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3. (Section 16.5) Consider √x² + y² + z² by √√32 = √² + y². Set-up the triple integral using: (a) rectangular (Cartesian) coordinates. (Do not evaluate). dV where solid D is bounded above by z² + y² +2²=4 and bounded below (b) cylindrical coordinates. (Do not evaluate).

3. (Section 16.5) Consider √x² + y² + z² by √√32 = √² + y². Set-up the triple integral using: (a) rectangular (Cartesian) coordinates. (Do not evaluate). dV where solid D is bounded above by z² + y² +2²=4 and bounded below (b) cylindrical coordinates. (Do not evaluate).

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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### Triple Integral Setup **Problem Statement:** Consider the triple integral \( \iiint_D \frac{1}{\sqrt{x^2 + y^2 + z^2}} \, dV \), where solid \( D \) is bounded above by \( x^2 + y^2 + z^2 = 4 \) and bounded below by \( \sqrt{3}z = \sqrt{x^2 + y^2} \). Set up the triple integral using: **(a) Rectangular (Cartesian) coordinates.** (Do not evaluate). --- **(b) Cylindrical coordinates.** (Do not evaluate). --- **(c) Spherical coordinates.** (Evaluate the integral). --- ### Explanation of Volume Boundaries: 1. **Upper Boundary:** - A sphere with radius 2 centered at the origin, as given by the equation \( x^2 + y^2 + z^2 = 4 \). 2. **Lower Boundary:** - A cone opening along the z-axis defined by \( \sqrt{3}z = \sqrt{x^2 + y^2} \). ### Integration in Spherical Coordinates: 1. **Conversion to Spherical Coordinates:** - \( x = \rho \sin \phi \cos \theta \) - \( y = \rho \sin \phi \sin \theta \) - \( z = \rho \cos \phi \) - The integrand becomes \( \frac{1}{\rho} \). 2. **Jacobian:** - The volume element \( dV \) becomes \( \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \). 3. **Domain (Boundary) in Spherical Coordinates:** - \( 0 \leq \theta < 2\pi \) - Cone: \( \phi = \frac{\pi}{6} \) to \( \phi = \frac{\pi}{2} \) - Sphere: \( 0 \leq \rho \leq 2 \) 4. **Integral:** \[ \int_0^{2\pi} \int_{\pi/6}^{\pi/2} \int_0^2 \frac{1}{\rho} \cdot \rho^2 \sin \phi \
Transcribed Image Text:### Triple Integral Setup **Problem Statement:** Consider the triple integral \( \iiint_D \frac{1}{\sqrt{x^2 + y^2 + z^2}} \, dV \), where solid \( D \) is bounded above by \( x^2 + y^2 + z^2 = 4 \) and bounded below by \( \sqrt{3}z = \sqrt{x^2 + y^2} \). Set up the triple integral using: **(a) Rectangular (Cartesian) coordinates.** (Do not evaluate). --- **(b) Cylindrical coordinates.** (Do not evaluate). --- **(c) Spherical coordinates.** (Evaluate the integral). --- ### Explanation of Volume Boundaries: 1. **Upper Boundary:** - A sphere with radius 2 centered at the origin, as given by the equation \( x^2 + y^2 + z^2 = 4 \). 2. **Lower Boundary:** - A cone opening along the z-axis defined by \( \sqrt{3}z = \sqrt{x^2 + y^2} \). ### Integration in Spherical Coordinates: 1. **Conversion to Spherical Coordinates:** - \( x = \rho \sin \phi \cos \theta \) - \( y = \rho \sin \phi \sin \theta \) - \( z = \rho \cos \phi \) - The integrand becomes \( \frac{1}{\rho} \). 2. **Jacobian:** - The volume element \( dV \) becomes \( \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \). 3. **Domain (Boundary) in Spherical Coordinates:** - \( 0 \leq \theta < 2\pi \) - Cone: \( \phi = \frac{\pi}{6} \) to \( \phi = \frac{\pi}{2} \) - Sphere: \( 0 \leq \rho \leq 2 \) 4. **Integral:** \[ \int_0^{2\pi} \int_{\pi/6}^{\pi/2} \int_0^2 \frac{1}{\rho} \cdot \rho^2 \sin \phi \
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