Section 15.3: Integration in Polar Coordinates

Suppose that we want to evaluate a double integral \(\iint_R f(x, y) dA\), where \(R\) is one of the regions shown in Figure 1.

In either case the description of \(R\) in terms of rectangular coordinates is rather complicated, but \(R\) is easily described using polar coordinates.

Recall from Figur2 that the polar coordinates \((r, \theta)\) of a point are related to the rectangular coordinates \((x, y)\) by the equations \[ r^2 = x^2 + y^2 \quad x = r \cos \theta \quad y = r \sin \theta \]

The regions in Figure 1 are special cases of a polar rectangle \[ R = \{(r, \theta) | a \le r \le b, \alpha \le \theta \le \beta\} \] which is shown in Figure 3.

If \(f\) is continuous on a polar rectangle \(R\) given by \(0 \le a \le r \le b, \alpha \le \theta \le \beta\), where \(0 \le \beta - \alpha \le 2\pi\), then \[ \iint_R f(x, y) dA = \int_\alpha^\beta \int_a^b f(r \cos \theta, r \sin \theta) r dr d\theta \tag{2} \]

The formula in (2) says that we convert from rectangular to polar coordinates in a double integral by writing \(x = r \cos \theta\) and \(y = r \sin \theta\), using the appropriate limits of integration for \(r\) and \(\theta\), and replacing \(dA\) by \(r dr d\theta\). Be careful not to forget the additional factor \(r\) on the right side of Formula 2. A classical method for remembering this is shown in Figure 5, where the “infinitesimal” polar rectangle can be thought of as an ordinary rectangle with dimensions \(r d\theta\) and \(dr\) and therefore has “area” \(dA = r dr d\theta\).

EXAMPLE 1 Evaluate \(\iint_R (3x + 4y^2) dA\), where \(R\) is the region in the upper half-plane bounded by the circles \(x^2 + y^2 = 1\) and \(x^2 + y^2 = 4\).

EXAMPLE 2 Find the volume of the solid bounded by the plane \(z = 0\) and the paraboloid \(z = 1 - x^2 - y^2\).

What we have done so far can be extended to the more complicated type of region shown in Figure 7. It’s similar to the type II rectangular regions considered in Section 15.2. In fact, by combining Formula 2 in this section with Formula 15.2.5, we obtain the following formula.

In particular, taking \(f(x, y) = 1, h_1(\theta) = 0,\) and \(h_2(\theta) = h(\theta)\) in this formula, we see that the area of the region \(D\) bounded by \(\theta = \alpha, \theta = \beta,\) and \(r = h(\theta)\) is \[ A(D) = \iint_D 1 dA = \int_\alpha^\beta \int_0^{h(\theta)} r dr d\theta = \int_\alpha^\beta \left[ \frac{r^2}{2} \right]_0^{h(\theta)} d\theta = \int_\alpha^\beta \frac{1}{2}[h(\theta)]^2 d\theta \] and this agrees with Formula 10.4.3.

EXAMPLE 3 Use a double integral to find the area enclosed by one loop of the four-leaved rose \(r = \cos 2\theta\).

EXAMPLE 4 Find the volume of the solid that lies under the paraboloid \(z = x^2 + y^2\), above the \(xy\)-plane, and inside the cylinder \(x^2 + y^2 = 2x\).

A region R is shown. Decide whether to use polar coordinates or rectangular coordinates and write \(\iint_R f(x, y) dA\) as an iterated integral, where \(f\) is an arbitrary continuous function on R.

A region R is shown. Decide whether to use polar coordinates or rectangular coordinates and write \(\iint_R f(x, y) dA\) as an iterated integral, where \(f\) is an arbitrary continuous function on R.

A region R is shown. Decide whether to use polar coordinates or rectangular coordinates and write \(\iint_R f(x, y) dA\) as an iterated integral, where \(f\) is an arbitrary continuous function on R.

A region R is shown. Decide whether to use polar coordinates or rectangular coordinates and write \(\iint_R f(x, y) dA\) as an iterated integral, where \(f\) is an arbitrary continuous function on R.

Sketch the region whose area is given by the integral \(\int_{\pi/4}^{3\pi/4} \int_1^2 r dr d\theta\) and evaluate the integral.

Sketch the region whose area is given by the integral \(\int_{\pi/2}^{\pi} \int_0^{2\sin\theta} r dr d\theta\) and evaluate the integral.

Evaluate the given integral by changing to polar coordinates: \(\iint_D x^2y dA\), where \(D\) is the top half of the disk with center the origin and radius 5.

Evaluate the given integral by changing to polar coordinates: \(\iint_R (2x - y) dA\), where \(R\) is the region in the first quadrant enclosed by the circle \(x^2 + y^2 = 4\) and the lines \(x = 0\) and \(y = x\).

Evaluate the given integral by changing to polar coordinates: \(\iint_R \sin(x^2 + y^2) dA\), where \(R\) is the region in the first quadrant between the circles with center the origin and radii 1 and 3.

Evaluate the given integral by changing to polar coordinates: \(\iint_R \frac{y^2}{x^2 + y^2} dA\), where \(R\) is the region that lies between the circles \(x^2 + y^2 = a^2\) and \(x^2 + y^2 = b^2\) with \(0 < a < b\).

Evaluate the given integral by changing to polar coordinates: \(\iint_D e^{-x^2-y^2} dA\), where \(D\) is the region bounded by the semicircle \(x = \sqrt{4 - y^2}\) and the y-axis.

Evaluate the given integral by changing to polar coordinates: \(\iint_D \cos\sqrt{x^2 + y^2} dA\), where \(D\) is the disk with center the origin and radius 2.

Evaluate the given integral by changing to polar coordinates: \(\iint_R \arctan(y/x) dA\), where \(R = \{(x, y) | 1 \le x^2 + y^2 \le 4, 0 \le y \le x\}\).

Evaluate the given integral by changing to polar coordinates: \(\iint_D x dA\), where \(D\) is the region in the first quadrant that lies between the circles \(x^2 + y^2 = 4\) and \(x^2 + y^2 = 2x\).

Use a double integral to find the area of the region. One loop of the rose \(r = \cos 3\theta\).

Use a double integral to find the area of the region enclosed by both of the cardioids \(r = 1 + \cos\theta\) and \(r = 1 - \cos\theta\).

Use a double integral to find the area of the region inside the circle \((x - 1)^2 + y^2 = 1\) and outside the circle \(x^2 + y^2 = 1\).

Use a double integral to find the area of the region inside the cardioid \(r = 1 + \cos\theta\) and outside the circle \(r = 3 \cos\theta\).

Use polar coordinates to find the volume of the solid under the paraboloid \(z = x^2 + y^2\) and above the disk \(x^2 + y^2 \le 25\).

Use polar coordinates to find the volume of the solid below the cone \(z = \sqrt{x^2 + y^2}\) and above the ring \(1 \le x^2 + y^2 \le 4\).

Use polar coordinates to find the volume of the solid below the plane \(2x + y + z = 4\) and above the disk \(x^2 + y^2 \le 1\).

Use polar coordinates to find the volume of the solid inside the sphere \(x^2 + y^2 + z^2 = 16\) and outside the cylinder \(x^2 + y^2 = 4\).

Use polar coordinates to find the volume of a sphere of radius \(a\).

Use polar coordinates to find the volume of the solid bounded by the paraboloid \(z = 1 + 2x^2 + 2y^2\) and the plane \(z = 7\) in the first octant.

Use polar coordinates to find the volume of the solid above the cone \(z = \sqrt{x^2 + y^2}\) and below the sphere \(x^2 + y^2 + z^2 = 1\).

Use polar coordinates to find the volume of the solid bounded by the paraboloids \(z = 6 - x^2 - y^2\) and \(z = 2x^2 + 2y^2\).

Use polar coordinates to find the volume of the solid inside both the cylinder \(x^2 + y^2 = 4\) and the ellipsoid \(4x^2 + 4y^2 + z^2 = 64\).

1. A cylindrical drill with radius \(r_1\) is used to bore a hole through the center of a sphere of radius \(r_2\). Find the volume of the ring-shaped solid that remains.
2. Express the volume in part (a) in terms of the height \(h\) of the ring. Notice that the volume depends only on \(h\), not on \(r_1\) or \(r_2\).

Evaluate the iterated integral by converting to polar coordinates: \(\int_{-2}^2 \int_0^{\sqrt{4-x^2}} e^{-x^2-y^2} dy dx\).

Evaluate the iterated integral by converting to polar coordinates: \(\int_0^a \int_{-\sqrt{a^2-y^2}}^{\sqrt{a^2-y^2}} (2x + y) dx dy\).

Evaluate the iterated integral by converting to polar coordinates: \(\int_0^{1/2} \int_{\sqrt{3}y}^{\sqrt{1-y^2}} xy^2 dx dy\).

Evaluate the iterated integral by converting to polar coordinates: \(\int_0^2 \int_0^{\sqrt{2x-x^2}} \sqrt{x^2 + y^2} dy dx\).

Express the double integral \(\iint_D e^{(x^2+y^2)^2} dA\) in terms of a single integral with respect to \(r\), where \(D\) is the disk with center the origin and radius 1. Then use your calculator to evaluate the integral correct to four decimal places.

Express the double integral \(\iint_D xy\sqrt{1 + x^2 + y^2} dA\) in terms of a single integral with respect to \(r\), where \(D\) is the portion of the disk \(x^2 + y^2 \le 1\) that lies in the first quadrant. Then use your calculator to evaluate the integral correct to four decimal places.

An agricultural sprinkler distributes water in a circular pattern of radius 100 ft. It supplies water to a depth of \(e^{-r}\) feet per hour at a distance of \(r\) feet from the sprinkler. (a) If \(0 < R \le 100\), what is the total amount of water supplied per hour to the region inside the circle of radius \(R\) centered at the sprinkler? (b) Determine an expression for the average amount of water per hour per square foot supplied to the region inside the circle of radius \(R\).

Find the average value of the function \(f(x, y) = 1/\sqrt{x^2 + y^2}\) on the annular region \(a^2 \le x^2 + y^2 \le b^2\), where \(0 < a < b\).

Let \(D\) be the disk with center the origin and radius \(a\). What is the average distance from points in \(D\) to the origin?

Use polar coordinates to combine the sum \[ \int_{1/\sqrt{2}}^1 \int_{\sqrt{1-x^2}}^x xy dy dx + \int_1^{\sqrt{2}} \int_0^x xy dy dx + \int_{\sqrt{2}}^2 \int_0^{\sqrt{4-x^2}} xy dy dx \] into one double integral. Then evaluate the double integral.

1. We define the improper integral (over the entire plane \(\mathbb{R}^2\)) \(I = \iint_{\mathbb{R}^2} e^{-(x^2+y^2)} dA = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} dy dx = \lim_{a \to \infty} \iint_{D_a} e^{-(x^2+y^2)} dA\) where \(D_a\) is the disk with radius \(a\) and center the origin. Show that \(\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} dA = \pi\).
2. An equivalent definition of the improper integral in part (a) is \(\iint_{\mathbb{R}^2} e^{-(x^2+y^2)} dA = \lim_{a \to \infty} \iint_{S_a} e^{-(x^2+y^2)} dA\) where \(S_a\) is the square with vertices \((\pm a, \pm a)\). Use this to show that \(\int_{-\infty}^\infty e^{-x^2} dx \int_{-\infty}^\infty e^{-y^2} dy = \pi\).
3. Deduce that \[ \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}. \]
4. By making the change of variable \(t = \sqrt{2}x\), show that \(\int_{-\infty}^\infty e^{-x^2/2} dx = \sqrt{2\pi}\). (This is a fundamental result for probability and statistics.)

Use the result of Exercise 40 part (c) to evaluate the following integrals. (a) \(\int_0^\infty x^2e^{-x^2} dx\) (b) \(\int_0^\infty \sqrt{x}e^{-x} dx\)