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Chapter 15: Problem 15

Sketch the solid whose volume is given by the integral and evaluate the integral. \(\int_{0}^{4} \int_{0}^{2 \pi} \int_{r}^{4} r d z d \theta d r\)

Short Answer

Expert verified
The volume of the solid is \( \frac{64\pi}{3} \).

Step by step solution

01

Understand the Integral Setup

The given integral is \[ \int_{0}^{4} \int_{0}^{2\pi} \int_{r}^{4} r \, dz \, d\theta \, dr \]This integral is presented in cylindrical coordinates. The order of integration is with respect to \(z\), then \(\theta\), and finally \(r\). This suggests that the volume is formed by rotating surfaces around the \(z\)-axis.
02

Identify the Limits of Integration

The limits for \(z\) vary from \(r\) to \(4\), the limits for \(\theta\) from \(0\) to \(2\pi\), and for \(r\) from \(0\) to \(4\). The lower bound for \(z\) being \(r\) suggests a slanted base starting from the \(z=r\) line upwards to the \(z=4\) plane.
03

Visualize the Solid

The solid is a volume of revolution formed by revolving the region between \(z=r\) and \(z=4\) around the \(z\)-axis. The upper boundary is a constant plane \(z=4\) and the lower boundary is sloped, starting from each \(r\) and extending vertically. The region extends for full rotation as \(\theta\) spans from \(0\) to \(2\pi\).
04

Integrate with Respect to z

Evaluate the innermost integral: \[ \int_{r}^{4} r \, dz \]. This simplifies to: \[ r(z) \bigg|_{r}^{4} = r(4 - r) \].
05

Integrate with Respect to 胃

Insert the result from step 4 into the next integral: \[ \int_{0}^{2\pi} r(4 - r) \, d\theta \]This integral simplifies to: \[ (r(4 - r))(\theta) \bigg|_{0}^{2\pi} = 2\pi r(4-r) \].
06

Integrate with Respect to r

Evaluate the outermost integral: \[ \int_{0}^{4} 2\pi r(4-r) \, dr \]This function can be expanded and integrated:\[ 2\pi \int_{0}^{4} (4r - r^2) \, dr = 2\pi \left( 2r^2 - \frac{r^3}{3} \right) \bigg|_{0}^{4} \].
07

Compute the Final Volume

Evaluate the final expression:\[ 2\pi \left( 2(4)^2 - \frac{(4)^3}{3} \right) = 2\pi \left( 32 - \frac{64}{3} \right) \]Calculating inside the brackets: \[ 32 - \frac{64}{3} = \frac{96}{3} - \frac{64}{3} = \frac{32}{3} \]Thus, the final volume is: \[ \frac{64\pi}{3} \].

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Key Concepts

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

Cylindrical Coordinates
Cylindrical coordinates are an extension of polar coordinates into three dimensions. They are particularly useful when dealing with problems involving circular symmetry around an axis, commonly the z-axis. In this system, a point in space is represented by three components: radius \( r \), angle \( \theta \), and height \( z \). These components relate to the Cartesian coordinates \( (x, y, z) \) as follows:
  • \( x = r \cos (\theta) \)
  • \( y = r \sin (\theta) \)
  • \( z = z \)
Cylindrical coordinates are especially powerful for evaluating triple integrals where the region of integration has rotational symmetry about one axis. The given integral in this problem, \( \int_{0}^{4} \int_{0}^{2\pi} \int_{r}^{4} r \, dz \, d\theta \, dr \), takes advantage of this situation by allowing the integral to be solved in a more flexible and intuitive manner than in Cartesian coordinates. By understanding cylindrical coordinates, we can effectively break down the solid into simpler segments, each defined by radius, angle, and height.
Volume of Revolution
The volume of revolution is a key concept in calculus used to find the volume of a solid formed by rotating a region around an axis. The region here is determined by the limits \( z = r \) to \( z = 4 \), which creates a sloped boundary rotating around the z-axis. This type of geometric formation lends itself perfectly to expressions in cylindrical coordinates.The reason this works so well is due to the symmetry created by the rotation. It simplifies the integral and allows for the formulation of a single triple integral that accounts for all aspects of the volume. In our example, after rotating around the z-axis through full circular angle from \( \theta = 0 \) to \( \theta = 2\pi \), the solid's volume is captured by the cylindrical coordinate integral calculation, leading to an efficient computation of the desired volume.
Integration Limits
Understanding integration limits is crucial when setting up a triple integral. These limits define the boundary conditions of the volume under consideration. For this particular problem, the limits were defined as follows:
  • For \( z \), the limits are from \( r \) to \( 4 \). This signifies that at any particular radius \( r \), the height of the solid begins at \( z = r \) and extends up to the plane \( z = 4 \).
  • For \( \theta \), the limits are from \( 0 \) to \( 2 \pi \), representing a complete rotation around the z-axis, effectively covering the entire circular symmetry.
  • For \( r \), the limits are from \( 0 \) to \( 4 \). These limits ensure that every possible radius from the center up to the edge of the base is accounted for during the integration.
Setting integration limits correctly is fundamental, as they dictate the size and shape of the volume being integrated over. Errors in these can lead to incorrect evaluation of the integral, yielding results that do not accurately reflect the intended physical volume.

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Most popular questions from this chapter

Evaluate the integral by making an appropriate change of variables. \(\iint_{R} e^{x+y} d A,\) where \(R\) is given by the inequality \(|x|+|y| \leqslant 1\)

(a) We define the improper integral (over the entire plane \(\mathbb{R}^{2}\) ) $$\begin{aligned} I &=\iint e^{-\left(x^{2}+y^{2}\right)} d A=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d y d x \\ &=\lim _{a \rightarrow \infty} \iint e^{-\left(x^{2}+y^{2}\right)} d A \end{aligned}$$ where \(D_{a}\) is the disk with radius \(a\) and center the origin. Show that $$\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-\left(x^{2}+y^{2}\right)} d A=\pi$$ (b) An equivalent definition of the improper integral in part (a) is $$\iint_{\mathbb{R}^{\prime}} e^{-\left(x^{2}+y^{2}\right)} d A=\lim _{a \rightarrow \infty} \iint_{S_{a}} e^{-\left(x^{2}+y^{2}\right)} d A$$ where \(S_{,}\) is the square with vertices \((\pm a, \pm a) .\) Use this to show that $$int_{-\infty}^{\infty} e^{-x^{2}} d x \int_{-\infty}^{\infty} e^{-y^{2}} d y=\pi$$ (c) Deduce that $$\int_{-\infty}^{\infty} e^{-x^{2}} d x=\sqrt{\pi}$$ (d) By making the change of variable \(t=\sqrt{2} x,\) show that $$\int_{-\infty}^{\infty} e^{-x^{2} / 2} d x=\sqrt{2 \pi}$$ (This is a fundamental result for probability and statistics.)

Graph the solid bounded by the plane \(x+y+z=1\) and the paraboloid \(z=4-x^{2}-y^{2}\) and find its exact volume. (Use your CAS to do the graphing, to find the equations of the boundary curves of the region of integration, and to evaluate the double integral.)

The latitude and longitude of a point \(P\) in the Northern Hemisphere are related to spherical coordinates \(\rho, \theta, \phi\) as follows. We take the origin to be the center of the earth and the positive \(z\) -axis to pass through the North Pole. The positive \(x\) -axis passes through the point where the prime meridian (the meridian through Greenwich, England) intersects the equator. Then the latitude of \(P\) is \(\alpha=90^{\circ}-\phi^{\circ}\) and the longitude is \(\beta=360^{\circ}-\theta^{\circ} .\) Find the great-circle distance from Los Angeles (lat. \(34.06^{\circ} \mathrm{N},\) long. \(118.25^{\circ} \mathrm{W}\) ) to Montreal (lat. \(45.50^{\circ} \mathrm{N},\) long. \(73.60^{\circ} \mathrm{W} ) .\) Take the radius of the earth to be 3960 \(\mathrm{mi} .\) (A great circle is the circle of intersection of a sphere and a plane through the center of the sphere.)

Sketch the solid whose volume is given by the iterated integral. $$ \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{2-2 z} d y d z d x $$
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