Using spherical coordinates calculate ſezsqrt(x²+y²)dV where E={(x,y,z) | x²+y²+z²<4, x>0,y>0,z>0}
Using spherical coordinates calculate ſezsqrt(x²+y²)dV where E={(x,y,z) | x²+y²+z²<4, x>0,y>0,z>0}
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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Question
![To compute the integral using spherical coordinates, follow the given steps. We will convert the given triple integral ∭_E z√(x² + y²) dV into spherical coordinates. Given:
\[ E = \{(x,y,z) \mid x^2 + y^2 + z^2 \leq 4, \; x>0, \; y>0, \; z>0 \} \]
\[ \text{Calculate:} \quad \iiint_E z\sqrt{x^2 + y^2} \, dV \]
### Steps to Solve:
1. **Convert to Spherical Coordinates:**
- In spherical coordinates, \( x = \rho \sin\phi \cos\theta \), \( y = \rho \sin\phi \sin\theta \), and \( z = \rho \cos\phi \).
- The expression \( z\sqrt{x^2 + y^2} \) becomes \( \rho \cos\phi \cdot \rho \sin\phi \).
2. **Jacobian Determinant:**
- The volume element \( dV \) in spherical coordinates is \( \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \).
3. **Set Boundaries:**
- Given \( x^2 + y^2 + z^2 \leq 4 \), we have \( \rho \leq 2 \).
- Since \( x>0, y>0, z>0 \), this corresponds to the first octant: \( 0 \leq \theta \leq \frac{\pi}{2} \), \( 0 \leq \phi \leq \frac{\pi}{2} \), \( 0 \leq \rho \leq 2 \).
4. **Integral Conversion:**
\[
\iiint_E z\sqrt{x^2 + y^2} \, dV = \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^2 (\rho \cos\phi \cdot \rho \sin\phi) \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta
\]
5](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6674a345-3819-404a-bb58-af4f58fff999%2Fbfcad45c-e699-42bf-84dd-350192a6dbfe%2Fevos5r29_processed.jpeg&w=3840&q=75)
Transcribed Image Text:To compute the integral using spherical coordinates, follow the given steps. We will convert the given triple integral ∭_E z√(x² + y²) dV into spherical coordinates. Given:
\[ E = \{(x,y,z) \mid x^2 + y^2 + z^2 \leq 4, \; x>0, \; y>0, \; z>0 \} \]
\[ \text{Calculate:} \quad \iiint_E z\sqrt{x^2 + y^2} \, dV \]
### Steps to Solve:
1. **Convert to Spherical Coordinates:**
- In spherical coordinates, \( x = \rho \sin\phi \cos\theta \), \( y = \rho \sin\phi \sin\theta \), and \( z = \rho \cos\phi \).
- The expression \( z\sqrt{x^2 + y^2} \) becomes \( \rho \cos\phi \cdot \rho \sin\phi \).
2. **Jacobian Determinant:**
- The volume element \( dV \) in spherical coordinates is \( \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \).
3. **Set Boundaries:**
- Given \( x^2 + y^2 + z^2 \leq 4 \), we have \( \rho \leq 2 \).
- Since \( x>0, y>0, z>0 \), this corresponds to the first octant: \( 0 \leq \theta \leq \frac{\pi}{2} \), \( 0 \leq \phi \leq \frac{\pi}{2} \), \( 0 \leq \rho \leq 2 \).
4. **Integral Conversion:**
\[
\iiint_E z\sqrt{x^2 + y^2} \, dV = \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^2 (\rho \cos\phi \cdot \rho \sin\phi) \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta
\]
5
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