Find volume of two intersecting figures

Questions
1

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.

2

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])
3

6873re03.png
6873re03.png675×534 46.4 KB

4

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:

araujo.png
araujo.png472×468 103 KB

5

7383re02.png
7383re02.png640×250 9.2 KB

6

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]

araujo.png
araujo.png472×468 103 KB

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]

araujo-2.png
araujo-2.png519×515 53 KB

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])}
7

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

8

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