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Find volume of two intersecting figures

Posted 8 years ago
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How can I calculate the volume of the intersection between a sphere xˆ2+yˆ2+zˆ2=4 and a cylinder xˆ2+yˆ2-2*x=0 Figure attached. Thanks.

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POSTED BY: Fabiano Araujo
7 Replies

The equation of the sphere is easy:

\[Rho] == 2

The equation of the cylinder is:

\[Rho] == 2 Cos[\[Phi]] Csc[\[Theta]]

See SphericalPlot3D for information about rho, theta, and phi. Theta and phi can change from one text to another.

Here is a plot:

SphericalPlot3D[{{2}, {2 Cos[\[Phi]] Csc[\[Theta]]}}, {\[Theta], 0, 
  Pi}, {\[Phi], 0, 2 Pi}]

Eric

POSTED BY: Eric Johnstone

What would be the triple integral with spherical coordinates that could solve it?

POSTED BY: Fabiano Araujo
Posted 8 years ago

This is the 3D contour plot of exactly the author's equations: {x^2 + y^2 + z^2 == 4, x^2 + y^2 - 2 x == 0}

ContourPlot3D[{x^2 + y^2 + z^2 == 4, x^2 + y^2 - 2 x == 0}, {x, -2, 
  2}, {y, -2, 2}, {z, -2, 2}, AxesOrigin -> {0, 0, 0}, 
 AxesStyle -> Thick]

enter image description here

Which gives the above plot!

To calculate the volume, you have to use solids equivalent to those equations:

Graphics3D[{Ball[{0, 0, 0}, 2], Cylinder[{{1, 0, -2}, {1, 0, 2}}, 1]},
  Axes -> True]

enter image description here

Which is different from Simon's plot due to the centre of the sphere not being the author's The volumes of the ball, the cylinder and their intersection can be perfectly calculated with Mathematica as I did in my first answer:

In[101]:= R1 = Ball[{0, 0, 0}, 2];
R2 = Cylinder[{{1, 0, -2}, {1, 0, 2}}, 1];
R = RegionIntersection[R1, R2];
Volume /@ {R1, R2, R}

Out[104]= {(32 \[Pi])/3, 4 \[Pi], 16/9 (-4 + 3 \[Pi])}
POSTED BY: Erik Mahieu

enter image description here

POSTED BY: Simon Cadrin
Posted 8 years ago

The above is not the correct plot of the two volumes of subject. The centre of the sphere is at {0,0,0} i.o at (0,01/2}. As can be seen from this plot: enter image description here

POSTED BY: Erik Mahieu

enter image description here

POSTED BY: Simon Cadrin
Posted 8 years ago

Use the geometric computation functions:

In[27]:= R1 = Ball[{0, 0, 0}, 2];
R2 = Cylinder[{{1, 0, -2}, {1, 0, 2}}, 1];
R = RegionIntersection[R1, R2];

In[32]:= RegionMeasure@R

Out[32]= 16/9 (-4 + 3 \[Pi])
POSTED BY: Erik Mahieu
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