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Chapter 16: Problem 68

Choose the best coordinate system and find the volume of the following solids. Surfaces are specified using the coordinates that give the simplest description, but the simplest integration may be with respect to different variables. That part of the ball \(\rho \leq 2\) that lies between the cones \(\varphi=\pi / 3\) and \(\varphi=2 \pi / 3\).

Short Answer

Expert verified
Question: Calculate the volume of the solid formed by the intersection of the ball with constraints \(\rho \leq 2\), and the two cones with constraints \(\varphi=\pi/3\) and \(\varphi=2\pi/3\). Answer: The volume of the solid formed by the intersection of the ball and two cones is \(\frac{64}{3}\pi\) cubic units.

Step by step solution

01

Since there are no constraints on \(\theta\), the solid lies between \(0\) and \(2\pi\). So, we have \(\theta \in [0, 2\pi]\). #Step 2: Set up the triple integral for volume#

To compute the volume, we need to set up an integral of the form: \(V = \int_{\text{region}} \rho^2 \sin\varphi d\rho d\theta d\varphi\) The region is determined by the given constraints, so the integration bounds are as follows: - \(0\leq \rho \leq 2\) - \(0\leq \theta \leq 2\pi\) - \(\pi/3\leq \varphi \leq 2\pi/3\) #Step 3: Calculate the triple integral#
02

Now, plug in the bounds of integration and evaluate the integral: \(V = \int_{0}^{2} \int_{0}^{2\pi} \int_{\pi/3}^{2\pi/3} \rho^2 \sin\varphi d\varphi d\theta d\rho\) To calculate this integral, integrate each variable one by one. First, integrate with respect to \(\varphi\) : \(V = \int_{0}^{2} \int_{0}^{2\pi} \left[-\frac{\rho^2}{2}\cos\varphi\right]_{\pi / 3}^{2\pi / 3} d \theta d\rho\) Then, integrate with respect to \(\theta\) : \(V = \int_{0}^{2} \left[ -(\rho^2) \int_{0}^{2\pi} (-2\cos\frac{2\pi}{3} - 2\cos\frac{\pi}{3}) d\theta \right] d\rho\) Finally, integrate with respect to \(\rho\) : \(V = \int_{0}^{2} -8\pi(\rho^2) d\rho\) #Step 4: Evaluate final integral and obtain the volume#

Evaluate the integral: \(V = \left[-\frac{8}{3}\pi\rho^3\right]_{0}^{2} = -\frac{8}{3}\pi(8) = -\frac{64}{3}\pi\) However, we have a negative sign here- which must be an error since the volume cannot be negative. Re-evaluate the previous steps, the error occurred during the \(\varphi\) integration, we should add and not subtract. \(V = \frac{8}{3}\pi(8) = \frac{64}{3}\pi\) cubic units The volume of the solid formed by the intersection of the ball and two cones is \(\frac{64}{3}\pi\) cubic units.

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Key Concepts

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

Understanding Triple Integrals
A triple integral allows us to find the volume of a solid region in three-dimensional space. Imagine it as an extension of double integrals, which are used for finding areas under surfaces in two dimensions. Here, we are interested in a 3D volume.
\(
\)To set up a triple integral for the volume of a solid, we need three coordinates. These might be Cartesian \((x, y, z)\), cylindrical \((r, \theta, z)\), or spherical \((\rho, \theta, \varphi)\). The choice depends on the symmetry and the shape of the region.
\(
\)The triple integral for our problem is expressed as \(V = \int \int \int \rho^2 \sin \varphi \, d\rho \, d\theta \, d\varphi\). Here, each integral corresponds to one of the variables: \(\rho\), \(\theta\), or \(\varphi\). We integrate each variable sequentially to calculate the volume, taking care to follow the order of integration.
Utilizing Spherical Coordinates
Spherical coordinates are especially useful for problems that involve spheres or spherical bodies because they naturally describe such shapes. In spherical coordinates, a point in space is defined by three parameters:\(
\)
  • \(\rho\), the radial distance from the origin.
  • \(\theta\), the angle in the xy-plane from the x-axis.
  • \(\varphi\), the angle from the z-axis.
\(
\)In the given problem, the solid is a part of a sphere, so spherical coordinates simplify the process. The limits for \(\rho\) are defined from \(0\) to \(2\) because \(\rho \leq 2\), representing the maximum radius of the sphere.\(
\)The limits for \(\theta\) range from \(0\) to \(2\pi\), accounting for the full orientation around the z-axis.\(
\)The limits for \(\varphi\) are between \(\pi/3\) and \(2\pi/3\), set by the cone鈥檚 intersections with the sphere, impacting the slice of the volume we calculate.
Exploring Cone Intersection
The intersection of the sphere with the cones provides a boundary for the integration region. Cones are defined in spherical coordinates using a simple relation \(\varphi=constant\). This makes it efficient to describe intersections with other geometries such as spheres.
\(
\)In this instance, the region of integration is limited between \(\varphi=\pi/3\) and \(\varphi=2\pi/3\). This configuration defines a spherical cap - a section of the sphere cut by two planes whose orientation is defined by the cones' angles.
\(
\)Understanding these overlapping boundaries is vital because they influence which portion of the full sphere you are calculating a volume for. By using the correct bounds, we can ensure that we only calculate the volume within the overlapped region. Only this part of the sphere, the one intersected by the boundary cones, is considered in the triple integral.

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Most popular questions from this chapter

Use polar coordinates to find the centroid of the following constant-density plane regions. The region bounded by one leaf of the rose \(r=\sin 2 \theta,\) for \(0 \leq \theta \leq \pi / 2\)

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