### Autor Tema: Volume of cone

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03 Febrero, 2022, 03:29 pm
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#### jacks

• $$\Large \color{#c88359}\pi\,\pi\,\pi\,\pi$$
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Evaluation of solid that lies inside the sphere

$\displaystyle x^2+y^2+z^2 =4z$ and between the cone $\displaystyle \phi =\frac{\pi}{4}$ and $\displaystyle \phi =\frac{\pi}{3}$

03 Febrero, 2022, 05:47 pm
Respuesta #1

#### Masacroso

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Evaluation of solid that lies inside the sphere

$\displaystyle x^2+y^2+z^2 =4z$ and between the cone $\displaystyle \phi =\frac{\pi}{4}$ and $\displaystyle \phi =\frac{\pi}{3}$

The sphere is centered in $(0,0,2)$ and have radius $2$ as

$\displaystyle{ x^2+y^2+z^2=4z\iff x^2+y^2+z^2-4z+4=4\iff x^2+y^2+(z-2)^2=4 }$

Therefore a point is inside the sphere when $0\leqslant x^2+y^2+(z-2)^2\leqslant 4$. Using spherical coordinates $(x,y,z)=r(\cos \phi \sin \theta ,\sin \phi \sin \theta ,\cos \theta )$ this gives

$\displaystyle{ 0\leqslant x^2+y^2+(z-2)^2\leqslant 4 \iff 0\leqslant r^2-4r\cos \theta +4\leqslant 4 }$

From the last expression we can see that, at most, $r=4$. Then if $r\in [0,4]$ we find from the same expression that $\cos \theta \in [r/4,1]$. Altogether this gives the conditions

$\displaystyle{ r\in [0,4],\,\theta \in [0,\arccos (r/4)] \text{ and }\phi \in [\pi/4,\pi/3] }$

Then, if my previous calculations are all right, the volume of the object is

$\displaystyle{ \int_{0}^4\int_{0}^{\arccos (r/4)}\int_{\pi/4}^{\pi/3}r^2 \sin \theta \,d \phi \,d \theta\,d r=\frac{\pi}{12}\int_{0}^4(1-r/4)\,d r=\frac{\pi }{6} }$

Addition: probably something is wrong as I used the conditions $\phi =\pi/3$ and $\phi =\pi/4$ as semi-planes. I dont see how to define cones from it.

04 Febrero, 2022, 03:40 am
Respuesta #2

#### delmar

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Hello

There are two ways A and B In A it is American spherical coordinate and B  it is International spherical coordinate. Solid corresponds to the dotted region, it is solid of revolution

A) american spherical coordinates $(R,\theta, \phi) \ \ \ 0\leq{R}<\infty, \ \ \ 0\leq{\theta}\leq{2 \pi}, \ \ \ \displaystyle\frac{- \pi}{2}\leq{\phi}\leq{\displaystyle\frac{\pi}{2}}$

the equation of the sphere is $x^2+y^2+z^2=4z$

$x=Rcos \phi \ cos \theta$

$y=R cos \phi \ sen \theta$

$z=R sen \phi$

substituting in the equation $R=4 sen \phi$

Then volume of the solid $V=\displaystyle\int_{\pi/4}^{\pi/3}\displaystyle\int_{0}^{2 \pi}\displaystyle\int_{0}^{4 sen \phi}R^2 \ cos \phi \ dR \ d \theta \ d \phi$

Similar for B)

regards

04 Febrero, 2022, 09:00 am
Respuesta #3

#### Masacroso

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Coming again to this problem: I was trying to solve it for cones of revolution around the $X$-axis instead of the $Z$-axis. This gives the conditions of the solid, as the intersection of the ball and the filled cone (as a volume of revolution around the positive $X$-semiaxis), as

$\displaystyle{ x^2+y^2+(z-2)^2\leqslant 4\quad \text{ and }\quad \sqrt{y^2+z^2}\leqslant x K,\quad \text{ where }K:=\tan \alpha \tag1 }$

There $\alpha \in[0,\pi/2)$ is the angle, respect to $X$-axis, that define the cone of revolution. Now, using the change of coordinates $(x,y,z)=(x, \rho \cos \beta ,\rho \sin \beta )$ then (1) can be re-written as

$\displaystyle{ x\in [0,2],\quad \rho \in[0,x K]\quad \text{ and }\quad \beta \in \left[\arcsin \left(\frac{x^2+\rho ^2}{4\rho }\right),\frac{\pi}{2}\right] \tag2 }$

However the integrals defined by (2) seems complicate to handle or intractable.

Wolfram Mathematica can compute the volume defined by (1) given an exact answer, but it does using a built-in function of volumes over defined regions. Trying to solve the integrals defined by (2) in Mathematica it seem unable to compute it. This means that (probably) it must be a clever way to compute the volume defined by (1) different to the use of cylindrical coordinates.