# 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 xÂˆ2+yÂˆ2+zÂˆ2=4 and a cylinder xÂˆ2+yÂˆ2-2*x=0 Figure attached. Thanks. Attachments:
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Posted 8 years ago
 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 8 years ago
 What would be the triple integral with spherical coordinates that could solve it?
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] 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] 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 8 years ago
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:
Posted 8 years ago
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])