8. Compute Jezsqrt(x²+y²)dV using spherical coordinates. E={(x,y,z) | x²+y²+z²<1, x>0,y>0,z>0}, that is, the first octant of the unit ball. Hints: (i) simplifying x²+y² after substitution of spherical variables cracks it open; (ii) first octant determines limits on o and 0.
8. Compute Jezsqrt(x²+y²)dV using spherical coordinates. E={(x,y,z) | x²+y²+z²<1, x>0,y>0,z>0}, that is, the first octant of the unit ball. Hints: (i) simplifying x²+y² after substitution of spherical variables cracks it open; (ii) first octant determines limits on o and 0.
Elementary Geometry For College Students, 7e
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ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
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Chapter7: Locus And Concurrence
Section7.2: Concurrence Of Lines
Problem 7E: Which lines or line segments or rays must be drawn or constructed in a triangle to locate its a...
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![### Problem Statement
Compute the triple integral \( \iiint_{E} z \sqrt{x^2 + y^2} \, dV \) using spherical coordinates. The region \( E \) is defined as:
\[ E = \{ (x, y, z) \mid x^2 + y^2 + z^2 \leq 1,\, x > 0,\, y > 0,\, z > 0 \} \]
This describes the first octant of the unit ball.
**Hints:**
1. Simplifying \( x^2 + y^2 \) after substitution of spherical variables cracks it open.
2. The first octant determines the limits on \( \phi \) and \( \theta \).
**Steps to solve:**
1. Substitute the spherical coordinates:
- \( x = \rho \sin \phi \cos \theta \)
- \( y = \rho \sin \phi \sin \theta \)
- \( z = \rho \cos \phi \)
2. Determine the limits for \( \rho \), \( \phi \), and \( \theta \):
- \( \rho \): \( 0 \leq \rho \leq 1 \)
- \( \phi \): \( 0 \leq \phi \leq \frac{\pi}{2} \)
- \( \theta \): \( 0 \leq \theta \leq \frac{\pi}{2} \)
3. The integrand \( z \sqrt{x^2 + y^2} \) becomes \( \rho \cos \phi \cdot \rho \sin \phi = \rho^2 \cos \phi \sin \phi \).
4. The volume element \( dV \) in spherical coordinates is \( \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \).
5. Set up and compute the integral:
\[ \iiint_E z \sqrt{x^2 + y^2} \, dV = \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^1 \rho^2 \cos \phi \sin \phi \cdot \rho^2 \sin \phi \, d\rho \, d\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F53b47ed7-4497-4fb9-9d0f-c41504626a91%2F540b4b48-286b-4f42-9d98-0b75453e7682%2F96qswq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Compute the triple integral \( \iiint_{E} z \sqrt{x^2 + y^2} \, dV \) using spherical coordinates. The region \( E \) is defined as:
\[ E = \{ (x, y, z) \mid x^2 + y^2 + z^2 \leq 1,\, x > 0,\, y > 0,\, z > 0 \} \]
This describes the first octant of the unit ball.
**Hints:**
1. Simplifying \( x^2 + y^2 \) after substitution of spherical variables cracks it open.
2. The first octant determines the limits on \( \phi \) and \( \theta \).
**Steps to solve:**
1. Substitute the spherical coordinates:
- \( x = \rho \sin \phi \cos \theta \)
- \( y = \rho \sin \phi \sin \theta \)
- \( z = \rho \cos \phi \)
2. Determine the limits for \( \rho \), \( \phi \), and \( \theta \):
- \( \rho \): \( 0 \leq \rho \leq 1 \)
- \( \phi \): \( 0 \leq \phi \leq \frac{\pi}{2} \)
- \( \theta \): \( 0 \leq \theta \leq \frac{\pi}{2} \)
3. The integrand \( z \sqrt{x^2 + y^2} \) becomes \( \rho \cos \phi \cdot \rho \sin \phi = \rho^2 \cos \phi \sin \phi \).
4. The volume element \( dV \) in spherical coordinates is \( \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \).
5. Set up and compute the integral:
\[ \iiint_E z \sqrt{x^2 + y^2} \, dV = \int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}} \int_0^1 \rho^2 \cos \phi \sin \phi \cdot \rho^2 \sin \phi \, d\rho \, d\
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