Ch. 6 - Applications of Integration

Briggs - Calculus: Early Transcendentals 3rd Edition

Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook

Briggs 3rd Edition

Ch. 6 - Applications of Integration

Problem 6.2.19

Chapter 6, Problem 6.2.19

Determine the area of the shaded region in the following figures.

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Identify the curves that bound the shaded region. The region is bounded by the line \(x = 2y\) and the curve \(x = y^2 - 3\).

Determine the points of intersection by setting the two expressions for \(x\) equal: \(2y = y^2 - 3\). Rearrange this to form a quadratic equation: \(y^2 - 2y - 3 = 0\).

Solve the quadratic equation \(y^2 - 2y - 3 = 0\) to find the \(y\)-values of the intersection points. These will be the limits of integration.

Set up the integral for the area of the shaded region with respect to \(y\). The area is given by the integral of the difference between the rightmost curve and the leftmost curve: \(\int_{y_1}^{y_2} [(2y) - (y^2 - 3)] \, dy\).

Evaluate the integral to find the area. This involves integrating the function \(2y - y^2 + 3\) over the interval from \(y_1\) to \(y_2\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Finding Points of Intersection

To determine the area between two curves, first find their points of intersection by solving their equations simultaneously. These points define the limits of integration and ensure the correct region is considered.

Setting Up the Integral for Area Between Curves

The area between two curves is found by integrating the difference of the functions over the interval defined by their intersection points. When curves are given as x in terms of y, integrate with respect to y, subtracting the left curve from the right.

Integrating with Respect to y

Since the curves are expressed as x = f(y), the integral for the area is set up with respect to y. This involves integrating the horizontal distance between the curves (right minus left) over the y-interval, which may differ from the usual vertical integration.

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Related Practice

Textbook Question

29–36. Position and velocity from acceleration Find the position and velocity of an object moving along a straight line with the given acceleration, initial velocity, and initial position. Use the Fundamental Theorem of Calculus (Theorems 6.1 and 6.2).

a(t) = e^−t; v(0) = 60; s(0) = 40

Textbook Question

Filling a spherical tank A spherical water tank with an inner radius of 8 m has its lowest point 2 m above the ground. It is filled by a pipe that feeds the tank at its lowest point (see figure). Neglecting the volume of the inflow pipe, how much work is required to fill the tank if it is initially empty?

Textbook Question

a(t) = cos2t; v(0) = 5; s(0) = 7

Textbook Question

Emptying a partially filled swimming pool If the water in the swimming pool in Exercise 35 is 2 m deep, then how much work is required to pump all the water to a level 3 m above the bottom of the pool?

Textbook Question

Why is the disk method a special case of the general slicing method?

Textbook Question

Work from force How much work is required to move an object from x=1 to x=3 (measured in meters) in the presence of a force (in N) given by F(x) = 2x² acting along the x-axis?

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