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Chapter 13: Problem 51

Use spherical coordinates to find the volume of the following solids. That part of the ball \(\rho \leq 4\) that lies between the planes \(z=2\) and \(z=2 \sqrt{3}\)

Short Answer

Expert verified
Based on the step-by-step solution provided, write a short answer for the following question: Find the volume of the solid lying inside a ball of radius 4 and between two planes parallel to the x-y plane, given by z = 2 and z = 2鈭�3. Answer: The volume of the solid lying inside the ball of radius 4 and between the given planes is \(V=\frac{64\pi}{3}\) cubic units.

Step by step solution

01

Identify the Limits of Integration for Spherical Coordinates

First, find the limits of integration for \(\rho\), \(\phi\), and \(\theta\). Given the equation for the ball, \(\rho \leq 4\), this indicates that \(\rho\) will range from 0 to 4. To find the limits of \(\phi\), we need to identify the intersection of the ball with the two planes. Plug in \(z=2\) and \(z=2\sqrt{3}\) into the spherical coordinates expression for z, \(z=\rho\cos\phi\). Solve for \(\phi\): 1. \(2 = \rho\cos\phi \Rightarrow \phi = \arccos(\frac{2}{\rho})\) 2. \(2\sqrt{3} = \rho\cos\phi \Rightarrow \phi = \arccos(\frac{2\sqrt{3}}{\rho})\) Since the entire range along the x-y plane is considered, and the solid lies between the two planes, \(\phi\) varies from \(\arccos(\frac{2}{\rho})\) to \(\arccos(\frac{2\sqrt{3}}{\rho})\) and \(\theta\) varies from \(0\) to \(2\pi\).
02

Set up the Triple Integral

Now that the limits of integration have been identified, set up the triple integral to find the volume of the given solid using the volume element \(dV=\rho^2\sin\phi\ d\rho\ d\phi\ d\theta\): \(V = \displaystyle\int_{0}^{2\pi} \int_{\arccos(\frac{2}{\rho})}^{\arccos(\frac{2\sqrt{3}}{\rho})} \int_{0}^{4} \rho^2\sin\phi\ d\rho\ d\phi\ d\theta\)
03

Integrate with Respect to \(\rho\)

First, integrate with respect to \(\rho\). Use the antiderivative formula for \(\rho\): \(\frac{1}{3}\rho^3\) Now, evaluate the limits of the integral from 0 to 4: \(V = \int_{0}^{2\pi} \int_{\arccos(\frac{2}{\rho})}^{\arccos(\frac{2\sqrt{3}}{\rho})} \frac{64}{3}\sin\phi\ d\phi\ d\theta\)
04

Integrate with Respect to \(\phi\)

Now, integrate with respect to \(\phi\): \(V = \int_{0}^{2\pi} \left[-\frac{64}{3}\cos\phi\right]_{\arccos(\frac{2}{\rho})}^{\arccos(\frac{2\sqrt{3}}{\rho})} d\theta\)
05

Evaluate the Integral with Respect to \(\phi\)

Evaluate the integral with respect to \(\phi\) by plugging in the limits of integration: \(V = \int_{0}^{2\pi} \left(-\frac{64}{3}\cos(\arccos(\frac{2\sqrt{3}}{\rho})) + \frac{64}{3}\cos(\arccos(\frac{2}{\rho}))\right) d\theta\) Simplify the expression inside the integral: \(V = \int_{0}^{2\pi} \left(\frac{64}{3}\left[\cos(\arccos(\frac{2}{\rho})) - \cos(\arccos(\frac{2\sqrt{3}}{\rho}))\right]\right) d\theta\) Notice that the range of \(\theta\) is from \(0\) to \(2\pi\). Therefore, this integral represents a full revolution around the z-axis.
06

Integrate with Respect to \(\theta\) and Calculate the Volume

Finally, integrate with respect to \(\theta\): \(V = \left[\frac{64}{3}\theta\right]_{0}^{2\pi}\left[\cos(\arccos(\frac{2}{\rho})) - \cos(\arccos(\frac{2\sqrt{3}}{\rho}))\right]\) Evaluate the integral to obtain the volume: \(V=\frac{128}{3}\pi\left[1-\frac{1}{2}\right] = \frac{128}{3}\pi\left[\frac{1}{2}\right] = \frac{64\pi}{3}\) Thus, the volume of the solid is \(\frac{64\pi}{3}\) cubic units.

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Key Concepts

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

Integral Calculus
Integral calculus helps us calculate areas, volumes, and other values when simple geometry isn't enough. It revolves around finding integrals, which are essentially sums of infinitely small parts. When we talk about "integral," we often refer to the definite integral. This gives us a numerical value that represents the area under a curve between two points. In mathematics, the integral is symbolized by the elongated S shape \(\int\). For three-dimensional regions like the one in our exercise, we use a triple integral to compute the volume.
Spherical coordinates simplify the integration process by fitting naturally around balls and spheres, aiding in calculations involving spherical symmetry. This approach converts rectangular coordinates into spherical ones, often making the computation of volumes of spherical objects more straightforward.
Limits of Integration
Limits of integration define the boundaries within which an integral is evaluated. In our exercise, these limits are crucial to determining the specific part of the sphere where the volume is calculated. In spherical coordinates, the limits are set for the radial distance \((\rho)\), the polar angle \(\phi\), and the azimuthal angle \(\theta\).
  • \(\rho\): Represents the radial distance, varying from the center of the sphere outward. Its limits are from 0 to 4, representing the radius.
  • \(\phi\): The angle between the positive z-axis and the line connecting the origin to a point. It helps identify which section of the sphere is being considered. Through this exercise, it ranges specifically due to the planes intercepting the sphere.
  • \(\theta\): Encompasses the revolution around the z-axis, typically set from 0 to \((2\pi)\), covering the full circular symmetry.
Triple Integral
The triple integral is a fundamental concept in calculus used to calculate functions over three-dimensional regions. Symbolized by three integral signs, it is particularly useful when determining volumes, as it allows us to add up an infinite number of tiny volume elements, precisely describing the space enclosed by a surface.
In the given example, evaluating a triple integral requires considering the volume element \(dV = \rho^2\sin\phi\d\rho\d\phi\d\theta\). Each component of the triple integral corresponds to an integral over one variable (usually the innermost one is solved first, followed by the others). This method provides the cumulative volume of the solid within the specified parameters.
Volume Calculation
Calculating the volume of a solid involves summing up all the small parts defined by the integration process. In spherical coordinates, this is simplified through the use of triple integrals. By integrating the volume element \(dV = \rho^2\sin\phi\d\rho\d\phi\d\theta\), we determine how individual elements converge to form the entire volume.
For our example, we carefully select the order of integration (\

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Most popular questions from this chapter

Use a change of variables to evaluate the following integrals. $$\begin{aligned} &\iiint_{D} x y d V ; D \text { is bounded by the planes } y-x=0\\\ &y-x=2, z-y=0, z-y=1, z=0, \text { and } z=3 \end{aligned}$$

Improper integrals arise in polar coordinates when the radial coordinate \(r\) becomes arbitrarily large. Under certain conditions, these integrals are treated in the usual way: $$\int_{\alpha}^{\beta} \int_{a}^{\infty} f(r, \theta) r d r d \theta=\lim _{b \rightarrow \infty} \int_{\alpha}^{\beta} \int_{a}^{b} f(r, \theta) r d r d \theta$$ Use this technique to evaluate the following integrals. $$\iint_{R} \frac{d A}{\left(x^{2}+y^{2}\right)^{5 / 2}} ; R=\\{(r, \theta): 1 \leq r < \infty, 0 \leq \theta \leq 2 \pi\\}$$

Find equations for the bounding surfaces, set up a volume integral, and evaluate the integral to obtain a volume formula for each region. Assume that \(a, b, c, r, R,\) and h are positive constants. Find the volume of an ellipsoid with axes of length \(2 a\) \(2 b,\) and \(2 c\)

Parabolic coordinates Let \(T\) be the transformation \(x=u^{2}-v^{2}\) \(y=2 u v\) a. Show that the lines \(u=a\) in the \(u v\) -plane map to parabolas in the \(x y\) -plane that open in the negative \(x\) -direction with vertices on the positive \(x\) -axis. b. Show that the lines \(v=b\) in the \(u v\) -plane map to parabolas in the \(x y\) -plane that open in the positive \(x\) -direction with vertices on the negative \(x\) -axis. c. Evaluate \(J(u, v)\) d. Use a change of variables to find the area of the region bounded by \(x=4-y^{2} / 16\) and \(x=y^{2} / 4-1\) e. Use a change of variables to find the area of the curved rectangle above the \(x\) -axis bounded by \(x=4-y^{2} / 16\) \(x=9-y^{2} / 36, x=y^{2} / 4-1,\) and \(x=y^{2} / 64-16\) f. Describe the effect of the transformation \(x=2 u v\) \(y=u^{2}-v^{2}\) on horizontal and vertical lines in the \(u v\) -plane.

\(A\) thin rod of length \(L\) has a linear density given by \(\rho(x)=2 e^{-x / 3}\) on the interval \(0 \leq x \leq L\). Find the mass and center of mass of the rod. How does the center of mass change as \(L \rightarrow \infty ?\)
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