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Chapter 15: Problem 72

Sphere and half-planes Find the volume of the region cut from the solid sphere \(\rho \leq a\) by the half-planes \(\theta=0\) and \(\theta=\pi / 6\) in the first octant.

Short Answer

Expert verified
The volume is \(\frac{a^3 \pi}{18}\).

Step by step solution

01

Understanding the Problem

We need to find the volume of the region within a sphere centered at the origin with radius \(a\), located between two half-planes defined by \(\theta=0\) and \(\theta=\pi/6\), and confined to the first octant.
02

Defining the Limits for Integration

The sphere is given in spherical coordinates as \(\rho \leq a\). In spherical coordinates, \(\theta\) varies from \(0\) to \(\pi/6\), and since the region is in the first octant, \(\phi\) varies from \(0\) to \(\pi/2\).
03

Setting up the Integral

The volume element in spherical coordinates is \(dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta\). We set up the triple integral: \[ V = \int_{\theta=0}^{\pi/6} \int_{\phi=0}^{\pi/2} \int_{\rho=0}^{a} \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \]
04

Integrating with Respect to \(\rho\)

Integrate the inner integral with respect to \(\rho\): \[ \int_{0}^{a} \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_0^a = \frac{a^3}{3} \]
05

Integrating with Respect to \(\phi\)

Substitute the result of the integration over \(\rho\) and integrate with respect to \(\phi\): \[ \int_{0}^{\pi/2} \frac{a^3}{3} \sin \phi \, d\phi = \frac{a^3}{3} \left[ -\cos \phi \right]_0^{\pi/2} = \frac{a^3}{3} \left(1 - 0\right) = \frac{a^3}{3} \]
06

Integrating with Respect to \(\theta\)

Finally, integrate with respect to \(\theta\): \[ \int_{0}^{\pi/6} \frac{a^3}{3} \, d\theta = \frac{a^3}{3} \left[ \theta \right]_0^{\pi/6} = \frac{a^3}{3} \cdot \frac{\pi}{6} = \frac{a^3 \pi}{18} \]
07

Result and Conclusion

The volume of the region cut from the sphere by the given half-planes in the first octant is \(\frac{a^3 \pi}{18}\).

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Key Concepts

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

Triple Integral
A triple integral is a way to compute a volume encompassed by a surface in three-dimensional space. It extends the concept of integration to functions of three variables, allowing us to sum up values over a region in three-dimensional space. This concept is crucial when evaluating the volume of a solid by summing infinitesimally small volume elements over a specified region. The triple integral is typically expressed in the form:\[ \int \int \int_R f(x, y, z) \, dx \, dy \, dz \]where \( f(x, y, z) \) is the function being integrated, and \( R \) is the region over which we are integrating.During spherical integration, the triple integral is transformed based on the geometry of the region by using spherical coordinates. Each integral accounts for one dimension of the space, and together they help compute the overall volume concerned using the limits set for the spherical region.
Volume of a Region
When discussing a region鈥檚 volume in 3D space, it refers to the amount of space enclosed within defined boundaries. Calculating volumes in spherical coordinates is often simpler when dealing with spheres or spherical regions, as these naturally fit the geometry.To calculate the volume of a region, you set up a triple integral which considers the entire enclosed space. This involves using the appropriate volume element, in this case, \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \), suited to spherical coordinates.The boundaries of the integral define the limits of the region being considered. By integrating over each variable, you measure how much space the sphere occupies, segment by segment, until all dimensions have been covered. This process results in the exact volume of the sphere confined within the boundaries outlined by the conditions.
Spherical Integration
Spherical integration employs the spherical coordinate system to evaluate integrals over spherical regions. It's particularly advantageous when analyzing volumes like those of spheres, as it takes advantage of symmetry to simplify calculations.The primary transformation to switch from Cartesian to spherical coordinates involves setting:
  • \( \rho \), which indicates the distance from the origin to the point.
  • \( \phi \), the angle subtended from the positive z-axis.
  • \( \theta \), the azimuthal angle in the xy-plane from the positive x-axis.
This system essentially breaks down a point into its radial distance and two angles, simplifying the process of setting integration limits. Spherical coordinates are ideal for this problem because the sphere's center coincides with the origin, making calculations more intuitive and straightforward.
First Octant
The first octant in three-dimensional space is the region where all three coordinates \(x\), \(y\), and \(z\) are positive. It is one of the eight divisions made by the x-, y-, and z-axes, essentially representing a corner section of the 3D space.In the context of this problem, the first octant restricts the domain of integration, ensuring the limits for \(\phi\) range from 0 to \(\frac{\pi}{2}\), and \(\theta\) ranges from 0 to \(\frac{\pi}{6}\). These angular limits ensure that the volume only accounts for the specified section of the sphere where all Cartesian coordinates remain positive.Understanding the first octant is crucial when setting integral bounds, as it helps in visualizing what portion of the sphere is being considered for volume calculations.

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Most popular questions from this chapter

Mass Find the mass of the solid bounded by the planes \(x+z=1, x-z=-1, y=0,\) and the surface \(y=\sqrt{z}\) . The density of the solid is \(\delta(x, y, z)=2 y+5 \mathrm{kg} / \mathrm{m}^{3}\).

The Parallel Axis Theorem Let \(L_{\mathrm{cm}}\) be a line through the center of mass of a body of mass \(m\) and let \(L\) be a parallel line \(h\) units away from \(L_{c . m .}\) The Parallel Axis Theorem says the moments of inertia \(I_{\mathrm{cm}}\) and \(I_{L}\) of the body about \(L_{\mathrm{cm}}\) and \(L\) satisfy the equation $$ I_{L}=I_{\mathrm{c.m.}}+m h^{2} $$ As in the two-dimensional case, the theorem gives a quick way to calculate one moment when the other moment and the mass are known. Proof of the Parallel Axis Theorem a. Show that the first moment of a body in space about any plane through the body's center of mass is zero. (Hint: Place the body's center of mass at the origin and let the plane be the \(y z\) -plane. What does the formula \(\overline{x}=M_{y z} / M\) then tell you?) b. To prove the Parallel Axis Theorem, place the body with its center of mass at the origin, with the line \(L_{c . m . \text { along the }}\) \(z\) z-axis and the line \(L\) perpendicular to the \(x y\) -plane at the point \((h, 0,0) .\) Let \(D\) be the region of space occupied by the body. Then, in the notation of the figure, $$ I_{L}=\iiint_{D}|\mathbf{v}-h \mathbf{i}|^{2} d m $$ Expand the integrand in this integral and complete the proof.

In Exercises 27 and \(28,\) sketch the solid whose volume is given by the specified integral. $$\int_{0}^{3} \int_{1}^{4}(7-x-y) d x d y$$

In polar coordinates, the average value of a function over a region \(R\) (Section 15.3\()\) is given by $$\frac{1}{\text { Area(R) }} \iint_{R} f(r, \theta) r d r d \theta$$ Volume of noncircular right cylinder The region enclosed by the lemniscate \(r^{2}=2 \cos 2 \theta\) is the base of a solid right cylinder whose top is bounded by the sphere \(z=\sqrt{2-r^{2}}\) . Find the cyl- inder's volume.

Find the volume of the solid cut from the square column \(|x|+|y| \leq 1\) by the planes \(z=0\) and \(3 x+z=3 .\)
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