Triple Integrals in Spherical Coordinates

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

$x = 2\sqrt{2}\cos\tfrac{\pi}{6} = \sqrt{6}$ and $y = 2\sqrt{2}\sin\tfrac{\pi}{6} = \sqrt{2}$.

$\varphi > \tfrac{\pi}{2}$ and $z < 0$ agree; $\theta = \tfrac{\pi}{6}$ and $x, y > 0$ agree.

$\rho = \sqrt{1 + 1 + 6} = 2\sqrt{2}$, and $\cos\varphi = \dfrac{\sqrt{6}}{2\sqrt{2}} = \dfrac{\sqrt{3}}{2}$, so $\varphi = \tfrac{\pi}{6}$.

$\tan\theta = \dfrac{-1}{1} = -1$, with reference angle $\tfrac{\pi}{4}$.

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

$\rho = \sqrt{3 + 0 + 1} = 2$, and $\cos\varphi = -\tfrac{1}{2}$, so $\varphi = \tfrac{2\pi}{3}$.

$y = 0$ and $x < 0$: the point is over the negative $x$-axis, so $\theta = \pi$.

$z < 0$ and $\varphi > \tfrac{\pi}{2}$ agree.

No full $x^2 + y^2 + z^2$ group.

$\rho^2\sin^2\varphi = 3\rho^2\cos^2\varphi$.

$\tan^2\varphi = 3$, so $\varphi = \tfrac{\pi}{3}$ or $\varphi = \tfrac{2\pi}{3}$.

$\rho^2 = 4z$.

$\rho^2 = 4\rho\cos\varphi$.

$\rho = 4\cos\varphi$, for $0 \le \varphi \le \tfrac{\pi}{2}$.

No full $x^2 + y^2 + z^2$ group.

$\rho^2\sin^2\varphi = 9$.

$\rho\sin\varphi = 3$, since $\rho \ge 0$ and $\sin\varphi \ge 0$. So $\rho = 3\csc\varphi$.

No $x^2 + y^2 + z^2$ group.

$\rho\cos\varphi = 2$.

$\rho = 2\sec\varphi$, for $0 \le \varphi < \tfrac{\pi}{2}$.

$0 \le \rho \le 3$. The cone is $\varphi = \tfrac{\pi}{4}$ and the $xy$-plane is $\varphi = \tfrac{\pi}{2}$, so $\tfrac{\pi}{4} \le \varphi \le \tfrac{\pi}{2}$, and $0 \le \theta \le 2\pi$.

$f = 1$: $V = \displaystyle\int_0^{2\pi}\int_{\pi/4}^{\pi/2}\int_0^3 \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$.

$9 \cdot \Big[-\cos\varphi\Big]_{\pi/4}^{\pi/2} \cdot 2\pi = 9 \cdot \tfrac{\sqrt{2}}{2} \cdot 2\pi = 9\sqrt{2}\,\pi$.

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

$z = \rho\cos\varphi$: $\displaystyle\int_0^{2\pi}\int_0^{\pi/2}\int_1^2 \rho^3\cos\varphi\sin\varphi\,d\rho\,d\varphi\,d\theta$.

$\displaystyle\int_1^2 \rho^3\,d\rho = \tfrac{15}{4}$ and $\displaystyle\int_0^{\pi/2}\cos\varphi\sin\varphi\,d\varphi = \tfrac{1}{2}$, so the total is $\tfrac{15}{4} \cdot \tfrac{1}{2} \cdot 2\pi = \tfrac{15\pi}{4}$.

All of space: $0 \le \rho < \infty$, $0 \le \varphi \le \pi$, $0 \le \theta \le 2\pi$.

$(x^2 + y^2 + z^2)^{3/2} = \rho^3$: $\displaystyle\int_0^{2\pi}\int_0^{\pi}\int_0^{\infty} e^{-\rho^3}\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$.

$\displaystyle\int_0^\infty \rho^2 e^{-\rho^3}\,d\rho = \tfrac{1}{3}$ and $\displaystyle\int_0^\pi \sin\varphi\,d\varphi = 2$, so the total is $\tfrac{1}{3} \cdot 2 \cdot 2\pi = \tfrac{4\pi}{3}$.

Expanding, the sphere is $\rho^2 = 4\rho\cos\varphi$, so $\rho = 4\cos\varphi$; the cone is $\varphi = \tfrac{\pi}{4}$. So $0 \le \rho \le 4\cos\varphi$, $0 \le \varphi \le \tfrac{\pi}{4}$, $0 \le \theta \le 2\pi$.

$f = 1$: $V = \displaystyle\int_0^{2\pi}\int_0^{\pi/4}\int_0^{4\cos\varphi} \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$.

Inner: $\tfrac{64}{3}\cos^3\varphi$. Then $\tfrac{64}{3}\displaystyle\int_0^{\pi/4}\cos^3\varphi\sin\varphi\,d\varphi = \tfrac{64}{3} \cdot \tfrac{3}{16} = 4$, and $4 \cdot 2\pi = 8\pi$.

$z = \tfrac{r}{\sqrt{3}}$ in $r^2 + z^2 = 4$ gives $\tfrac{4r^2}{3} = 4$, so $r = \sqrt{3}$ (at $z = 1$). And $\tan\varphi = \sqrt{3}$, so $\varphi = \tfrac{\pi}{3}$.

$\displaystyle\int_0^{2\pi}\int_0^{\sqrt{3}}\int_{r/\sqrt{3}}^{\sqrt{4 - r^2}} z\,r\,dz\,dr\,d\theta$.

$\displaystyle\int_0^{2\pi}\int_0^{\pi/3}\int_0^2 (\rho\cos\varphi)\,\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$.

Spherical factors: $\displaystyle\int_0^2 \rho^3\,d\rho = 4$ and $\displaystyle\int_0^{\pi/3}\cos\varphi\sin\varphi\,d\varphi = \tfrac{3}{8}$, so $2\pi \cdot 4 \cdot \tfrac{3}{8} = 3\pi$.

$z = r$ meets $z = 3$ at $r = 3$. The cone is $\varphi = \tfrac{\pi}{4}$.

$V = \displaystyle\int_0^{2\pi}\int_0^3\int_r^3 r\,dz\,dr\,d\theta$.

The plane is $\rho\cos\varphi = 3$: $V = \displaystyle\int_0^{2\pi}\int_0^{\pi/4}\int_0^{3\sec\varphi}\rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$.

Cylindrical: $2\pi\displaystyle\int_0^3 (3r - r^2)\,dr = 2\pi\left(\tfrac{27}{2} - 9\right) = 9\pi$. Check: a cone of radius $3$ and height $3$ has volume $\tfrac{1}{3}\pi(9)(3) = 9\pi$.

$0 \le z \le \sqrt{4 - x^2 - y^2}$ over the quarter disk $x, y \ge 0$, $x^2 + y^2 \le 4$: the part of the ball of radius $2$ in the first octant.

$0 \le \rho \le 2$, $0 \le \varphi \le \tfrac{\pi}{2}$, $0 \le \theta \le \tfrac{\pi}{2}$.

$\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}\int_0^2 \rho \cdot \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$.

$\displaystyle\int_0^2 \rho^3\,d\rho \cdot \int_0^{\pi/2}\sin\varphi\,d\varphi \cdot \tfrac{\pi}{2} = 4 \cdot 1 \cdot \tfrac{\pi}{2} = 2\pi$.

Between the lower half of the unit sphere and $z = 0$, over the whole unit disk: the lower half ball.

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

$z^2 = \rho^2\cos^2\varphi$: $\displaystyle\int_0^{2\pi}\int_{\pi/2}^{\pi}\int_0^1 \rho^4\cos^2\varphi\sin\varphi\,d\rho\,d\varphi\,d\theta$.

$\tfrac{1}{5} \cdot \Big[-\tfrac{\cos^3\varphi}{3}\Big]_{\pi/2}^{\pi} \cdot 2\pi = \tfrac{1}{5} \cdot \tfrac{1}{3} \cdot 2\pi = \tfrac{2\pi}{15}$.

Above the cone $z = \sqrt{x^2 + y^2}$ and below the sphere of radius $\sqrt{2}$. They meet at $r = 1$, the edge of the unit-disk shadow.

$0 \le \rho \le \sqrt{2}$, $0 \le \varphi \le \tfrac{\pi}{4}$, $0 \le \theta \le 2\pi$.

$\dfrac{1}{\rho} \cdot \rho^2\sin\varphi = \rho\sin\varphi$: $\displaystyle\int_0^{2\pi}\int_0^{\pi/4}\int_0^{\sqrt{2}} \rho\sin\varphi\,d\rho\,d\varphi\,d\theta$.

$\displaystyle\int_0^{\sqrt{2}} \rho\,d\rho = 1$ and $\displaystyle\int_0^{\pi/4}\sin\varphi\,d\varphi = 1 - \tfrac{\sqrt{2}}{2}$, so the total is $2\pi\left(1 - \tfrac{\sqrt{2}}{2}\right) = \pi(2 - \sqrt{2})$.