Autor Tema: Volume of cone

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03 Febrero, 2022, 03:29 pm
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jacks

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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.