5.5 Triple integrals in cylindrical and spherical coordinates  (Page 9/12)

$f\left(x,y,z\right)=x^2+y^2,$ $E=\left\{\left(x,y,z\right)|0\le x^2+y^2\le 4,y\ge 0,0\le z\le 3-x\right\}$

$f\left(x,y,z\right)=x,$ $E=\left\{\left(x,y,z\right)|1\le y^2+z^2\le 9,0\le x\le 1-y^2-z^2\right\}$

$f\left(x,y,z\right)=y,$ $E=\left\{\left(x,y,z\right)|1\le x^2+z^2\le 9,0\le y\le 1-x^2-z^2\right\}$

$E$ is above the $xy$ -plane, inside the cylinder $x^2+y^2=1,$ and below the plane $z=1.$

$E$ is below the plane $z=1$ and inside the paraboloid $z=x^2+y^2.$

$E$ is bounded by the circular cone $z=\sqrt{x^2+y^2}$ and $z=1.$

$E$ is located above the $xy$ -plane, below $z=1,$ outside the one-sheeted hyperboloid $x^2+y^2-z^2=1,$ and inside the cylinder $x^2+y^2=2.$

$E$ is located inside the cylinder $x^2+y^2=1$ and between the circular paraboloids $z=1-x^2-y^2$ and $z=x^2+y^2.$

$E$ is located inside the sphere $x^2+y^2+z^2=1,$ above the $xy$ -plane, and inside the circular cone $z=\sqrt{x^2+y^2}.$

$E$ is located outside the circular cone $x^2+y^2={\left(z-1\right)}^2$ and between the planes $z=0$ and $z=2.$

$E$ is located outside the circular cone $z=1-\sqrt{x^2+y^2},$ above the $xy$ -plane, below the circular paraboloid, and between the planes $z=0\text{and}z=2.$

[T] Use a computer algebra system (CAS) to graph the solid whose volume is given by the iterated integral in cylindrical coordinates ${\displaystyle \underset{\text{−}\pi \text{/}2}{\overset{\pi \text{/}2}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{r^2}{\overset{r}{\int }}rdzdrd\theta }}}.$ Find the volume $V$ of the solid. Round your answer to four decimal places.

[T] Use a CAS to graph the solid whose volume is given by the iterated integral in cylindrical coordinates ${\displaystyle \underset{0}{\overset{\pi \text{/}2}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{r^4}{\overset{r}{\int }}rdzdrd\theta }}}.$ Find the volume $V$ of the solid Round your answer to four decimal places.

Convert the integral ${\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{\text{−}\sqrt{1-y^2}}{\overset{\sqrt{1-y^2}}{\int }}{\displaystyle \underset{x^2+y^2}{\overset{\sqrt{x^2+y^2}}{\int }}xzdzdxdy}}}$ into an integral in cylindrical coordinates.

Convert the integral ${\displaystyle \underset{0}{\overset{2}{\int }}{\displaystyle \underset{0}{\overset{x}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}\left(xy+z\right)}}}dzdxdy$ into an integral in cylindrical coordinates.

$f(x,y,z)=1,$ $B=\left\{\left(x,y,z\right)|x^2+y^2+z^2\le 90,z\ge 0\right\}$

$f(x,y,z)=1-\sqrt{x^2+y^2+z^2},$ $B=\left\{\left(x,y,z\right)|x^2+y^2+z^2\le 9,y\ge 0,z\ge 0\right\}$

$f(x,y,z)=\sqrt{x^2+y^2},$ $B$ is bounded above by the half-sphere $x^2+y^2+z^2=9$ with $z\ge 0$ and below by the cone $2z^2=x^2+y^2.$

$f(x,y,z)=z,$ $B$ is bounded above by the half-sphere $x^2+y^2+z^2=16$ with $z\ge 0$ and below by the cone $2z^2=x^2+y^2.$

Show that if $F\left(\rho ,\theta ,\phi \right)=f\left(\rho \right)g\left(\theta \right)h\left(\phi \right)$ is a continuous function on the spherical box $B=\left\{\left(\rho ,\theta ,\phi \right)|a\le \rho \le b,\alpha \le \theta \le \beta ,\gamma \le \phi \le \psi \right\},$ then

1. A function $F$ is said to have spherical symmetry if it depends on the distance to the origin only, that is, it can be expressed in spherical coordinates as $F\left(x,y,z\right)=f(\rho ),$ where $\rho =\sqrt{x^2+y^2+z^2}.$ Show that
   
   ${\displaystyle \underset{B}{\iiint }F(x,y,z)dV}=2\pi {\displaystyle \underset{a}{\overset{b}{\int }}{\rho }^{2}f(\rho )d\rho },$
   
   where $B$ is the region between the upper concentric hemispheres of radii $a$ and $b$ centered at the origin, with $0<a<b$ and $F$ a spherical function defined on $B.$
2. Use the previous result to show that ${\displaystyle \underset{B}{\iiint }\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2}dV}=21\pi ,$ where
   
   $B=\left\{\left(x,y,z\right)|1\le x^2+y^2+z^2\le 2,z\ge 0\right\}.$

1. Let $B$ be the region between the upper concentric hemispheres of radii a and b centered at the origin and situated in the first octant, where $0<a<b.$ Consider F a function defined on B whose form in spherical coordinates $\left(\rho ,\theta ,\phi \right)$ is $F\left(x,y,z\right)=f(\rho )\text{cos}\phi .$ Show that if $g(a)=g(b)=0$ and ${\displaystyle \underset{a}{\overset{b}{\int }}h(\rho )d\rho =0},$ then
   
   ${\displaystyle \underset{B}{\iiint }F(x,y,z)dV}=\frac{{\pi }^2}{4}\left[ah(a)-bh(b)\right],$
   
   where $g$ is an antiderivative of $f$ and $h$ is an antiderivative of $g.$
2. Use the previous result to show that ${\displaystyle \underset{B}{\iiint }\frac{z\text{cos}\sqrt{x^2+y^2+z^2}}{\sqrt{x^2+y^2+z^2}}dV}=\frac{3{\pi }^2}{2},$ where $B$ is the region between the upper concentric hemispheres of radii $\pi$ and $2\pi$ centered at the origin and situated in the first octant.