Math Problem Statement

Using triple integral, find the volume of the sphere whose radius is ‘a’ units.

Solution

The problem asks us to find the volume of a sphere with radius aa using a triple integral. Let’s solve it.


Step-by-Step Solution:

We use spherical coordinates for simplicity in evaluating the triple integral. In spherical coordinates:

  1. x=ρsinϕcosθx = \rho \sin\phi \cos\theta
  2. y=ρsinϕsinθy = \rho \sin\phi \sin\theta
  3. z=ρcosϕz = \rho \cos\phi
  4. The Jacobian determinant (volume element) is ρ2sinϕdρdϕdθ\rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.

The sphere's radius is aa, so the region is bounded by:

0ρa,0ϕπ,0θ2π.0 \leq \rho \leq a, \quad 0 \leq \phi \leq \pi, \quad 0 \leq \theta \leq 2\pi.

The volume of the sphere is given by:

V=02π0π0aρ2sinϕdρdϕdθ.V = \int_{0}^{2\pi} \int_{0}^{\pi} \int_{0}^{a} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta.


Step 1: Evaluate the innermost integral (ρ\rho):

0aρ2dρ=[ρ33]0a=a33.\int_{0}^{a} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_{0}^{a} = \frac{a^3}{3}.


Step 2: Evaluate the second integral (ϕ\phi):

0πsinϕdϕ=[cosϕ]0π=cos(π)(cos(0))=2.\int_{0}^{\pi} \sin\phi \, d\phi = \left[ -\cos\phi \right]_{0}^{\pi} = -\cos(\pi) - (-\cos(0)) = 2.


Step 3: Evaluate the outermost integral (θ\theta):

02π1dθ=[θ]02π=2π.\int_{0}^{2\pi} 1 \, d\theta = \left[\theta\right]_{0}^{2\pi} = 2\pi.


Final Step: Multiply the results:

V=a3322π=4πa33.V = \frac{a^3}{3} \cdot 2 \cdot 2\pi = \frac{4\pi a^3}{3}.


Final Answer:

The volume of the sphere is:

4πa33\boxed{\frac{4\pi a^3}{3}}


Would you like me to break down any of the steps further or provide additional explanations?

Related Questions:

  1. How is the Jacobian determinant derived in spherical coordinates?
  2. What are the limits for ρ\rho, ϕ\phi, and θ\theta in spherical coordinates for a sphere?
  3. Can you explain the difference between Cartesian and spherical coordinate systems?
  4. How would the solution change for a hemisphere instead of a sphere?
  5. What is the physical significance of the volume element ρ2sinϕ\rho^2 \sin\phi?

Tip:

When solving triple integrals in spherical coordinates, always carefully define the limits of integration and include the Jacobian determinant.

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Math Problem Analysis

Mathematical Concepts

Triple Integrals
Spherical Coordinates
Volume of a Sphere

Formulas

x = ρ sin(φ) cos(θ)
y = ρ sin(φ) sin(θ)
z = ρ cos(φ)
Jacobian determinant = ρ² sin(φ) dρ dφ dθ
Volume integral: V = ∫∫∫ ρ² sin(φ) dρ dφ dθ

Theorems

Triple Integral Theorem
Conversion to Spherical Coordinates

Suitable Grade Level

Undergraduate (Calculus III or Advanced Mathematics)