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Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoBriggs 3rd EditionCh. 6 - Applications of IntegrationProblema 6.R.21
Capitolo 6, Problema 6.R.21

14–25. {Use of Tech} Areas of regions Determine the area of the given region.


The region bounded by y = x²,y = 2x²−4x, and y = 0

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First, identify the curves that bound the region: \( y = x^2 \), \( y = 2x^2 - 4x \), and \( y = 0 \). We want to find the area enclosed by these curves.
Next, find the points of intersection between the curves to determine the limits of integration. Start by setting \( y = x^2 \) equal to \( y = 2x^2 - 4x \) and solve for \( x \): \[ x^2 = 2x^2 - 4x \]
Rearrange the equation to isolate terms: \[ 0 = 2x^2 - 4x - x^2 = x^2 - 4x \] Then factor: \[ x(x - 4) = 0 \] So the intersection points are at \( x = 0 \) and \( x = 4 \).
Determine which curve is on top (greater \( y \) value) between \( x = 0 \) and \( x = 4 \) by testing a point in the interval, for example \( x = 1 \). This will help set up the integral for the area between the curves.
Set up the integral for the area between the curves and above \( y = 0 \) as: \[ \text{Area} = \int_{0}^{4} \left( \text{top function} - \text{bottom function} \right) \, dx \] where the top and bottom functions are determined from the previous step.

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Finding Points of Intersection

To determine the area bounded by curves, first find where the curves intersect by setting their equations equal. These intersection points define the limits of integration and help identify the region enclosed by the curves.
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Critical Points

Setting Up Definite Integrals for Area

The area between curves is found by integrating the difference of the functions over the interval defined by their intersection points. Specifically, integrate the upper function minus the lower function with respect to x to calculate the enclosed area.
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Percorso guidato
05:43
Definition of the Definite Integral

Understanding the Role of the x-axis (y=0)

Since the region is bounded by y=0 (the x-axis), it acts as a boundary line. Recognizing when the x-axis forms part of the boundary helps in correctly setting up the integral and determining which parts of the curves contribute to the enclosed area.
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Disk Method Using y-Axis
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43–55. Volumes of solids Choose the general slicing method, the disk/washer method, or the shell method to answer the following questions.


The region bounded by the curves y = sec x and y=2, for 0 ≤ x ≤ π/3, is revolved about the x-axis. What is the volume of the solid that is generated? 

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Comparing volumes Let R be the region bounded by y=1/x^p and the x-axis on the interval [1, a], where p>0 and a>1 (see figure). Let Vₓ and Vᵧ be the volumes of the solids generated when R is revolved about the x- and y-axes, respectively.


d. Find a general expression for Vᵧ in terms of a and p. Note that p=2 is a special case. What is Vᵧ when p=2?

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43–55. Volumes of solids Choose the general slicing method, the disk/washer method, or the shell method to answer the following questions.


What is the volume of the solid whose base is the region in the first quadrant bounded by y = √x,y = 2-x, and the x-axis, and whose cross sections perpendicular to the base and parallel to the y-axis are semicircles?

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Position, displacement, and distance A projectile is launched vertically from the ground at t=0, and its velocity in flight (in m/s) is given by v(t)=20−10t. Find the position, displacement, and distance traveled after t seconds, for 0≤t≤4.

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82–84. Fluid Forces Suppose the following plates are placed on a vertical wall so that the top of the plate is 2 m below the surface of a pool that is filled with water. Compute the force on each plate.


A circular plate with a radius of 2 m

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27–33. Multiple regions The regions R₁,R₂, and R₃ (see figure) are formed by the graphs of y = 2√x,y = 3−x,and x=3.


Find the volume of the solid obtained by revolving region R₂ about the y-axis.

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