f(x,y), and the bounds of the integrals are determined by the region R.

Some regions R are easy to describe using rectangular coordinates -- that is, with equations of the form y=f(x), x=a, etc. However, some regions are easier to handle if we represent their boundaries with polar equations of the form r=f(θ), θ=α, etc.

The basic form of the double integral is Rf(x,y) dA. We interpret this integral as follows: over the region R, sum up lots of products of heights (given by f(xi,yi)) and areas (given by ΔAi). That is, dA represents "a little bit of area.'' In rectangular coordinates, we can describe a small rectangle as having area dx dy or dydx -- the area of a rectangle is simply length×width -- a small change in x times a small change in y. Thus we replace dA in the double integral with dxdy or dydx.

13.19.PNG

FIGURE 3.4.1

Now consider representing a region R with polar coordinates. Consider Figure 3.4.1a. Let R be the region in the first quadrant bounded by the curve. We can approximate this region using the natural shape of polar coordinates: portions of sectors of circles. In the figure, one such region is shaded, shown again in part (b) of the figure.

As the area of a sector of a circle with radius r, subtended by an angle θ, is A=12r2θ, we can find the area of the shaded region. The whole sector has area 12r22Δθ, whereas the smaller, unshaded sector has area 12r21Δθ. The area of the shaded region is the difference of these areas:
ΔAi=12r22Δθ12r21Δθ=12(r22r21)(Δθ)=r2+r12(r2r1)Δθ.

Note that (r2+r1)/2 is just the average of the two radii.

To approximate the region R, we use many such subregions; doing so shrinks the difference r2r1 between radii to 0 and shrinks the change in angle Δθ also to 0. We represent these infinitesimal changes in radius and angle as dr and dθ, respectively. Finally, as dr is small, r2r1, and so (r2+r1)/2r1. Thus, when dr and dθ are small,
ΔAiridrdθ.

Taking a limit, where the number of subregions goes to infinity and both r2r1 and Δθ go to 0, we get dA=rdrdθ.

So to evaluate Rf(x,y) dA, replace dA with rdrdθ. Convert the function z=f(x,y) to a function with polar coordinates with the substitutions x=rcosθ, y=rsinθ. Finally, find bounds g1(θ)rg2(θ) and αθβ that describe R. This is the key principle of this section, so we restate it here as a Key Idea.

Key Idea: Evaluating Double Integrals with Polar Coordinates

Let R be a plane region bounded by the polar equations αθβ and g1(θ)rg2(θ). Then
Rf(x,y) dA=βαg2(θ)g1(θ)f(rcosθ,rsinθ) rdrdθ.

Examples will help us understand this Key Idea.

Example 3.4.1: Evaluating a double integral with polar coordinates

Find the signed volume under the plane z=4x2y over the circle with equation x2+y2=1.

Solution

The bounds of the integral are determined solely by the region R over which we are integrating. In this case, it is a circle with equation x2+y2=1. We need to find polar bounds for this region. It may help to review polar coordinates earlier in this text; bounds for this circle are 0r1 and 0θ2π.

We replace f(x,y) with f(rcosθ,rsinθ). That means we make the following substitutions:

4x2y4rcosθ2rsinθ.

Finally, we replace dA in the double integral with rdrdθ. This gives the final iterated integral, which we evaluate:

Rf(x,y) dA=2π010(4rcosθ2rsinθ)rdrdθ=2π010(4rr2(cosθ2sinθ))drdθ=2π0(2r213r3(cosθ2sinθ))|10dθ=2π0(213(cosθ2sinθ))dθ=(2θ13(sinθ+2cosθ))|2π0=4π12.566.

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FIGURE 3.4.2

The surface and region R are shown in Figure 3.4.2.

Example 3.4.2: Evaluating a double integral with polar coordinates

Find the volume under the paraboloid z=4(x2)2y2 over the region bounded by the circles (x1)2+y2=1 and (x2)2+y2=4.

Solution

At first glance, this seems like a very hard volume to compute as the region R (shown in Figure 3.4.3a) has a hole in it, cutting out a strange portion of the surface, as shown in part (b) of the figure. However, by describing R in terms of polar equations, the volume is not very difficult to compute.

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FIGURE 3.4.3

It is straightforward to show that the circle (x1)2+y2=1 has polar equation r=2cosθ, and that the circle (x2)2+y2=4 has polar equation r=4cosθ. Each of these circles is traced out on the interval 0θπ. The bounds on r are 2cosθr4cosθ.

Replacing x with rcosθ in the integrand, along with replacing y with rsinθ, prepares us to evaluate the double integral Rf(x,y) dA:

Rf(x,y) dA=π04cosθ2cosθ(4(rcosθ2)2(rsinθ)2)rdrdθ=π04cosθ2cosθ(r3+4r2cosθ)drdθ=π0(14r4+43r3cosθ)|4cosθ2cosθdθ=π0([14(256cos4θ)+43(64cos4θ)] [14(16cos4θ)+43(8cos4θ)])dθ=π0443cos4θdθ.

To integrate cos4θ, rewrite it as cos2θcos2θ and employ the power-reducing formula twice:

cos4θ=cos2θcos2θ=12(1+cos(2θ))12(1+cos(2θ))=14(1+2cos(2θ)+cos2(2θ))=14(1+2cos(2θ)+12(1+cos(4θ)))=38+12cos(2θ)+18cos(4θ).

Picking up from where we left off above, we have

=π0443cos4θdθ=π0443(38+12cos(2θ)+18cos(4θ))dθ=443(38θ+14sin(2θ)+132sin(4θ))|π0=112π17.279.

While this example was not trivial, the double integral would have been much harder to evaluate had we used rectangular coordinates.

Example 3.4.3: Evaluating a double integral with polar coordinates

Find the volume under the surface f(x,y)=1x2+y2+1 over the sector of the circle with radius a centered at the origin in the first quadrant, as shown in Figure 3.4.4.

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FIGURE 3.4.4

Solution

The region R we are integrating over is a circle with radius a, restricted to the first quadrant. Thus, in polar, the bounds on R are 0ra, 0θπ/2. The integrand is rewritten in polar as

1x2+y2+11r2cos2θ+r2sin2θ+1=1r2+1.

We find the volume as follows:

Rf(x,y) dA=π/20a0rr2+1drdθ=π/2012(ln|r2+1|)|a0dθ=π/2012ln(a2+1)dθ=(12ln(a2+1)θ)|π/20=π4ln(a2+1).

Figure 3.4.4 shows that f shrinks to near 0 very quickly. Regardless, as a grows, so does the volume, without bound.

Note: Previous work has shown that there is finite area under 1x2+1 over the entire x-axis. However, Example 3.4.3 shows that there is infinite volume under 1x2+y2+1 over the entire xy-plane.

Example 3.4.4: Finding the volume of a sphere

Find the volume of a sphere with radius a.

Solution
The sphere of radius a, centered at the origin, has equation x2+y2+z2=a2; solving for z, we have z=a2x2y2. This gives the upper half of a sphere. We wish to find the volume under this top half, then double it to find the total volume.

The region we need to integrate over is the circle of radius a, centered at the origin. Polar bounds for this equation are 0ra, 0θ2π.

All together, the volume of a sphere with radius a is:

2Ra2x2y2 dA=22π0a0a2(rcosθ)2(rsinθ)2rdrdθ=22π0a0ra2r2drdθ.

We can evaluate this inner integral with substitution. With u=a2r2, du=2rdr. The new bounds of integration are u(0)=a2 to u(a)=0. Thus we have:

=2π00a2(u1/2)dudθ=2π0(23u3/2)|0a2dθ=2π0(23a3)dθ=(23a3θ)|2π0=43πa3.

Generally, the formula for the volume of a sphere with radius r is given as 4/3πr3; we have justified this formula with our calculation.

Example 3.4.5: Finding the volume of a solid

A sculptor wants to make a solid bronze cast of the solid shown in Figure 3.4.5, where the base of the solid has boundary, in polar coordinates, r=cos(3θ), and the top is defined by the plane z=1x+0.1y. Find the volume of the solid.

13.23.PNG

FIGURE 3.4.5

Solution
From the outset, we should recognize that knowing how to set up this problem is probably more important than knowing how to compute the integrals. The iterated integral to come is not "hard'' to evaluate, though it is long, requiring lots of algebra. Once the proper iterated integral is determined, one can use readily--available technology to help compute the final answer.

The region R that we are integrating over is bound by 0rcos(3θ), for 0θπ (note that this rose curve is traced out on the interval [0,π], not [0,2π]). This gives us our bounds of integration. The integrand is z=1x+0.1y; converting to polar, we have that the volume V is:

V=Rf(x,y) dA=π0cos(3θ)0(1rcosθ+0.1rsinθ)rdrdθ.

Distributing the r, the inner integral is easy to evaluate, leading to

π0(12cos2(3θ)13cos3(3θ)cosθ+0.13cos3(3θ)sinθ)dθ.

This integral takes time to compute by hand; it is rather long and cumbersome. The powers of cosine need to be reduced, and products like cos(3θ)cosθ need to be turned to sums using the Product To Sum formulas in the back cover of this text.

We rewrite 12cos2(3θ) as 14(1+cos(6θ)). We can also rewrite 13cos3(3θ)cosθ as:

13cos3(3θ)cosθ=13cos2(3θ)cos(3θ)cosθ=131+cos(6θ)2(cos(4θ)+cos(2θ)).

This last expression still needs simplification, but eventually all terms can be reduced to the form acos(mθ) or asin(mθ) for various values of a and m.

We forgo the algebra and recommend the reader employ technology, such as WolframAlpha, to compute the numeric answer. Such technology gives:

π0cos(3θ)0(1rcosθ+0.1rsinθ)rdrdθ=π40.785u3.

Since the units were not specified, we leave the result as almost 0.8 cubic units (meters, feet, etc.) Should the artist want to scale the piece uniformly, so that each rose petal had a length other than 1, she should keep in mind that scaling by a factor of k scales the volume by a factor of k3.

We have used iterated integrals to find areas of plane regions and volumes under surfaces. Just as a single integral can be used to compute much more than "area under the curve,'' iterated integrals can be used to compute much more than we have thus far seen. The next two sections show two, among many, applications of iterated integrals.

 


This page titled 3.4: Double Integration with Polar Coordinates is shared under a CC BY license and was authored, remixed, and/or curated by OpenStax.

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