Math Problem Statement

Convert the integral \( \int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{4-x^2-y^2}} \frac{x^2 y}{\sqrt{x^2 + y^2}} \, dz \, dy \, dx \) to spherical coordinates (do not evaluate it).

Solution

To convert the given integral to spherical coordinates, we follow these steps:

Step 1: Spherical Coordinates Relations

The spherical coordinates (ρ,θ,ϕ)(\rho, \theta, \phi) are related to Cartesian coordinates (x,y,z)(x, y, z) as: x=ρsinϕcosθ,y=ρsinϕsinθ,z=ρcosϕ,x = \rho \sin\phi \cos\theta, \quad y = \rho \sin\phi \sin\theta, \quad z = \rho \cos\phi, with the corresponding volume element: dxdydz=ρ2sinϕdρdϕdθ.dx \, dy \, dz = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.

Step 2: Identifying the Boundaries

The given Cartesian integral is: 0101x204x2y2x2yx2+y2dzdydx.\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{4-x^2-y^2}} \frac{x^2 y}{\sqrt{x^2 + y^2}} \, dz \, dy \, dx.

  • Outer integral (xx): xx ranges from 00 to 11.
  • Middle integral (yy): For a fixed xx, yy ranges from 00 to 1x2\sqrt{1-x^2}.
  • Inner integral (zz): For fixed xx and yy, zz ranges from 00 to 4x2y2\sqrt{4-x^2-y^2}.

These limits represent a region in the first octant of a sphere of radius 22.

Step 3: Converting the Limits to Spherical Coordinates

  • ρ\rho: The radial distance ranges from 00 to 22, as the sphere has radius 22.
  • θ\theta: The azimuthal angle ranges from 00 to π2\frac{\pi}{2}, since the region is in the first octant.
  • ϕ\phi: The polar angle ranges from 00 to π2\frac{\pi}{2}, again due to the first-octant restriction.

Step 4: Converting the Integrand

The integrand x2yx2+y2\frac{x^2 y}{\sqrt{x^2 + y^2}} becomes: x2y=(ρsinϕcosθ)2(ρsinϕsinθ),x^2 y = (\rho \sin\phi \cos\theta)^2 (\rho \sin\phi \sin\theta), and x2+y2=ρsinϕ\sqrt{x^2 + y^2} = \rho \sin\phi. Substituting, the integrand in spherical coordinates is: x2yx2+y2=(ρsinϕcosθ)2(ρsinϕsinθ)ρsinϕ=ρ2sin2ϕcos2θsinθ.\frac{x^2 y}{\sqrt{x^2 + y^2}} = \frac{(\rho \sin\phi \cos\theta)^2 (\rho \sin\phi \sin\theta)}{\rho \sin\phi} = \rho^2 \sin^2\phi \cos^2\theta \sin\theta.

Step 5: Volume Element

The volume element dxdydzdx \, dy \, dz becomes ρ2sinϕdρdϕdθ\rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.

Step 6: The Final Integral

Substituting everything into spherical coordinates: 0101x204x2y2x2yx2+y2dzdydx\int_0^1 \int_0^{\sqrt{1-x^2}} \int_0^{\sqrt{4-x^2-y^2}} \frac{x^2 y}{\sqrt{x^2 + y^2}} \, dz \, dy \, dx becomes: 0π20π202ρ4sin3ϕcos2θsinθdρdϕdθ.\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^2 \rho^4 \sin^3\phi \cos^2\theta \sin\theta \, d\rho \, d\phi \, d\theta.

Let me know if you'd like additional details or have further questions!


Follow-Up Questions:

  1. How are spherical coordinates derived from Cartesian coordinates?
  2. What are the general integration limits for spherical coordinates?
  3. Why does the volume element include ρ2sinϕ\rho^2 \sin\phi?
  4. How does the first-octant restriction influence the limits?
  5. How would this integral change if the region was in the full sphere?

Tip:

Always carefully analyze the geometry of the region when converting integrals to spherical coordinates.

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Math Problem Analysis

Mathematical Concepts

Triple Integrals
Spherical Coordinates
Coordinate System Transformations

Formulas

Spherical coordinates: \( x = \rho \sin\phi \cos\theta, y = \rho \sin\phi \sin\theta, z = \rho \cos\phi \)
Volume element in spherical coordinates: \( dx \, dy \, dz = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \)

Theorems

Conversion of integrals to spherical coordinates
Geometric limits in spherical regions

Suitable Grade Level

Undergraduate (Calculus III / Multivariable Calculus)