Answer:
we get:
= 1/2 ʃ02π ʃ14 ( √u -1)du dθ
= ʃ02π 11/12 dθ
= 11/6 π
Step-by-step explanation:
Given:
Radius = 2 units
Plane 1 unit from center of sphere.
The volume of D as an triple integral in spherical, cylindrical and rectangular coordinates are:
Spherical:
ʃ02π ʃ0π/3 ʃsecΦ2 p2 sinΦ dp dΦ dθ
Cylindrical:
ʃ02π ʃ0√3 ʃ1√4-r2 r dz dr dθ
Rectangular:
ʃ-√3√3 ʃ-√3-x2√3-x2 ʃ1√4-x2-y2 1dz dy dx
Solving the integral by using cylindrical coordinates:
ʃ02π ʃ0√3 ʃ1√4-r2 r dz dr dθ = ʃ02π ʃ0√3 r ( √(4-r2) -1) dr dθ
put u = 4-r2, by substituting,
we get:
= 1/2 ʃ02π ʃ14 ( √u -1)du dθ
= ʃ02π 11/12 dθ
= 11/6 π
