Use spherical coordinates to evaluate the
triple integral x2 + y2 + z2 DV, where E
is the ball: x2 + y2 + z2 = 16.
1/S22+

Use spherical coordinates to evaluate the triple integral x2 y2 z2 DV where E is the ball x2 y2 z2 16 1S22 class=

Respuesta :

Answer:

4096π / 5

Step-by-step explanation:

∫∫∫ (x² + y² + z²) dV

In spherical coordinates, x² + y² + z² = r², and dV = r² sin φ dr dθ dφ.

E is the range 0 ≤ r ≤ 4, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π.

∫₀ᵖⁱ∫₀²ᵖⁱ∫₀⁴ (r²) (r² sin φ dr dθ dφ)

∫₀ᵖⁱ∫₀²ᵖⁱ∫₀⁴ (r⁴ sin φ) dr dθ dφ

Evaluate the first integral.

∫₀ᵖⁱ∫₀²ᵖⁱ (⅕ r⁵ sin φ)|₀⁴ dθ dφ

∫₀ᵖⁱ∫₀²ᵖⁱ (¹⁰²⁴/₅ sin φ) dθ dφ

¹⁰²⁴/₅ ∫₀ᵖⁱ∫₀²ᵖⁱ (sin φ) dθ dφ

Evaluate the second integral.

¹⁰²⁴/₅ ∫₀ᵖⁱ (θ sin φ)|₀²ᵖⁱ dφ

¹⁰²⁴/₅ ∫₀ᵖⁱ (2π sin φ) dφ

²⁰⁴⁸/₅ π ∫₀ᵖⁱ sin φ dφ

Evaluate the third integral.

²⁰⁴⁸/₅ π (-cos φ)|₀ᵖⁱ

²⁰⁴⁸/₅ π (-cos π + cos 0)

²⁰⁴⁸/₅ π (1 + 1)

⁴⁰⁹⁶/₅ π

Space

Answer:

[tex]\displaystyle \iiint_E {x^2 + y^2 + z^2} \, dV = \frac{4096 \pi}{5}[/tex]

General Formulas and Concepts:
Calculus

Integration

  • Integrals

Integration Rule [Reverse Power Rule]:
[tex]\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C[/tex]

Integration Rule [Fundamental Theorem of Calculus 1]:
[tex]\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)[/tex]

Integration Property [Multiplied Constant]:
[tex]\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx[/tex]

Integration Property [Addition/Subtraction]:
[tex]\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx[/tex]

Multivariable Calculus

Triple Integrals

Cylindrical Coordinate Conversions:

  • [tex]\displaystyle x = r \cos \theta[/tex]
  • [tex]\displaystyle y = r \sin \theta[/tex]
  • [tex]\displaystyle z = z[/tex]
  • [tex]\displaystyle r^2 = x^2 + y^2[/tex]
  • [tex]\displaystyle \tan \theta = \frac{y}{x}[/tex]

Spherical Coordinate Conversions:

  • [tex]\displaystyle r = \rho \sin \phi[/tex]
  • [tex]\displaystyle x = \rho \sin \phi \cos \theta[/tex]
  • [tex]\displaystyle z = \rho \cos \phi[/tex]
  • [tex]\displaystyle y = \rho \sin \phi \sin \theta[/tex]
  • [tex]\displaystyle \rho = \sqrt{x^2 + y^2 + z^2}[/tex]

Integral Conversion [Spherical Coordinates]:
[tex]\displaystyle \iiint_T {f( \rho, \phi, \theta )} \, dV = \iiint_T {\rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex]

Step-by-step explanation:

*Note:

Recall that φ is bounded by 0 ≤ φ ≤ 0.5π from the z-axis to the x-axis.

I will not show/explain any intermediate calculus steps as there isn't enough space.

Step 1: Define

Identify given.

[tex]\displaystyle \iiint_E {x^2 + y^2 + z^2} \, dV[/tex]

[tex]\displaystyle \text{Region E (Ball):} \ x^2 + y^2 + z^2 \leq 16[/tex]

Step 2: Integrate Pt. 1

Find ρ bounds.

  1. [Ball] Substitute in Spherical Coordinate Conversions:
    [tex]\displaystyle \rho^2 \leq 16[/tex]
  2. Solve:
    [tex]\displaystyle \rho \leq 4[/tex]
  3. Define limits:
    [tex]\displaystyle 0 \leq \rho \leq 4[/tex]

Find θ bounds.

  1. [Ball] Substitute in z = 0:
    [tex]\displaystyle x^2 + y^2 \leq 16[/tex]
  2. [Circle] Graph [See 2nd Attachment]
  3. [Graph] Identify limits [Unit Circle]:
    [tex]\displaystyle 0 \leq \theta \leq 2 \pi[/tex]

Find φ bounds.

  1. [Circle] Substitute in Cylindrical Coordinate Conversions:
    [tex]\displaystyle r^2 \leq 16[/tex]
  2. Solve:
    [tex]\displaystyle r \leq 4[/tex]
  3. Substitute in Spherical Coordinate Conversions:
    [tex]\displaystyle \rho \sin \phi \leq 4[/tex]
  4. Solve:
    [tex]\displaystyle \phi \leq \frac{\pi}{2}, \frac{3 \pi}{2}[/tex]
  5. Define limits:
    [tex]\displaystyle 0 \leq \phi \leq \pi[/tex]

Step 3: Integrate Pt. 2

  1. [Integrals] Convert [Integral Conversion - Spherical Coordinates]:
    [tex]\displaystyle \iiint_E {x^2 + y^2 + z^2} \, dV = \iiint_E {(x^2 + y^2 + z^2) \rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex]
  2. [dρ Integrand] Rewrite [Spherical Coordinate Conversions]:
    [tex]\displaystyle \iiint_E {x^2 + y^2 + z^2} \, dV = \iiint_E {\rho^4 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex]
  3. [Integrals] Substitute in region E:
    [tex]\displaystyle \iiint_E {x^2 + y^2 + z^2} \, dV = \int\limits^{2 \pi}_0 \int\limits^{\frac{\pi}{2}}_0 \int\limits^4_0 {\rho^4 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex]

We evaluate this spherical integral by using the integration rules, properties, and methods listed above:

[tex]\displaystyle \begin{aligned}\iiint_E {x^2 + y^2 + z^2} \, dV & = \int\limits^{2 \pi}_0 \int\limits^{\pi}_0 \int\limits^4_0 {\rho^4 \sin \phi} \, d\rho \, d\phi \, d\theta \\& = \int\limits^{2 \pi}_0 \int\limits^{\pi}_0 {\frac{\rho^5}{5} \sin \phi \bigg| \limits^{\rho = 4}_{\rho = 0}} \, d\phi \, d\theta \\& = \int\limits^{2 \pi}_0 \int\limits^{\pi}_0 {\frac{1024}{5} \sin \phi} \, d\phi \, d\theta\end{aligned}[/tex]

[tex]\displaystyle \begin{aligned}\iiint_E {x^2 + y^2 + z^2} \, dV & = \int\limits^{2 \pi}_0 {\frac{-1024}{5} \cos \phi \bigg| \limits^{\phi = \pi}_{\phi = 0}} \, d\theta \\& = \int\limits^{2 \pi}_0 {\frac{2048}{5}} \, d\theta \\& = \frac{2048}{5} \theta \bigg| \limits^{\theta = 2 \pi}_{\theta = 0} \\& = \frac{4096 \pi}{5}\end{aligned}[/tex]

∴ we have evaluated the given integral in spherical coordinates.

---

Learn more about spherical coordinates: https://brainly.com/question/17166987

Learn more about multivariable calculus: https://brainly.com/question/4746216

---

Topic: Multivariable Calculus

Unit: Triple Integrals Applications

Ver imagen Space
Ver imagen Space