Triple Integrals in Cylindrical Coordinates

$z = -1$.

$x = 4\cos\tfrac{2\pi}{3} = -2$ and $y = 4\sin\tfrac{2\pi}{3} = 2\sqrt{3}$.

$\tfrac{2\pi}{3}$ is in quadrant II, and $x < 0$, $y > 0$ agree.

$z = 5$.

$r = \sqrt{3 + 1} = 2$, and $\tan\theta = -\tfrac{1}{\sqrt{3}}$, with reference angle $\tfrac{\pi}{6}$.

$x > 0$, $y < 0$: quadrant IV, so $\theta = \tfrac{11\pi}{6}$ (or $-\tfrac{\pi}{6}$).

$z = 2$.

$r = \sqrt{9 + 27} = 6$, and $\tan\theta = \sqrt{3}$, with reference angle $\tfrac{\pi}{3}$.

$x < 0$, $y < 0$: quadrant III, so $\theta = \pi + \tfrac{\pi}{3} = \tfrac{4\pi}{3}$.

$z = 9 - (x^2 + y^2) = 9 - r^2$.

No $x$ or $y$ is left.

$z = 9 - r^2$, a paraboloid opening down.

$r^2 = 6y$.

$r^2 = 6r\sin\theta$.

$r = 6\sin\theta$: the cylinder over the circle of radius $3$ centered at $(0, 3)$.

No $x^2 + y^2$ group: the sign between them is minus.

$z = r^2\cos^2\theta - r^2\sin^2\theta$.

$z = r^2\cos 2\theta$.

$r^2 \le z \le 8 - r^2$.

They meet where $r^2 = 8 - r^2$, so $r = 2$: $0 \le r \le 2$, $0 \le \theta \le 2\pi$.

$f = 1$: $V = \displaystyle\int_0^{2\pi}\int_0^2\int_{r^2}^{8 - r^2} r\,dz\,dr\,d\theta$.

$\displaystyle\int_0^{2\pi}\int_0^2 (8 - 2r^2)\,r\,dr\,d\theta = 2\pi(16 - 8) = 16\pi$.

$0 \le z \le 4 - r^2$.

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

$\displaystyle\int_0^{2\pi}\int_0^1\int_0^{4 - r^2} z\,r\,dz\,dr\,d\theta$.

Inner: $\tfrac{r(4 - r^2)^2}{2}$. Then $\displaystyle\int_0^1 \tfrac{r(4 - r^2)^2}{2}\,dr = \tfrac{37}{12}$, and $2\pi \cdot \tfrac{37}{12} = \tfrac{37\pi}{6}$.

$0 \le z \le r^2$.

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

$f = 1$: $V = \displaystyle\int_{-\pi/2}^{\pi/2}\int_0^{2\cos\theta}\int_0^{r^2} r\,dz\,dr\,d\theta$.

$\displaystyle\int_0^{2\cos\theta} r^3\,dr = 4\cos^4\theta$, and $\displaystyle\int_{-\pi/2}^{\pi/2} 4\cos^4\theta\,d\theta = 4 \cdot \tfrac{3\pi}{8} = \tfrac{3\pi}{2}$.

$0 \le z \le r\cos\theta$.

The top is above the bottom only where $x \ge 0$: $-\tfrac{\pi}{2} \le \theta \le \tfrac{\pi}{2}$, $0 \le r \le 3$.

$f = 1$: $V = \displaystyle\int_{-\pi/2}^{\pi/2}\int_0^3\int_0^{r\cos\theta} r\,dz\,dr\,d\theta$.

The $z$-integral gives $r^2\cos\theta$. Then $\displaystyle\int_{-\pi/2}^{\pi/2}\cos\theta\,d\theta \cdot \int_0^3 r^2\,dr = 2 \cdot 9 = 18$.

$0 \le z \le 4 - r^2$, $0 \le r \le 2$, $0 \le \theta \le 2\pi$.

Symmetric around the $z$-axis, constant density: $\bar x = \bar y = 0$.

$V = \displaystyle\int_0^{2\pi}\int_0^2 (4 - r^2)\,r\,dr\,d\theta = 2\pi(8 - 4) = 8\pi$.

$\displaystyle\iiint z\,dV = 2\pi\int_0^2 \tfrac{(4 - r^2)^2}{2}\,r\,dr = \tfrac{32\pi}{3}$, so $\bar z = \dfrac{32\pi/3}{8\pi} = \dfrac{4}{3}$.

$1 \le r \le 3$, $0 \le \theta \le 2\pi$, $0 \le z \le 2$, and $\rho = r^2$.

$\bar x = \bar y = 0$.

$M = \displaystyle\int_0^{2\pi}\int_1^3\int_0^2 r^2 \cdot r\,dz\,dr\,d\theta = 2\pi \cdot 2 \cdot \tfrac{81 - 1}{4} = 80\pi$.

$\displaystyle\iiint z\rho\,dV = 2\pi \cdot \tfrac{81 - 1}{4} \cdot 2 = 80\pi$, so $\bar z = \dfrac{80\pi}{80\pi} = 1$, the middle height.

$0 \le r \le 1$, $0 \le \theta \le 2\pi$, $0 \le z \le 3$.

$\bar x = \bar y = 0$.

$M = \displaystyle\int_0^{2\pi}\int_0^1\int_0^3 z\,r\,dz\,dr\,d\theta = 2\pi \cdot \tfrac{1}{2} \cdot \tfrac{9}{2} = \tfrac{9\pi}{2}$.

$\displaystyle\iiint z \cdot z\,dV = 2\pi \cdot \tfrac{1}{2} \cdot 9 = 9\pi$, so $\bar z = \dfrac{9\pi}{9\pi/2} = 2$.

$0 \le z \le 4 - (x^2 + y^2) = 4 - r^2$.

$-2 \le x \le 2$ with $|y| \le \sqrt{4 - x^2}$ is the whole disk of radius $2$: $0 \le r \le 2$, $0 \le \theta \le 2\pi$.

$\displaystyle\int_0^{2\pi}\int_0^2\int_0^{4 - r^2} r \cdot r\,dz\,dr\,d\theta$.

$2\pi\displaystyle\int_0^2 r^2(4 - r^2)\,dr = 2\pi\left(\tfrac{32}{3} - \tfrac{32}{5}\right) = \tfrac{128\pi}{15}$.

$r^2 \le z \le 1$.

$0 \le x \le 1$, $0 \le y \le \sqrt{1 - x^2}$: the quarter disk in quadrant I, so $0 \le r \le 1$, $0 \le \theta \le \tfrac{\pi}{2}$.

$\displaystyle\int_0^{\pi/2}\int_0^1\int_{r^2}^{1} r\cos\theta \cdot r\,dz\,dr\,d\theta$.

$\displaystyle\int_0^{\pi/2}\cos\theta\,d\theta \cdot \int_0^1 r^2(1 - r^2)\,dr = 1 \cdot \left(\tfrac{1}{3} - \tfrac{1}{5}\right) = \tfrac{2}{15}$.