How to Find the Volume of a Bounded Region with Sphere and Cone Equations

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SUMMARY

The volume of the region bounded above by the unit sphere defined by the equation x²+y²+z²=1 and below by the cone z=sqrt(x²+y²) can be calculated using a triple integral in spherical coordinates. The correct setup involves integrating from 0 to 2π for φ, from 0 to π/4 for θ, and from 0 to 1 for r, resulting in the integral ∫₀²π ∫₀^(π/4) ∫₀¹ (r² sinθ dr dθ dφ). This approach effectively captures the volume between the two surfaces.

PREREQUISITES
  • Understanding of spherical coordinates and their application in volume calculations.
  • Familiarity with the equations of a sphere and a cone.
  • Knowledge of triple integrals and their setup in calculus.
  • Ability to interpret and plot three-dimensional graphs.
NEXT STEPS
  • Study the application of triple integrals in cylindrical coordinates.
  • Learn about the derivation and use of spherical coordinates in volume calculations.
  • Explore examples of volume calculations involving bounded regions in calculus.
  • Review the properties and equations of conic sections and their geometric interpretations.
USEFUL FOR

Students studying calculus, particularly those focusing on multivariable calculus and volume calculations, as well as educators seeking to clarify concepts related to bounded regions in three-dimensional space.

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Homework Statement



Find the volume of a region bounded above by the unit sphere x^2+y^2+z^2=1 and below by the cone z=sqrt(x^2+y^2). I am really confuse here.. ><


Homework Equations


Sphere: x^2+y^2+z^2=1
Cone: z=sqrt(x^2+y^2)

The Attempt at a Solution


I had plot the graph of the sphere and cone and i am pretty confuse to use. Just want to have some help.
 
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Volume requires a triple integral. How would you set this up? have you ever used polar/cylindrical/spherical coordinates?
 
Tea, i try using polar coordinates..

In the spherical, the equation of the sphere is r = 1

And the cone is rcosθ=√(r^2 sin^2 θ cos^2 φ + r^2 sin^2 θ sin^2 φ)=rsinθ

If we divide both sides by rcosθ, then we get

tanθ=1,θ=π/4

So, we have the triple integral:

∫0-2pi ∫0-(π/4) ∫0-1 (r^2 sinθdrdθdφ)

I am not sure if this is correct..
 

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