Consider the triple integral of f(x,y,z)=(z²)(x²+y²+z²)¹/² over the solid region S where z is bounded by 0 and (4-x²-y²)¹/2, y is bounded by -(4-x2) ¹/2 and (4-x²)¹/2 and x is bounded by -2 and 2. Convert I this to spherical coordinates. (You need not integrate)
Consider the triple integral of f(x,y,z)=(z²)(x²+y²+z²)¹/² over the solid region S where z is bounded by 0 and (4-x²-y²)¹/2, y is bounded by -(4-x2) ¹/2 and (4-x²)¹/2 and x is bounded by -2 and 2. Convert I this to spherical coordinates. (You need not integrate)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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![### Converting Cartesian Triple Integral to Spherical Coordinates
Consider the triple integral of \( f(x,y,z) = (z^2)(x^2 + y^2 + z^2)^{1/2} \) over the solid region \( S \), where the bounds are defined as follows:
- \( z \) is bounded by 0 and \((4 - x^2 - y^2)^{1/2} \)
- \( y \) is bounded by \(-(4 - x^2)^{1/2} \) and \((4 - x^2)^{1/2} \)
- \( x \) is bounded by -2 and 2
Convert this integral into spherical coordinates. (Note: You do not need to integrate.)
### Explanation
When converting a triple integral from Cartesian to spherical coordinates, the following transformations are used:
\[ x = \rho \sin\phi \cos\theta \]
\[ y = \rho \sin\phi \sin\theta \]
\[ z = \rho \cos\phi \]
and the Jacobian determinant for spherical coordinates is:
\[ dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \]
Given function \( f(x,y,z) = z^2(x^2 + y^2 + z^2)^{1/2} \):
In spherical coordinates, this becomes:
\[ f(\rho, \phi, \theta) = (\rho \cos\phi)^2 (\rho^2)^{1/2} = \rho^3 \cos^2 \phi \]
### Bound Transformations
The bounds change as follows in spherical coordinates:
- \( \rho \) (radial distance): This will range from 0 to 2.
- \( \phi \) (polar angle): This will range from 0 to \( \pi/2 \). The upper bound on \( z \) implies semi-sphere.
- \( \theta \) (azimuthal angle): This will range from 0 to \( 2\pi \) for the entire rotation around the z-axis.
### Integral in Spherical Coordinates
The transformed integral is:
\[ \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{2} \rho^3 \cos^2\phi](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5b8d4444-b0a9-4092-aeb7-60143570d673%2F8254c2fe-7fc7-41b1-b50b-1a3bb6f1d69b%2Fgnwhp59_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Converting Cartesian Triple Integral to Spherical Coordinates
Consider the triple integral of \( f(x,y,z) = (z^2)(x^2 + y^2 + z^2)^{1/2} \) over the solid region \( S \), where the bounds are defined as follows:
- \( z \) is bounded by 0 and \((4 - x^2 - y^2)^{1/2} \)
- \( y \) is bounded by \(-(4 - x^2)^{1/2} \) and \((4 - x^2)^{1/2} \)
- \( x \) is bounded by -2 and 2
Convert this integral into spherical coordinates. (Note: You do not need to integrate.)
### Explanation
When converting a triple integral from Cartesian to spherical coordinates, the following transformations are used:
\[ x = \rho \sin\phi \cos\theta \]
\[ y = \rho \sin\phi \sin\theta \]
\[ z = \rho \cos\phi \]
and the Jacobian determinant for spherical coordinates is:
\[ dV = \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta \]
Given function \( f(x,y,z) = z^2(x^2 + y^2 + z^2)^{1/2} \):
In spherical coordinates, this becomes:
\[ f(\rho, \phi, \theta) = (\rho \cos\phi)^2 (\rho^2)^{1/2} = \rho^3 \cos^2 \phi \]
### Bound Transformations
The bounds change as follows in spherical coordinates:
- \( \rho \) (radial distance): This will range from 0 to 2.
- \( \phi \) (polar angle): This will range from 0 to \( \pi/2 \). The upper bound on \( z \) implies semi-sphere.
- \( \theta \) (azimuthal angle): This will range from 0 to \( 2\pi \) for the entire rotation around the z-axis.
### Integral in Spherical Coordinates
The transformed integral is:
\[ \int_{0}^{2\pi} \int_{0}^{\pi/2} \int_{0}^{2} \rho^3 \cos^2\phi
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