/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 The equation is \(z=10-r^{2}\). ... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 52

The equation is \(z=10-r^{2}\). $$\begin{aligned} V &=\int_{0}^{2 \pi} \int_{0}^{3} \int_{1}^{10-r^{2}} r d z d r d \theta=\int_{0}^{2 \pi} \int_{0}^{3} r\left(9-r^{2}\right) d r d \theta=\left.\int_{0}^{2 \pi}\left(\frac{9}{2} r^{2}-\frac{1}{4} r^{4}\right)\right|_{0} ^{3} d \theta \\ &=\int_{0}^{2 \pi} \frac{81}{4} d \theta=\frac{81 \pi}{2}. \end{aligned}$$

Short Answer

Expert verified
The volume is \( \frac{81\pi}{2} \).

Step by step solution

01

Understand the integral setup

The volume, denoted by V, is calculated using a cylindrical coordinate triple integral over specific ranges. The limits for the variables are \( r: [0, 3] \), \( \theta: [0, 2\pi] \), and \( z: [1, 10-r^2] \).
02

Integrate with respect to z

Given the equation \( z = 10 - r^2 \), for the inner integration over \( z \) from 1 to \( 10 - r^2 \), the integral becomes \( \int_{1}^{10-r^{2}} r \, dz \). Evaluating this, we get \( r(z) \bigg|_{1}^{10-r^{2}} = r((10 - r^2) - 1) = r(9 - r^2) \).
03

Integrate with respect to r

Substitute the result from the previous step into the next integral: \( \int_{0}^{3} r(9 - r^2) \, dr \). This becomes \( \int_{0}^{3} (9r - r^3) \, dr \). The antiderivative is \( \frac{9}{2}r^2 - \frac{1}{4}r^4 \). Evaluate this from 0 to 3.
04

Evaluate definite integral for r

Substitute the limits 0 and 3 into \( \frac{9}{2}r^2 - \frac{1}{4}r^4 \). Calculate: \((\frac{9}{2} \cdot 3^2 - \frac{1}{4} \cdot 3^4) - (\frac{9}{2} \cdot 0^2 - \frac{1}{4} \cdot 0^4) = \frac{81}{2} - \frac{81}{4} = \frac{81}{4} \).
05

Integrate with respect to \( \theta \)

The integration with respect to \( \theta \) is now \( \int_{0}^{2 \pi} \frac{81}{4} \, d\theta \). Since \( \frac{81}{4} \) is a constant, the integral becomes \( \frac{81}{4}[\theta]_{0}^{2\pi} = \frac{81}{4} \cdot 2\pi = \frac{81\pi}{2} \).
06

Final Step: Conclusion

The final volume is calculated as \( \frac{81\pi}{2} \). This result reflects the volume of the region defined by the given integrals and limits.

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Key Concepts

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

Cylindrical Coordinates
To understand cylindrical coordinates, think of them as a way to extend polar coordinates into three dimensions. We have three key components: radial distance (\( r \) ), angle (\( \theta \) ), and height (\( z \) ). These coordinates are particularly useful for problems involving symmetry around a central axis, like cylinders or cones.
  • The radial distance \( r \) represents the distance from the z-axis.
  • The angle \( \theta \) is the angular position from the positive x-axis and ranges from \( 0 \) to \( 2\pi \).
  • The height \( z \) indicates the vertical position, similar to the way y-coordinates operate in Cartesian coordinates.
Cylindrical coordinates are converted to Cartesian coordinates using the formulas: \( x = r \cos(\theta) \), \( y = r \sin(\theta) \), and \( z = z \).
In the given exercise, they allow the computation of volume under a curved surface, which, in turn, simplifies integrating over complex 3D shapes.
Volume Calculation
Calculating volume using triple integrals is like building a 3D picture of the space. You slice a big volume into tiny pieces, calculate the volume of each piece, and then sum them up. This slicing process is what integration does in three dimensions.
In the given problem, the volume is bounded by the surface \( z = 10 - r^2 \).
  • The surface acts as the upper limit of the volume.
  • For the cylinder with base in \( r \) and height \( z \) range 1 to \( 10 - r^2 \), each section of volume is calculated with respect to \( z \).
The resulting integration process involves evaluating the volume piece-wise and combining them to obtain the total volume of the region.
Definite Integral
Definite integrals play a crucial role in determining exact values over a specified interval. They measure the accumulation of quantities鈥攈ere, the small volume elements鈥攆rom the start to the end of the integration range.
  • In this exercise, each definite integral has specific limits defining the range over which integration occurs.
  • The integral over \( z \) is evaluated first, because it represents volume slices bounded by \( 1 \) and \( 10 - r^2 \).
  • Once integrated with respect to \( z \), the remaining definite integrals are evaluated with respect to \( r \) and \( \theta \).
The power of definite integration lies in providing precise volume measurements by considering the contribution of smaller sections one by one.
Limits of Integration
The limits of integration dictate over which region you apply each integral. They define the bounds of your integration problem and are fundamental in understanding the area or volume involved. In the given problem, the limits were carefully chosen to match the region's boundaries in the cylindrical coordinate system.
  • For \( r \) (radial distance), the limits are from \( 0 \) to \( 3 \), demonstrating the radial span from the origin outwards.
  • For \( \theta \) (angle), the full circle is covered, with limits \( 0 \) to \( 2\pi \).
  • The \( z \) (height) limits are from a fixed lower surface at \( 1 \) to the curved upper surface \( 10 - r^2 \).
Understanding these limits helps visualize and calculate the exact portion of 3D space considered in the integration.

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Most popular questions from this chapter

Solving \(x=2 u-4 v, y=3 u+v\) for \(u\) and \(v\) we obtain \(u=\frac{1}{14} x+\frac{2}{7} y, v=-\frac{3}{14} x+\frac{1}{7} y .\) The \(x y\) corner points (-4,1),(0,0),(2,3) correspond to the \(u v\) -points (0,1),(0,0),(1,0) \(\frac{\partial(x, y)}{\partial(u, v)}=\left|\begin{array}{cc}2 & -4 \\ 3 & 1\end{array}\right|=14\) $$\begin{aligned} \iint_{R} y d A &=\iint_{S}(3 u+v)(14) d A^{\prime}=14 \int_{0}^{1} \int_{0}^{1-u}(3 u+v) d v d u=\left.14 \int_{0}^{1}\left(3 u v+\frac{1}{2} v^{2}\right)\right|_{0} ^{1-u} d u \\ &=14 \int_{0}^{1}\left(\frac{1}{2}+2 u-\frac{5}{2} u^{2}\right) d u=\left.\left(7 u+14 u^{2}-\frac{35}{3} u^{3}\right)\right|_{0} ^{1}=\frac{28}{3} \end{aligned}$$

The equations are \(\rho=2, \theta=\pi / 4,\) and \(\theta=\pi / 3\). $$\begin{aligned} \int_{\pi / 4}^{\pi / 3} \int_{0}^{\pi / 2} \int_{0}^{2} \rho^{2} \sin \phi d \rho d \phi d \theta &=\left.\int_{\pi / 4}^{\pi / 3} \int_{0}^{\pi / 2} \frac{1}{3} \rho^{3} \sin \phi\right|_{0} ^{2} d \phi d \theta \\ &=\int_{\pi / 4}^{\pi / 3} \int_{0}^{\pi / 2} \frac{8}{3} \sin \phi d \phi d \theta=\frac{8}{3} \int_{\pi / 4}^{\pi / 3}-\left.\cos \phi\right|_{0} ^{\pi / 2} d \theta \\ &=\frac{8}{3} \int_{\pi / 4}^{\pi / 3}(0+1) d \theta=\frac{2 \pi}{9} \end{aligned}$$

Under the transformation \(u=y+z, v=-y+z, w=x-y\) the parallelepiped \(D\) is mapped to the parallelepiped \(E: 1 \leq u \leq 3,-1 \leq v \leq 1,0 \leq w \leq 3\) $$\frac{\partial(u, v, w)}{\partial(x, y, z)}=\left|\begin{array}{ccc} 0 & 1 & 1 \\ 0 & -1 & 1 \\ 1 & -1 & 0 \end{array}\right|=2 \Longrightarrow \frac{\partial(x, y, z)}{\partial(u, v, w)}=\frac{1}{2}$$ $$\begin{aligned} \iint_{D}(4 z+2 x-2 y) d V &=\iint_{E}(2 u+2 v+2 w) \frac{1}{2} d V^{\prime}=\frac{1}{2} \int_{0}^{3} \int_{-1}^{1} \int_{1}^{3}(2 u+2 v+2 w) d u d v d w \\ &=\left.\frac{1}{2} \int_{0}^{3} \int_{-1}^{1}\left(u^{2}+2 u v+2 u w\right)\right|_{1} ^{3} d v d w=\frac{1}{2} \int_{0}^{3} \int_{-1}^{1}(8+4 v+4 w) d v d w \\ &=\left.\int_{0}^{3}\left(4 v+v^{2}+2 v w\right)\right|_{-1} ^{1} d w=\int_{0}^{3}(8+4 w) d w=\left.\left(8 w+2 w^{2}\right)\right|_{0} ^{3}=42 \end{aligned}$$

$$\rho^{2}=4 \rho \cos \phi ; \rho=4 \cos \phi$$

\(\operatorname{div} \mathbf{F}=y^{2}+x^{2} .\) Using spherical coordinates, we have \(x^{2}+y^{2}=\rho^{2} \sin ^{2} \phi\) and \(z=\rho \cos \phi\) or \(\rho=z \sec \phi .\) Then $$\begin{aligned}\iint_{S} \mathbf{F} \cdot \mathbf{n} d S &=\iiint_{D}\left(x^{2}+y^{2}\right) d S=\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \int_{2 \sec \phi}^{4 \sec \phi} \rho^{2} \sin ^{2} \phi \rho^{2} \sin \phi d \rho d \phi d \theta \\ &=\left.\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \frac{1}{5} \rho^{5} \sin ^{3} \phi\right|_{2 \sec \phi} ^{4 \sec \phi} d \phi d \theta=\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \frac{992}{5} \sec ^{5} \phi \sin ^{3} \phi d \phi d \theta \\\ &=\int_{0}^{2 \pi} \int_{0}^{\pi / 4} \frac{992}{5} \tan ^{3} \phi \sec ^{2} \phi d \phi d \theta=\left.\frac{992}{5} \int_{0}^{2 \pi} \frac{1}{4} \tan ^{4} \phi\right|_{0} ^{\pi / 4} d \theta=\frac{992}{5} \int_{0}^{2 \pi} \frac{1}{4} d \theta=\frac{496 \pi}{5}\end{aligned}$$
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