To integrate in polar coordinates, we must first be fluent in describing regions using them. Recall that a point $(x,y)$ can be described by a radius $r$ (distance from origin) and an angle $\theta$.
Convert the Cartesian equation $x^2 + y^2 - 2x = 0$ to polar form.
We substitute the conversion formulas. Replace $x^2+y^2$ with $r^2$ and $x$ with $r\cos\theta$.
$$ r^2 - 2(r\cos\theta) = 0 \implies r(r - 2\cos\theta) = 0 $$This gives two possibilities: $r=0$ (the origin) or $r=2\cos\theta$. This equation describes a circle of radius 1 centered at the Cartesian point (1,0).
Describe the region in the first quadrant bounded by the circle $x^2+y^2=4$, the line $y=x$, and the y-axis in polar coordinates.
The circle $x^2+y^2=4$ is $r=2$, so $0 \le r \le 2$. The line $y=x$ is $\theta=\pi/4$. The y-axis is $\theta=\pi/2$. Since we are between these lines, the angle is bounded by $\pi/4 \le \theta \le \pi/2$.
The region is $D = \{ (r, \theta) \mid 0 \le r \le 2, \pi/4 \le \theta \le \pi/2 \}$. This is a sector of a circle.
Describe the region that lies above the x-axis and inside the circle $x^2+y^2=9$ using polar coordinates.
We switch to polar coordinates when the region or the integrand is "round". The simplest case is integrating over a polar rectangle, where $a \le r \le b$ and $\alpha \le \theta \le \beta$.
In Cartesian coordinates, $dA = dx \, dy$. This represents a tiny rectangle with a fixed area, no matter where it is in the $xy$-plane.
This is not true for polar coordinates. A "polar rectangle" is a small patch bounded by two radii and two arcs. Think about its area:
The area of the patch, $dA$, depends on its distance $r$ from the origin. As $r$ increases, the arc length (given by $r \, d\theta$) increases, stretching the patch. The area is approximately (arc length) $\times$ (radial width) $\approx (r \, d\theta) \times (dr)$.
This gives us the crucial formula: $dA = r \, dr \, d\theta$. The $r$ is a scaling factor that accounts for the geometry of the polar system.
To convert from a Cartesian integral to a polar integral, we apply three changes:
$$ \iint_D f(x,y) \, dA = \int_{\alpha}^{\beta} \int_{a}^{b} f(r \cos\theta, r \sin\theta) \, \boldsymbol{r} \, dr \, d\theta $$1. The function $f(x,y)$ is rewritten in terms of $r$ and $\theta$.
2. The area element $dA$ is replaced with the polar area element $r\,dr\,d\theta$.
3. The limits of integration describe the region $D$ in polar coordinates.
The crucial step is understanding the area element. A small polar "rectangle" is a sector of an annulus. Its area $\Delta A$ is the difference between two circular sectors:
$$ \Delta A = \frac{1}{2}(r+\Delta r)^2 \Delta \theta - \frac{1}{2}r^2 \Delta \theta = (r\Delta r + \frac{1}{2}(\Delta r)^2)\Delta \theta $$For infinitesimally small changes, the $(\Delta r)^2$ term vanishes much faster than the $r\Delta r$ term, leaving $\Delta A \approx r\Delta r \Delta \theta$. In the limit, this becomes the differential area element:
$$ dA = r \, dr \, d\theta $$Find the volume of the solid under the plane $z = 4-x$ and above the disk $x^2+y^2 \le 4$.
The region is a disk of radius 2, described by $0 \le r \le 2$ and $0 \le \theta \le 2\pi$. The function is $z=4-r\cos\theta$.
$$ V = \int_0^{2\pi} \int_0^2 (4-r\cos\theta) r \, dr \, d\theta = \int_0^{2\pi} \int_0^2 (4r-r^2\cos\theta) \, dr \, d\theta $$Inner integral: $\int_0^2 (4r-r^2\cos\theta)dr = [2r^2 - \frac{r^3}{3}\cos\theta]_0^2 = 8 - \frac{8}{3}\cos\theta$.
Outer integral: $\int_0^{2\pi} (8 - \frac{8}{3}\cos\theta)d\theta = [8\theta - \frac{8}{3}\sin\theta]_0^{2\pi} = (16\pi - 0) - (0-0) = 16\pi$.
Find the volume of the solid bounded by the paraboloid $z=x^2+y^2$, the xy-plane, and the cylinders $x^2+y^2=1$ and $x^2+y^2=4$.
The region is an annulus (a washer shape) between radii 1 and 2. This is a polar rectangle: $1 \le r \le 2$ and $0 \le \theta \le 2\pi$. The function is $z = r^2$.
$$ V = \int_0^{2\pi} \int_1^2 (r^2) r \, dr \, d\theta = \int_0^{2\pi} \int_1^2 r^3 \, dr \, d\theta $$Inner integral: $\int_1^2 r^3 dr = [\frac{1}{4}r^4]_1^2 = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}$.
Outer integral: $\int_0^{2\pi} \frac{15}{4} d\theta = \frac{15}{4}[\theta]_0^{2\pi} = \frac{15\pi}{2}$.
Set up the integral to find the volume under the cone $z=\sqrt{x^2+y^2}$ over the region in the first quadrant between the circles $x^2+y^2=1$ and $x^2+y^2=9$.
More often, the radius $r$ is bounded by functions of the angle $\theta$. Imagine a radar sweep: as $\theta$ turns from $\alpha$ to $\beta$, a ray from the origin enters the region at an inner curve $r=h_1(\theta)$ and exits at an outer curve $r=h_2(\theta)$.
For a region defined as:
$$ D = \{ (r, \theta) \mid \alpha \le \theta \le \beta, \quad h_1(\theta) \le r \le h_2(\theta) \} $$The integral over this region is given by:
$$ \iint_D f(x,y) \, dA = \int_{\alpha}^{\beta} \int_{h_1(\theta)}^{h_2(\theta)} f(r \cos\theta, r \sin\theta) \, r \, dr \, d\theta $$Find the area of one petal of the rose given by $r=\cos(2\theta)$.
A petal begins and ends when $r=0$, so we solve $\cos(2\theta)=0 \implies 2\theta = \pm \pi/2 \implies \theta = \pm \pi/4$. A ray from the origin enters at $r=0$ and leaves at the curve $r=\cos(2\theta)$.
$$ A = \iint_D 1 \, dA = \int_{-\pi/4}^{\pi/4} \int_0^{\cos(2\theta)} 1 \cdot r \, dr \, d\theta $$Inner integral: $\int_0^{\cos(2\theta)} r \, dr = \left[ \frac{1}{2}r^2 \right]_0^{\cos(2\theta)} = \frac{1}{2}\cos^2(2\theta)$.
Outer integral: Use $\cos^2(u) = \frac{1}{2}(1+\cos(2u))$.
$$ \frac{1}{2} \int_{-\pi/4}^{\pi/4} \cos^2(2\theta) \, d\theta = \frac{1}{4} \int_{-\pi/4}^{\pi/4} (1+\cos(4\theta)) \, d\theta = \frac{1}{4} \left[ \theta + \frac{1}{4}\sin(4\theta) \right]_{-\pi/4}^{\pi/4} = \frac{\pi}{8}. $$Find the volume of the solid under the plane $z=y$ and above the region D bounded by the cardioid $r=1+\cos\theta$ in the upper half-plane.
The region is described by $0 \le \theta \le \pi$. A ray from the origin enters at $r=0$ and exits at the cardioid $r=1+\cos\theta$. The function is $z=y=r\sin\theta$.
$$ V = \int_0^{\pi} \int_0^{1+\cos\theta} (r\sin\theta) r \, dr \, d\theta = \int_0^{\pi} \sin\theta \left( \int_0^{1+\cos\theta} r^2 \, dr \right) d\theta $$Inner integral: $\int_0^{1+\cos\theta} r^2 \, dr = [\frac{1}{3}r^3]_0^{1+\cos\theta} = \frac{1}{3}(1+\cos\theta)^3$.
Outer integral: $\frac{1}{3} \int_0^{\pi} (1+\cos\theta)^3 \sin\theta \, d\theta$. Let $u=1+\cos\theta, du=-\sin\theta d\theta$.
$$ -\frac{1}{3} \int_{2}^{0} u^3 du = \frac{1}{3} \int_{0}^{2} u^3 du = \frac{1}{3} [\frac{1}{4}u^4]_0^2 = \frac{1}{12}(16) = \frac{4}{3}. $$Set up the integral for the area of the region inside the cardioid $r=1+\sin\theta$ and outside the circle $r=1$.
Evaluate by first converting to polar coordinates: $\int_{-1}^1 \int_0^{\sqrt{1-x^2}} \sin(x^2+y^2) \, dy \, dx$.
Find the volume of the solid bounded by the cone $z=\sqrt{x^2+y^2}$ and the paraboloid $z=x^2+y^2$.
Evaluate $\iint_D \frac{1}{(x^2+y^2)^{3/2}} \, dA$ where $D$ is the region in the first quadrant between the circles $x^2+y^2=a^2$ and $x^2+y^2=b^2$ for $0 < a < b$.
Use polar coordinates when the region $D$ or the integrand $f(x,y)$ is "round."
Convert $f(x,y)$ to $f(r\cos\theta, r\sin\theta)$ and use the correct limits.
$$ \iint_D f(x,y) \, dA = \int_{\alpha}^{\beta} \int_{h_1(\theta)}^{h_2(\theta)} f(r \cos\theta, r \sin\theta) \, \boldsymbol{r} \, dr \, d\theta $$DO NOT FORGET THE $r$! The area element is $dA = \boldsymbol{r} \, dr \, d\theta$, not just $dr \, d\theta$. This $r$ is a crucial scaling factor, not just part of the function.