Respuesta :
Answer:
[tex]\displaystyle \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV = \frac{\pi (17e^5 - 2)}{2}[/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]
Integration Method [Integration by Parts]:
[tex]\displaystyle \int {u} \, dv = uv - \int {v} \, du[/tex]
- [IBP] LIPET: Logs, Inverses, Polynomials, Exponentials, Trig
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 {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV[/tex]
[tex]\displaystyle \text{Region E:} \ x^2 + y^2 + z^2 = 25 \ \text{bounded by first octant}[/tex]
Step 2: Integrate Pt. 1
Find ρ bounds.
- [Sphere] Substitute in Spherical Coordinate Conversions:
[tex]\displaystyle \rho^2 = 25[/tex] - Solve:
[tex]\displaystyle \rho = 5[/tex] - Define limits:
[tex]\displaystyle 0 \leq \rho \leq 5[/tex]
Find θ bounds.
- [Sphere] Substitute in z = 0:
[tex]\displaystyle x^2 + y^2 = 25[/tex] - [Circle] Graph [See 2nd Attachment]
- [Graph] Identify limits [Unit Circle]:
[tex]\displaystyle 0 \leq \theta \leq \frac{\pi}{2}[/tex]
Find φ bounds.
- [Circle] Substitute in Cylindrical Coordinate Conversions:
[tex]\displaystyle r^2 = 25[/tex] - Solve:
[tex]\displaystyle r = 5[/tex] - Substitute in Spherical Coordinate Conversions:
[tex]\displaystyle \rho \sin \phi = 5[/tex] - Solve:
[tex]\displaystyle \phi = \frac{\pi}{2}[/tex] - Define limits:
[tex]\displaystyle 0 \leq \phi \leq \frac{\pi}{2}[/tex]
Step 3: Integrate Pt. 2
- [Integrals] Convert [Integral Conversion - Spherical Coordinates]:
[tex]\displaystyle \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV = \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}} \rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex] - [dρ Integrand] Rewrite [Spherical Coordinate Conversions]:
[tex]\displaystyle \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV = \iiint_E {e^{\rho} \rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta[/tex] - [Integrals] Substitute in region E:
[tex]\displaystyle \iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV = \int\limits^{\frac{\pi}{2}}_0 \int\limits^{\frac{\pi}{2}}_0 \int\limits^5_0 {e^{\rho} \rho^2 \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 {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV & = \int\limits^{\frac{\pi}{2}}_0 \int\limits^{\frac{\pi}{2}}_0 \int\limits^5_0 {e^{\rho} \rho^2 \sin \phi} \, d\rho \, d\phi \, d\theta \\ & = \int\limits^{\frac{\pi}{2}}_0 \int\limits^{\frac{\pi}{2}}_0 {\bigg[ (\rho^2 - 2 \rho + 2) e^{\rho} \sin \phi \bigg] \bigg| \limits^{\rho = 5}_{\rho = 0}} \, d\phi \, d\theta\end{aligned}[/tex]
[tex]\displaystyle \begin{aligned}\iiint_E {e^{\sqrt{x^2 + y^2 + z^2}}} \, dV & = \int\limits^{\frac{\pi}{2}}_0 \int\limits^{\frac{\pi}{2}}_0 {(17e^5 - 2) \sin \phi} \, d\phi \, d\theta \\& = \int\limits^{\frac{\pi}{2}}_0 {\bigg[ -(17e^5 - 2) \cos \phi \bigg] \bigg| \limits^{\phi = \frac{\pi}{2}}_{\phi = 0}} \, d\theta \\& = \int\limits^{\frac{\pi}{2}}_0 {17e^5 - 2} \, d\theta \\& = (17e^5 - 2) \theta \bigg| \limits^{\theta = \frac{\pi}{2}}_{\theta = 0} \\& = \frac{\pi (17e^5 - 2)}{2}\end{aligned}[/tex]
∴ the given integral equals [tex]\displaystyle \bold{\frac{\pi (17e^5 - 2)}{2}}[/tex].
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Learn more about spherical coordinates: https://brainly.com/question/16415822
Learn more about multivariable calculus: https://brainly.com/question/4746216
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Topic: Multivariable Calculus
Unit: Triple Integrals Applications
