Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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![**Problem Statement:**
Evaluate the triple integral
\[
\iiint\limits_{E} \frac{1}{x^2 + y^2 + z^2}\, dV
\]
where \( E \) lies between the spheres \( x^2 + y^2 + z^2 = 25 \) and \( x^2 + y^2 + z^2 = 49 \) in the first octant.
**Explanation:**
This problem requires evaluating a triple integral over a region \( E \) that is confined between two concentric spheres. The inner sphere has a radius of \(\sqrt{25} = 5\) and the outer sphere has a radius of \(\sqrt{49} = 7\). The region of integration is restricted to the first octant, meaning that \( x, y, \) and \( z \) are all non-negative.
The expression within the integral, \( \frac{1}{x^2 + y^2 + z^2} \), is a function of the distance to the origin, indicating that spherical coordinates might be useful for simplification.
**Approach:**
1. **Convert to Spherical Coordinates:**
- In spherical coordinates, \( x = \rho \sin \phi \cos \theta \), \( y = \rho \sin \phi \sin \theta \), \( z = \rho \cos \phi \).
- The differential volume element is \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \).
2. **Set up the Integral Bounds:**
- Radius (\( \rho \)) ranges from 5 to 7.
- Angle (\( \phi \)) ranges from 0 to \(\frac{\pi}{2}\) due to the first octant restriction.
- Angle (\( \theta \)) ranges from 0 to \(\frac{\pi}{2}\).
3. **Evaluate the Integral:**
- Substitute into the integral and evaluate over the specified bounds.](https://content.bartleby.com/qna-images/question/21e8a523-579f-40b9-8f0a-805524283a8d/c740f2cd-7d5d-48c3-97fa-d644afe85ac1/8id6g5b_processed.png)
Transcribed Image Text:**Problem Statement:**
Evaluate the triple integral
\[
\iiint\limits_{E} \frac{1}{x^2 + y^2 + z^2}\, dV
\]
where \( E \) lies between the spheres \( x^2 + y^2 + z^2 = 25 \) and \( x^2 + y^2 + z^2 = 49 \) in the first octant.
**Explanation:**
This problem requires evaluating a triple integral over a region \( E \) that is confined between two concentric spheres. The inner sphere has a radius of \(\sqrt{25} = 5\) and the outer sphere has a radius of \(\sqrt{49} = 7\). The region of integration is restricted to the first octant, meaning that \( x, y, \) and \( z \) are all non-negative.
The expression within the integral, \( \frac{1}{x^2 + y^2 + z^2} \), is a function of the distance to the origin, indicating that spherical coordinates might be useful for simplification.
**Approach:**
1. **Convert to Spherical Coordinates:**
- In spherical coordinates, \( x = \rho \sin \phi \cos \theta \), \( y = \rho \sin \phi \sin \theta \), \( z = \rho \cos \phi \).
- The differential volume element is \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \).
2. **Set up the Integral Bounds:**
- Radius (\( \rho \)) ranges from 5 to 7.
- Angle (\( \phi \)) ranges from 0 to \(\frac{\pi}{2}\) due to the first octant restriction.
- Angle (\( \theta \)) ranges from 0 to \(\frac{\pi}{2}\).
3. **Evaluate the Integral:**
- Substitute into the integral and evaluate over the specified bounds.
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- 4. Find дw b. dz a. дх at the point (x, y, z) = (0, 1, T) if w = x² + y? + z? and y sin z + z sin x = 0.arrow_forwardUse spherical coordinates to evaluate (2²+² +2²) ¹/4 dV, where E is the solid E 2 that lies between x² + y² +2²=1 and 2 2 x² + y² +²=4 given in the figure. 2 ZA x² + y² + z² = 4 x² + y² +2²=1arrow_forward1 Evaluate = √√√₂²² + 3² + 2² -dV, where E lies between the spheres z² + y² + E x² + y² + 16 in the first octant. = 4 andarrow_forward
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