15.8 triple integrals in spherical coordinates: problem 3 (1 point) the region w is the cone shown below…

Explanation:

Step1: Analyze Cartesian - coordinates

The cone has its vertex at the origin $(0,0,0)$ and top at $z = 7\sqrt{3}$. The equation of a cone with vertex at the origin and semi - vertical angle $\alpha=\frac{\pi}{3}$ is $z=\sqrt{x^{2}+y^{2}}\cot\alpha$. Since $\alpha = \frac{\pi}{3}$, $\cot\alpha=\frac{\sqrt{3}}{3}$, so $z=\frac{\sqrt{3}}{3}\sqrt{x^{2}+y^{2}}$. For the $z$ - limits, $z$ ranges from $z=\frac{\sqrt{3}}{3}\sqrt{x^{2}+y^{2}}$ to $z = 7\sqrt{3}$. To find the $x$ and $y$ limits, when $z = 7\sqrt{3}$, we have $7\sqrt{3}=\frac{\sqrt{3}}{3}\sqrt{x^{2}+y^{2}}$, then $\sqrt{x^{2}+y^{2}}=21$. So $x$ ranges from $x=- 21$ to $x = 21$ and for a given $x$, $y$ ranges from $y=-\sqrt{441 - x^{2}}$ to $y=\sqrt{441 - x^{2}}$. $a=-21$, $b = 21$, $c=-\sqrt{441 - x^{2}}$, $d=\sqrt{441 - x^{2}}$, $e=\frac{\sqrt{3}}{3}\sqrt{x^{2}+y^{2}}$, $f = 7\sqrt{3}$ Volume $=\int_{-21}^{21}\int_{-\sqrt{441 - x^{2}}}^{\sqrt{441 - x^{2}}}\int_{\frac{\sqrt{3}}{3}\sqrt{x^{2}+y^{2}}}^{7\sqrt{3}}dzdydx$

Step2: Analyze Cylindrical - coordinates

In cylindrical coordinates, $x = r\cos\theta$, $y = r\sin\theta$, $z = z$ and $dV=rdzdrd\theta$. The equation of the cone is $z=\frac{\sqrt{3}}{3}r$. The top of the cone is at $z = 7\sqrt{3}$. The range of $\theta$ is from $0$ to $2\pi$, the range of $r$ is from $0$ to $21$ (since when $z = 7\sqrt{3}$, $7\sqrt{3}=\frac{\sqrt{3}}{3}r$ gives $r = 21$), and the range of $z$ is from $z=\frac{\sqrt{3}}{3}r$ to $z = 7\sqrt{3}$. $a = 0$, $b=2\pi$, $c = 0$, $d = 21$, $e=\frac{\sqrt{3}}{3}r$, $f = 7\sqrt{3}$ Volume $=\int_{0}^{2\pi}\int_{0}^{21}\int_{\frac{\sqrt{3}}{3}r}^{7\sqrt{3}}rdzdrd\theta$

Step3: Analyze Spherical - coordinates

In spherical coordinates, $x=\rho\sin\varphi\cos\theta$, $y = \rho\sin\varphi\sin\theta$, $z=\rho\cos\varphi$ and $dV=\rho^{2}\sin\varphi d\rho d\varphi d\theta$. The semi - vertical angle of the cone is $\varphi=\frac{\pi}{3}$. The top of the cone is at $z = 7\sqrt{3}$, and since $z=\rho\cos\varphi$, when $\varphi=\frac{\pi}{3}$ and $z = 7\sqrt{3}$, we have $\rho=\frac{z}{\cos\varphi}=\frac{7\sqrt{3}}{\frac{1}{2}} = 14\sqrt{3}$. The range of $\theta$ is from $0$ to $2\pi$, the range of $\varphi$ is from $0$ to $\frac{\pi}{3}$, and the range of $\rho$ is from $0$ to $14\sqrt{3}$. $a = 0$, $b=2\pi$, $c = 0$, $d=\frac{\pi}{3}$, $e = 0$, $f = 14\sqrt{3}$ Volume $=\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{0}^{14\sqrt{3}}\rho^{2}\sin\varphi d\rho d\varphi d\theta$

Answer:

(a) $a=-21$, $b = 21$, $c=-\sqrt{441 - x^{2}}$, $d=\sqrt{441 - x^{2}}$, $e=\frac{\sqrt{3}}{3}\sqrt{x^{2}+y^{2}}$, $f = 7\sqrt{3}$, Volume $=\int_{-21}^{21}\int_{-\sqrt{441 - x^{2}}}^{\sqrt{441 - x^{2}}}\int_{\frac{\sqrt{3}}{3}\sqrt{x^{2}+y^{2}}}^{7\sqrt{3}}dzdydx$ (b) $a = 0$, $b=2\pi$, $c = 0$, $d = 21$, $e=\frac{\sqrt{3}}{3}r$, $f = 7\sqrt{3}$, Volume $=\int_{0}^{2\pi}\int_{0}^{21}\int_{\frac{\sqrt{3}}{3}r}^{7\sqrt{3}}rdzdrd\theta$ (c) $a = 0$, $b=2\pi$, $c = 0$, $d=\frac{\pi}{3}$, $e = 0$, $f = 14\sqrt{3}$, Volume $=\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{0}^{14\sqrt{3}}\rho^{2}\sin\varphi d\rho d\varphi d\theta$