9.11 Double Integrals in Polar Coordinates

INTRODUCTION

A double integral, which may be difficult or even impossible to evaluate in rectangular xy-coordinates, may become more tractable when expressed in a different coordinate system. In this section we examine double integrals in polar rθ-coordinates.

Polar Rectangles

Suppose R is a region in the plane bounded by the graphs of the polar equations r = g1(θ), r = g2(θ), and the rays θ = α, θ = β, and f is a function of r and θ that is continuous on R. In order to define the double integral of f over R, we use rays and concentric circles to partition the region into a grid of polar rectangles or subregions Rk. See FIGURE 9.11.1(a) and 9.11.1(b). The area ΔAk of typical subregion Rk, shown in Figure 9.11.1(c), is the difference of the areas of two circular sectors: ΔAk = Δθk – Δθk. Now ΔAk can be written as

2 graphs and a visual representation. Graph a. Caption. Region R is bound by polar graphs and rays. Graph. 2 curves and 2 rays are graphed on a polar coordinate system. The first curve is labeled r = g subscript 1(theta). The second curve is labeled r = g subscript 2(theta), and is graphed above the first curve. The first ray forms an angle labeled alpha with the horizontal polar axis. The second ray forms an angle labeled beta with the horizontal polar axis. The region on the graph, between the 2 curves and the 2 rays, is shaded and labeled R. Graph b. Caption. Subregion R subscript k. Graph. 2 curves, several rays and concentric arcs are graphed on a polar coordinate system. The first curve is labeled r = g subscript 1(theta). The second curve is labeled r = g subscript 2(theta), and is graphed above the first curve. The ray closest to the horizontal polar axis forms an angle labeled alpha with it. The ray farthest from the horizontal polar axis forms an angle labeled beta with it. The region on the graph, between the 2 curves and the 2 rays, is shaded. Several rays and concentric arcs, with their centers at the origin, partition the shaded region into a grid of polar rectangles. One of the polar rectangles is highlighted and labeled R subscript k, and contains a point at the center. Visual representation c. Caption. Enlargement of R subscript k. Figure. An enlarged view of a polar rectangle, bound by 2 consecutive rays and concentric arcs is shown, along with the origin of the polar graph. The polar rectangle contains a point at the center. The length between the first concentric arc and the origin is labeled r subscript k. The length between the second concentric arc and the origin is labeled r subscript k+1. The length between the origin and the point is labeled 1 over 2 ((r subscript k + r subscript k+1). The length between the 2 concentric arcs is labeled delta r subscript k. The angle formed by the consecutive rays is labeled delta theta subscript k.

FIGURE 9.11.1 Rk in (b) and (c) is called a polar rectangle

where Δrk = rk+1 – rk and = (rk+1 + rk) denotes the average radius. By choosing (, ) on each Rk, the double integral of f over R is

The double integral is then evaluated by means of the iterated integral:

.(1)

On the other hand, if the region R is as given in FIGURE 9.11.2, the double integral of f over R is then

.(2)

A graph. Caption. 2 curves and 2 circular and concentric arcs are graphed on a polar coordinate system. The first curve is labeled theta = h subscript 1(r). The second curve is labeled theta = h subscript 2(r), and is graphed above the first curve. The first circular arc is labeled r = a and is closer to the origin. The second circular arc is farther to the origin and is labeled r = b. The region on the graph, between the 2 curves and the 2 circular arcs, is shaded and labeled R.

FIGURE 9.11.2 R bounded by polar graphs and circular arcs

EXAMPLE 1 Center of Mass

Find the center of mass of the lamina that corresponds to the region bounded by one leaf of the rose r = 2 sin 2θ in the first quadrant if the density at a point P in the lamina is directly proportional to the distance from the pole.

SOLUTION

By varying θ from 0 to π/2, we obtain the graph in FIGURE 9.11.3. Now, d(O, P) = |r|. Hence, ρ(r, θ) = k|r|, where k is a constant of proportionality. From (9) of Section 9.10, we have

A graph. A closed curve, resembling a petal, and 2 rays, are graphed in polar coordinate system. The closed curve, labeled r = 2 sin 2 theta, begins at the origin, goes up and to the right with increasing steepness initially, then changes direction and continues going up to the left with decreasing steepness, reaches a high point, goes down to the left with increasing steepness, and ends at the origin. The region inside the closed curve is shaded. The 2 rays graphed inside the shaded region, begin at the origin and are upward sloping. Several concentric arcs, with their centers at the origin, partition the shaded region between the 2 arcs into several polar rectangles.

FIGURE 9.11.3 Lamina in Example 1

Since x = r cos θ, we can write as

Similarly, by using x = r sin θ, we find*

Here the rectangular coordinates of the center of mass are

≡

Change of Variables: Rectangular to Polar Coordinates

In some instances a double integral f(x, y) dA that is difficult or even impossible to evaluate using rectangular coordinates may be readily evaluated when a change of variables is used. If we assume that f is continuous on the region R and if R can be described in polar coordinates as 0 ≤ g1(θ) ≤ r ≤ g2(θ), α ≤ θ ≤ β, 0 < β – α ≤ 2π, then

. (3)

Equation (3) is particularly useful when f contains the expression x2 + y2 since, in polar coordinates, we can now write

EXAMPLE 2 Changing an Integral to Polar Coordinates

Use polar coordinates to evaluate

SOLUTION

From x ≤ y ≤ , 0 ≤ x ≤ 2, we have sketched the region R of integration in FIGURE 9.11.4. Since x2 + y2 = r2, the polar description of the circle x2 + y2 = 8 is r = . Hence, in polar coordinates, the region of R is given by 0 ≤ r ≤ , π/4 ≤ θ ≤ π/2. With 1/(5 + x2 + y2) = 1/(5 + r2), the original integral becomes

≡

A graph. A circle and a closed curve are graphed on an x y coordinate plane. The circle is labeled x^2 + y^2 = 8, and its center is at the origin. The closed curve consists of 2 lines and a curve. The first line, labeled y = x, begins at the marked origin, goes up to the right and ends on the circle at the marked point (2, 2). The curve, labeled y = sqrt(8 minus x^2), begins on the circle at the marked point (2, 2), goes along the circle counter clockwise, and ends on the y axis at a marked point. The second line begins on the y axis at the marked point, goes vertically down along the y axis, and ends at the marked origin. The region inside the closed curve is shaded and labeled R.

FIGURE 9.11.4 Region R in Example 2

EXAMPLE 3 Volume

Find the volume of the solid that is under the hemisphere z = and above the region bounded by the graph of the circle x2 + y2 – y = 0.

SOLUTION

From FIGURE 9.11.5, we see that V = dA. In polar coordinates the equations of the hemisphere and the circle become, respectively, z = and r = sin θ. Now, using symmetry, we have

≡

A graph. A hemisphere labeled z = sqrt(1 minus x^2 minus y^2), is graphed on a three dimensional x y z coordinate system. The base of the hemisphere is on the x y plane, and its center is at the origin. A circle, labeled x^2 + y^2 minus y = 0, is graphed on the x y plane. The circle passes through the origin, and its center is on the y axis. A fence along the circle intersects the top surface of the hemisphere following a closed curve. 2 rays graphed inside the circle. A rectangular prism, with its base between the 2 arcs, and its top face on the surface of the hemisphere, connects the surface and the region inside the circle in the x y plane. The region inside the rectangular prism is shaded.

FIGURE 9.11.5 Solid in Example 3

Area

Note that in (1) if f(r, θ) = 1, then the area of the region R in Figure 9.11.1(a) is given by

The same observation holds for (2) and Figure 9.11.2 when f(r, θ) = 1.

REMARKS

The reader is invited to reexamine Example 3. The graph of the circle r = sin θ is obtained by varying θ from 0 to π. However, carry out the iterated integration

and see if the result is the incorrect answer π/3. What goes wrong?

9.11 Exercises Answers to selected odd-numbered problems begin on page ANS-25.

In Problems 1–4, use a double integral in polar coordinates to find the area of the region bounded by the graphs of the given polar equations.

  1. r = 3 + 3 sin θ
  2. r = 2 + cos θ
  3. r = 2 sin θ, r = 1, common area
  4. r = 8 sin 4θ, one petal

In Problems 5–10, find the volume of the solid bounded by the graphs of the given equations.

  1. One petal of r = 5 cos 3θ, z = 0, z = 4
  2. x2 + y2 = 4, z =
  3. Between x2 + y2 = 1 and x2 + y2 = 9,
    z =
  4. z = + y2 = 25, z = 0
  5. r = 1 + cos θ, z = y, z = 0, first octant
  6. r = cos θ, z = 2 + x2 + y2, z = 0

In Problems 11–16, find the center of mass of the lamina that has the given shape and density.

  1. r = 1, r = 3, x = 0, y = 0, first quadrant; ρ(r, θ) = k (constant)
  2. r = cos θ; density at point P directly proportional to the distance from the pole
  3. y = x, y = 0, x = 3; ρ(r, θ) = r2
  4. r = 4 cos 2θ, petal on the polar axis; ρ(r, θ) = k (constant)
  5. Outside r = 2 and inside r = 2 + 2 cos θ, y = 0, first quadrant; density at a point P inversely proportional to the distance from the pole
  6. r = 2 + 2 cos θ, y = 0, first and second quadrants; ρ(r, θ) = k (constant)

In Problems 17–20, find the indicated moment of inertia of the lamina that has the given shape and density.

  1. r = a; ρ(r, θ) = k (constant); Ix
  2. r = a; ρ(r, θ) = ; Ix
  3. Outside r = a and inside r = 2a cos θ; density at a point P inversely proportional to the cube of the distance from the pole; Iy
  4. Outside r = 1 and inside r = 2 sin 2θ, first quadrant; ρ(r, θ) = sec2θ; Iy

In Problems 21–24, find the polar moment of inertia


of the lamina that has the given shape and density.

  1. r = a; ρ(r, θ) = k (constant) [Hint: Use Problem 17 and the fact that Ix = Iy.]
  2. r = θ, 0 ≤ θ ≤ π, y = 0; density at a point P proportional to the distance from the pole
  3. rθ = 1, ≤ θ ≤ 1, r = 1, r = 3, y = 0; density at a point P inversely proportional to the distance from the pole [Hint: Integrate first with respect to θ.]
  4. r = 2a cos θ; ρ(r, θ) = k (constant)

In Problems 25–32, evaluate the given iterated integral by changing to polar coordinates.

  9. The liquid hydrogen tank in the space shuttle has the form of a right circular cylinder with a semi-ellipsoidal cap at each end. The radius of the cylindrical part of the tank is 4.2 m. Find the volume of the tank shown in FIGURE 9.11.6.
    A liquid hydrogen tank inside a space shuttle has the form of a right cylindrical cylinder, with identical semi-ellipsoidal caps at each end. The heights of the semi-ellipsoidal caps measure 5.15 meters. The height of the right cylinder measures 19.3 meters.

    FIGURE 9.11.6 Fuel tank in Problem 33

  10. Evaluate (x + y) dA over the region shown in FIGURE 9.11.7.
    A graph. 2 circles are graphed on a polar coordinate system. The first circle has a radius r = 2, and its center is at the origin. The second circle, labeled r = 2 sin theta, with its diameter equal to the vertical radius of the first circle on the positive vertical polar axis. The region bound by the 2 circles and the horizontal polar axis, in the first quadrant, is shaded and labeled R.

    FIGURE 9.11.7 Region R for Problem 34

Discussion Problems

  1. The improper integral dx is important in the theory of probability, statistics, and other areas of applied mathematics. If I denotes the integral, then

    Discuss how to use polar coordinates to evaluate the last integral. Find the value of I.

 

* We could have argued to the fact that Mx = My and hence = from the fact that the lamina and the density function are symmetric about the ray θ = π/4.