Double Integrals in Polar Coordinates

$r = \sqrt{3 + 1} = 2$.

$\left|\tfrac{y}{x}\right| = \tfrac{1}{\sqrt3}$: reference angle $\tfrac{\pi}{6}$.

$x < 0$, $y > 0$: second quadrant, $\theta = \pi - \tfrac{\pi}{6} = \tfrac{5\pi}{6}$.

$2\cos\tfrac{5\pi}{6} = -\sqrt3$ and $2\sin\tfrac{5\pi}{6} = 1$.

$r = 4$.

$x = 0$: the point is on the $y$-axis, so skip $\tan\theta$.

Negative $y$-axis: $\theta = \tfrac{3\pi}{2}$.

$4\cos\tfrac{3\pi}{2} = 0$ and $4\sin\tfrac{3\pi}{2} = -4$.

$r = \sqrt{4 + 4} = 2\sqrt2$.

$\left|\tfrac{y}{x}\right| = 1$: reference angle $\tfrac{\pi}{4}$.

$x > 0$, $y < 0$: fourth quadrant, $\theta = \tfrac{7\pi}{4}$.

All pairs: $\left(2\sqrt2, \tfrac{7\pi}{4} + 2n\pi\right)$ or $\left(-2\sqrt2, \tfrac{3\pi}{4} + 2n\pi\right)$. Check: $-2\sqrt2\cos\tfrac{3\pi}{4} = 2$, $-2\sqrt2\sin\tfrac{3\pi}{4} = -2$.

A piece of a ring between radii $1$ and $3$, from the $x$-axis up to the line $y = x$.

$0 \le \theta \le \tfrac{\pi}{4}$.

$1 \le r \le 3$.

$\displaystyle\int_0^{\pi/4}\int_1^3 f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta$.

$(x - 2)^2 + y^2 \le 4$: center $(2, 0)$, radius $2$, touching the origin.

It lies in $x \ge 0$: $-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}$.

$r^2 \le 4r\cos\theta$, so $0 \le r \le 4\cos\theta$.

$\displaystyle\int_{-\pi/2}^{\pi/2}\int_0^{4\cos\theta} f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta$.

The triangle with vertices $(0, 0)$, $(2, 0)$, $(2, 2)$.

From $y = 0$ up to $y = x$: $0 \le \theta \le \tfrac{\pi}{4}$.

From the origin out to the line $x = 2$: $r\cos\theta = 2$, so $0 \le r \le \dfrac{2}{\cos\theta}$.

$\displaystyle\int_0^{\pi/4}\int_0^{2/\cos\theta} f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta$.

$0 \le r \le 1$, $0 \le \theta \le 2\pi$.

$e^{-r^2}$.

$\displaystyle\int_0^{2\pi}\int_0^1 e^{-r^2}\,r\,dr\,d\theta$.

$\displaystyle\int_0^1 re^{-r^2}\,dr = \tfrac12(1 - e^{-1})$; times $2\pi$: $\pi(1 - e^{-1})$.

$1 \le r \le 2$, $0 \le \theta \le \tfrac{\pi}{2}$.

$xy = r^2\cos\theta\sin\theta$.

$\displaystyle\int_0^{\pi/2}\int_1^2 r^3\cos\theta\sin\theta\,dr\,d\theta$.

$\displaystyle\int_1^2 r^3\,dr = \tfrac{15}{4}$ and $\displaystyle\int_0^{\pi/2}\sin\theta\cos\theta\,d\theta = \tfrac12$: product $\tfrac{15}{8}$.

$0 \le r < \infty$, $0 \le \theta \le 2\pi$.

$\dfrac{1}{(1 + r^2)^3}$.

$\displaystyle\int_0^{2\pi}\int_0^\infty \dfrac{r}{(1 + r^2)^3}\,dr\,d\theta$.

$u = 1 + r^2$: $\displaystyle\int_0^\infty \dfrac{r}{(1 + r^2)^3}\,dr = \lim_{R\to\infty}\left[-\dfrac{1}{4(1 + r^2)^2}\right]_0^R = \tfrac14$; times $2\pi$: $\tfrac{\pi}{2}$.

$r^2 \le 2r\cos\theta$: $-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}$, $0 \le r \le 2\cos\theta$.

$x = r\cos\theta$.

$\displaystyle\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta} r^2\cos\theta\,dr\,d\theta$.

Inner: $\cos\theta \cdot \tfrac83\cos^3\theta = \tfrac83\cos^4\theta$, so $\displaystyle\int_{-\pi/2}^{\pi/2} \tfrac83\cos^4\theta\,d\theta = \tfrac83 \cdot \tfrac{3\pi}{8} = \pi$.

$-3 \le x \le 3$, $0 \le y \le \sqrt{9 - x^2}$: the upper half of the disk of radius $3$.

$0 \le r \le 3$, $0 \le \theta \le \pi$.

$\sqrt{x^2 + y^2} = r$, times $r$: $r^2$.

$\displaystyle\int_0^\pi\int_0^3 r^2\,dr\,d\theta = \pi \cdot 9 = 9\pi$.

$0 \le x \le 1$, $x \le y \le \sqrt{2 - x^2}$: between the line $y = x$ and the circle of radius $\sqrt2$, which meet at $(1, 1)$.

$\tfrac{\pi}{4} \le \theta \le \tfrac{\pi}{2}$, $0 \le r \le \sqrt2$.

$(r\cos\theta + r\sin\theta)\,r = r^2(\cos\theta + \sin\theta)$.

$\displaystyle\int_0^{\sqrt2} r^2\,dr = \tfrac{2\sqrt2}{3}$ and $\displaystyle\int_{\pi/4}^{\pi/2} (\cos\theta + \sin\theta)\,d\theta = 1$: product $\tfrac{2\sqrt2}{3}$.

$y = \sqrt{6x - x^2}$ is the top half of $(x - 3)^2 + y^2 = 9$: the upper half of the disk centered at $(3, 0)$.

$x^2 + y^2 = 6x$ is $r = 6\cos\theta$: $0 \le \theta \le \tfrac{\pi}{2}$, $0 \le r \le 6\cos\theta$.

$\dfrac{1}{r} \cdot r = 1$.

$\displaystyle\int_0^{\pi/2}\int_0^{6\cos\theta} 1\,dr\,d\theta = \int_0^{\pi/2} 6\cos\theta\,d\theta = 6$.

$9 - r^2 = 0$ at $r = 3$.

$9 - r^2$.

$0 \le r \le 3$, $0 \le \theta \le 2\pi$.

$2\pi\displaystyle\int_0^3 (9 - r^2)\,r\,dr = 2\pi\left(\tfrac{81}{2} - \tfrac{81}{4}\right) = \tfrac{81\pi}{2}$.

$r = 6 - r^2$, so $r^2 + r - 6 = 0$ and $r = 2$.

$6 - r^2 - r$.

$0 \le r \le 2$, $0 \le \theta \le 2\pi$.

$2\pi\displaystyle\int_0^2 (6r - r^2 - r^3)\,dr = 2\pi\left(12 - \tfrac83 - 4\right) = \tfrac{32\pi}{3}$.

$\sqrt{4 - r^2} = 1$ at $r = \sqrt3$.

$\sqrt{4 - r^2} - 1$.

$0 \le r \le \sqrt3$, $0 \le \theta \le 2\pi$.

$2\pi\displaystyle\int_0^{\sqrt3} \left(r\sqrt{4 - r^2} - r\right)\,dr = 2\pi\left(\tfrac73 - \tfrac32\right) = \tfrac{5\pi}{3}$.