Problem 13. Integrate f(x, y, z) = 12xz over the region in the first octant (x, y, z ≥0) above the parabolic cylinder z = y² and below the pa- raboloid z = 8 - 2x² - y². Answer:
Problem 13. Integrate f(x, y, z) = 12xz over the region in the first octant (x, y, z ≥0) above the parabolic cylinder z = y² and below the pa- raboloid z = 8 - 2x² - y². Answer:
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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1.13
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![### Problem 13
Integrate \( f(x,y,z) = 12xz \) over the region in the first octant \((x, y, z \geq 0)\) above the parabolic cylinder \( z = y^2 \) and below the paraboloid \( z = 8 - 2x^2 - y^2 \).
**Answer:** ________
**Explanation:**
For this problem, we need to set up and solve a triple integral to find the volume in question. Here's how we could approach it:
1. **Understand the Boundaries:**
- The region is in the first octant (\( x, y, z \geq 0 \)).
- The lower boundary is defined by the parabolic cylinder \( z = y^2 \).
- The upper boundary is defined by the paraboloid \( z = 8 - 2x^2 - y^2 \).
2. **Determine the Boundaries in X and Y:**
- To find the intersection of \( z = y^2 \) and \( z = 8 - 2x^2 - y^2 \), we set \( y^2 = 8 - 2x^2 - y^2 \).
- Solving for \( y \) and \( x \), we get \( 2y^2 = 8 - 2x^2 \) which simplifies to \( y^2 = 4 - x^2 \).
- The region in the \( xy \)-plane is thus bounded by \( y = 0 \) to \( y = \sqrt{4-x^2} \), with \( x \) ranging from \( 0 \) to \( 2 \) (the positive root of \( x^2 = 4 \)).
3. **Set Up the Triple Integral:**
- The triple integral can be written as:
\[
\iiint\limits_{R} 12xz \, dV
\]
- In Cartesian coordinates, this becomes:
\[
\int_{0}^{2} \int_{0}^{\sqrt{4-x^2}} \int_{y^2}^{8-2x^2-y^2} 12xz \, dz \, dy \, dx.
\]
**Note:** This guide is for setting up](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbbff2935-77bb-4550-bfd1-d595e6271f30%2F9173265c-63e0-468f-9249-17ef7de1a2d4%2Fm0dmg1f_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem 13
Integrate \( f(x,y,z) = 12xz \) over the region in the first octant \((x, y, z \geq 0)\) above the parabolic cylinder \( z = y^2 \) and below the paraboloid \( z = 8 - 2x^2 - y^2 \).
**Answer:** ________
**Explanation:**
For this problem, we need to set up and solve a triple integral to find the volume in question. Here's how we could approach it:
1. **Understand the Boundaries:**
- The region is in the first octant (\( x, y, z \geq 0 \)).
- The lower boundary is defined by the parabolic cylinder \( z = y^2 \).
- The upper boundary is defined by the paraboloid \( z = 8 - 2x^2 - y^2 \).
2. **Determine the Boundaries in X and Y:**
- To find the intersection of \( z = y^2 \) and \( z = 8 - 2x^2 - y^2 \), we set \( y^2 = 8 - 2x^2 - y^2 \).
- Solving for \( y \) and \( x \), we get \( 2y^2 = 8 - 2x^2 \) which simplifies to \( y^2 = 4 - x^2 \).
- The region in the \( xy \)-plane is thus bounded by \( y = 0 \) to \( y = \sqrt{4-x^2} \), with \( x \) ranging from \( 0 \) to \( 2 \) (the positive root of \( x^2 = 4 \)).
3. **Set Up the Triple Integral:**
- The triple integral can be written as:
\[
\iiint\limits_{R} 12xz \, dV
\]
- In Cartesian coordinates, this becomes:
\[
\int_{0}^{2} \int_{0}^{\sqrt{4-x^2}} \int_{y^2}^{8-2x^2-y^2} 12xz \, dz \, dy \, dx.
\]
**Note:** This guide is for setting up
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