Use spherical coordinates to find the smaller volume cut out from the sphere 𝑥^2+𝑦^2+𝑧2=4by the plane z= 1.

Question

Use spherical coordinates to find the smaller volume cut out from the sphere 𝑥^2+𝑦^2+𝑧2=4by the plane z= 1.

Did I set up the limit and integration correctly [0,2π]∫() [0,2π]∫() [sec(𝜑),2]∫()p^2 sin(𝜑)d𝜌𝑑𝜑𝑑𝜃 ?

Answer 1

Use spherical coordinates to find the smaller volume cut out from the sphere 𝑥^2 +𝑦^2 +𝑧^2= 4by the plane z= 1.

Answer 2

since the plane intersects the sphere when θ = π/3 that would make the integral

∫[0,2π]∫[0,π/3]∫[0,2] r^2 sinθ dr dθ dφ

google can provide other examples and videos

Answer 3

Can you explain how you got θ = π/3? The only part I don't get it.

Answer 4

To find the smaller volume cut out from the sphere x^2 + y^2 + z^2 = 4 by the plane z = 1, we can indeed use spherical coordinates.

The correct setup for the triple integral would be:

∫(0, 2π) ∫(0, π/4) ∫(0, 𝑟) 𝑝^2 sin(𝜑) d𝑝 d𝜑 d𝜃

The limits for each variable are as follows:

- 𝜃: The angle in the xy-plane. It ranges from 0 to 2π since it represents a complete revolution.
- 𝜑: The angle from the positive z-axis. In this case, we are interested in the region above the plane z = 1, so 𝜑 ranges from 0 to π/4.
- 𝑝: The radial distance from the origin. Since we want to find the smaller volume cut out from the sphere, 𝑝 ranges from 0 to 𝑟, where 𝑟 is the radius of the sphere. In this case, 𝑟 = 2.

Therefore, the correct setup for the triple integral is:

∫(0, 2π) ∫(0, π/4) ∫(0, 2) 𝑝^2 sin(𝜑) d𝑝 d𝜑 d𝜃

Make sure to compute the integral correctly to find the desired volume.