Use Spherical Coordinates. (a) Find The Volume Of The Solid That Lies Above The Cone = /3 And Below The
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Introduction
In the realm of multivariable calculus, calculating the volume of complex three-dimensional shapes often requires advanced coordinate systems. Spherical coordinates provide a powerful tool for solving problems involving spheres, cones, and other symmetric solids. This article explores the application of spherical coordinates to find the volume of a specific solid: the region lying above a cone defined by a particular angle and below another surface. We will systematically analyze the problem, convert the boundaries into spherical coordinates, set up the integral, and compute the volume.
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Understanding the Problem
What is Given?
The problem states: "Find the volume of the solid that lies above the cone = /3 and below the..." (assuming the second boundary is a sphere or another surface). The key elements are:
- The cone is defined by an angular boundary, specifically, the angle \(\theta = \pi/3\).
- The "above" and "below" refer to the spatial regions between the cone and some other surface, which, based on typical problems, could be a sphere or a plane.
Clarification of Boundaries
Given the partial statement, the common scenario in spherical coordinates problems is:
- The upper boundary: a sphere of radius \( R \).
- The lower boundary: the cone given by \(\theta = \pi/3\), which divides the space into a conical region.
The goal is to find the volume of the region:
- Above the cone \(\theta = \pi/3\),
- Below the sphere \( r = R \).
Assumptions for Complete Clarity
Since the original problem is incomplete, the typical complete problem involves:
> Find the volume of the solid lying above the cone \(\theta = \pi/3\) and below the sphere \( r = R \).
This is a standard problem encountered in multivariable calculus, and we'll proceed with this interpretation.
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Spherical Coordinates: An Overview
Definition and Transformation
Spherical coordinates \((\rho, \phi, \theta)\) are related to Cartesian coordinates \((x, y, z)\) by:
\[
\begin{cases}
x = \rho \sin \phi \cos \theta \\
y = \rho \sin \phi \sin \theta \\
z = \rho \cos \phi
\end{cases}
\]
where:
- \(\rho \geq 0\) is the distance from the origin,
- \(\phi \in [0, \pi]\) is the polar angle measured from the positive \(z\)-axis,
- \(\theta \in [0, 2\pi]\) is the azimuthal angle in the \(xy\)-plane.
Volume Element in Spherical Coordinates
The differential volume element is:
\[
dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta
\]
This expression is central to setting up integrals for volume calculations.
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Setting Up the Surface Boundaries
The Sphere \( r = R \)
The sphere boundary in spherical coordinates:
\[
\rho = R
\]
The Cone \(\theta = \pi/3\)
In standard spherical coordinates, the cone boundary is often expressed in terms of \(\phi\) or \(\theta\).
Important Note: In the context of spherical coordinates:
- \(\phi\) (polar angle): angle from the positive z-axis,
- \(\theta\) (azimuthal angle): angle in the xy-plane.
Given that the problem involves the cone \(\theta = \pi/3\), this suggests the cone opens along the \(z\)-axis with an angular boundary in the azimuthal plane.
However, in many problems involving cones, the cone is defined by an angle from the \(z\)-axis, often written as \(\phi = \alpha\).
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Clarifying the Cone's Definition
Cone Defined by \(\phi = \pi/3\)
If the cone is symmetric about the \(z\)-axis, its boundary is given by:
\[
\phi = \alpha
\]
where \(\alpha\) is the half-angle of the cone.
In this case:
- The region above the cone corresponds to \(\phi \leq \pi/3\),
- The region below the cone corresponds to \(\phi \geq \pi/3\).
Given the partial statement, and typical problem structures, it is more consistent that the cone is defined by \(\phi = \pi/3\), rather than \(\theta\).
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Final Assumption for the Problem
Assumption:
- The cone is defined by \(\phi = \pi/3\),
- The sphere boundary is at \(\rho = R\),
- The region of interest is above the cone (i.e., \(\phi \leq \pi/3\)),
- The volume is contained within the sphere (\(0 \leq \rho \leq R\)).
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Computing the Volume
Step 1: Setting the Integration Limits
- \(\theta\): Since the problem involves a symmetric cone around the \(z\)-axis, \(\theta\) spans the full azimuthal angle:
\[
0 \leq \theta \leq 2\pi
\]
- \(\phi\): For the region above the cone, the polar angle varies from:
\[
0 \leq \phi \leq \pi/3
\]
- \(\rho\): From the origin out to the sphere:
\[
0 \leq \rho \leq R
\]
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Step 2: Setting Up the Triple Integral
The volume \(V\) is:
\[
V = \int{0}^{2\pi} \int{0}^{\pi/3} \int_{0}^{R} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta
\]
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Step 3: Computing the Integral
The integral separates into three parts:
\[
V = \left( \int{0}^{2\pi} d\theta \right) \times \left( \int{0}^{\pi/3} \sin \phi \, d\phi \right) \times \left( \int_{0}^{R} \rho^2 \, d\rho \right)
\]
Calculate each:
- Azimuthal integral:
\[
\int_{0}^{2\pi} d\theta = 2\pi
\]
- Polar angle integral:
\[
\int{0}^{\pi/3} \sin \phi \, d\phi = -\cos \phi \big|{0}^{\pi/3} = -\cos (\pi/3) + \cos 0 = -\frac{1}{2} + 1 = \frac{1}{2}
\]
- Radial integral:
\[
\int{0}^{R} \rho^2 \, d\rho = \frac{\rho^3}{3} \big|{0}^{R} = \frac{R^3}{3}
\]
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Step 4: Final Expression for Volume
Putting it all together:
\[
V = 2\pi \times \frac{1}{2} \times \frac{R^3}{3} = \pi \times \frac{R^3}{3} = \frac{\pi R^3}{3}
\]
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Summary of Results
- The volume of the solid above the cone \(\phi = \pi/3\) and below the sphere \(r = R\) is:
\[
\boxed{V = \frac{\pi R^3}{3}}
\]
- This result applies when the cone is defined by \(\phi = \pi/3\) (i.e., a cone opening downward from the z-axis).
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Additional Considerations
Variations in the Boundary Conditions
- If the boundary is a different surface, such as a paraboloid or an oblique plane, the integral limits and the coordinate transformation would change accordingly.
- For cones defined by \(\theta = \text{constant}\), the problem reduces to an angular wedge, and the limits in \(\theta\) would be from the cone's azimuthal angle to another boundary.
- For cones symmetric about the \(z\)-axis, the most natural boundary is often expressed in terms of \(\phi\).
Application of Spherical Coordinates in Real-World Problems
- Astrophysics: Calculating volumes of celestial bodies or regions of space.
- Engineering: Analyzing the volume of conical tanks or regions with spherical symmetry.
- Mathematics: Solving integrals involving spheres and cones in multivariable calculus courses.
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Conclusion
Using spherical coordinates simplifies the process of calculating volumes bounded by spheres and cones due to the symmetry of these shapes. By converting the boundaries into the spherical coordinate system, setting appropriate limits, and integrating over the defined region, we obtain precise volume measurements efficiently. The key steps involve understanding the geometric boundaries,
