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In the following exercises, evaluate the triple integral \(\iiint_{B} f(x, y, z) d V\) over the solid \(B\). $$ f(x, y, z)=1, B=\left\\{(x, y, z) \mid x^{2}+y^{2}+z^{2} \leq 90, z \geq 0\right\\} $$

Short Answer

Expert verified
The value of the triple integral is \(60 \sqrt{90} \pi\).

Step by step solution

01

Interpret the region of integration

The solid region \( B \) is defined by the inequality \( x^2 + y^2 + z^2 \leq 90 \) and \( z \geq 0 \). This represents the upper half of a sphere centered at the origin with radius \( \sqrt{90} \).
02

Set up the integral with spherical coordinates

Since \( B \) is half of a sphere, it's convenient to use spherical coordinates. In these coordinates, \( x = \rho \sin \phi \cos \theta \), \( y = \rho \sin \phi \sin \theta \), and \( z = \rho \cos \phi \). The Jacobian of this transformation is \( \rho^2 \sin \phi \).
03

Determine the bounds of integration

\( \rho \) ranges from 0 to \( \sqrt{90} \). \( \phi \) ranges from 0 to \( \frac{\pi}{2} \) because \( z \geq 0 \). \( \theta \) ranges from 0 to \( 2\pi \) to cover the entire azimuthal angle around the z-axis.
04

Write the triple integral in spherical coordinates

The triple integral becomes: \[\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{2}} \int_{0}^{\sqrt{90}} 1 \cdot \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\]
05

Evaluate the innermost integral

Evaluate the integral with respect to \( \rho \): \[\int_{0}^{\sqrt{90}} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_{0}^{\sqrt{90}} = \frac{(\sqrt{90})^3}{3} = \frac{90 \sqrt{90}}{3}\]
06

Evaluate the integral with respect to \( \phi \)

Substituting back for the \( \rho \) integral result:\[\int_{0}^{\frac{\pi}{2}} \sin \phi \, d\phi = \left[ -\cos \phi \right]_{0}^{\frac{\pi}{2}} = 1\]
07

Evaluate the outermost integral

Finally, evaluate the remaining integral with respect to \( \theta \):\[\int_{0}^{2\pi} 1 \, d\theta = \left[ \theta \right]_{0}^{2\pi} = 2\pi\]
08

Calculate the final result

Combine the results from steps 5, 6, and 7:\[\frac{90 \sqrt{90}}{3} \times 1 \times 2\pi = 60 \sqrt{90} \pi\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spherical Coordinates
To understand triple integrals in spherical coordinates, imagine slicing through a sphere in three dimensions, much like peeling an orange. In spherical coordinates, we represent points using
  • \( \rho \), the radius: distance from the origin
  • \( \phi \), the polar angle: angle from the positive z-axis
  • \( \theta \), the azimuthal angle: rotation around the z-axis
Switching from Cartesian to spherical coordinates often simplifies integration, especially in cases involving symmetrical shapes such as spheres. The mathematical relationships are:
  • \( x = \rho \sin \phi \cos \theta \)
  • \( y = \rho \sin \phi \sin \theta \)
  • \( z = \rho \cos \phi \)
When a function or region of integration is symmetrical around an axis, spherical coordinates can make the integration process more straightforward, reducing complex geometric regions into simpler, radial components. In spherical coordinates, the volume element \( dV \) is represented by \( \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \), owing to the Jacobian transformation.
Region of Integration
In solving a triple integral, one must first determine the bounds and limits of integration that define the region, often referred to as the region of integration. This region, denoted as \( B \) in our problem, is given as the upper half of a sphere. More specifically, it is bound by \( x^2 + y^2 + z^2 \leq 90 \) and \( z \geq 0 \). These inequalities define a 3D solid that is symmetric around the center and only extends above the xy-plane.
This region is crucial because it determines the values that the variables within the integral can take. By interpreting this as a hemisphere, the transition to spherical coordinates becomes natural. For our integration, we use \( \rho \) from 0 to \( \sqrt{90} \) (the radius), \( \phi \) from 0 to \( \frac{\pi}{2} \) (ensuring \( z \geq 0 \)), and \( \theta \) from 0 to \( 2\pi \) (to cover the complete circular rotation on the xy-plane). This gives us a complete understanding of the space through which we are integrating.
Jacobian Transformation
When integrating in different coordinate systems, such as moving from Cartesian to spherical coordinates, adjustments must be made to account for changes in the volume element size. This is where the Jacobian transformation comes into play.
The Jacobian is a determinant used in changing variables to account for how the coordinate system skews or compresses the volume. In spherical coordinates, the Jacobian transformation factor is \( \rho^2 \sin \phi \).
This ensures the volume element \( dV \) in Cartesian coordinates \( (dx \, dy \, dz) \) is correctly represented in spherical coordinates as \( (\rho^2 \sin \phi \, d\rho \, d\phi \, d\theta) \).
The Jacobian specifically does the following:
  • It scales the volume element correctly to match the geometry of the system.
  • Accounts for stretching and skewing that occurs when transforming to curvilinear coordinates.
It is essential to apply this transformation to solve triple integrals correctly when transitioning from Cartesian to spherical coordinates, ensuring an accurate calculation of volume or mass within a defined region.

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Most popular questions from this chapter

A solid \(Q\) has a volume given by \(\iint_{D} \int_{a}^{b} d A d z\), where \(D\) is the projection of the solid onto the \(x y\) -plane and \(a

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