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Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
All textbooksHass 15th EditionCh. 8 - Techniques of IntegrationProblem 8.6.59b
Chapter 8, Problem 8.6.59b

Centroid:
Find the centroid of the region cut from the first quadrant by the curve
y = 1/√(x + 1) and the line x = 3.

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1
Identify the region bounded by the curve \(y = \frac{1}{\sqrt{x + 1}}\), the vertical line \(x = 3\), and the coordinate axes in the first quadrant. This means the region is bounded by \(x = 0\), \(x = 3\), \(y = 0\), and the curve itself.
Set up the expressions for the area \(A\) of the region using the integral: \(A = \int_0^3 \frac{1}{\sqrt{x + 1}} \, dx\). This integral will give the total area under the curve from \(x=0\) to \(x=3\).
Find the coordinates of the centroid \((\bar{x}, \bar{y})\) using the formulas: \(\bar{x} = \frac{1}{A} \int_0^3 x \cdot \frac{1}{\sqrt{x + 1}} \, dx\) and \(\bar{y} = \frac{1}{A} \int_0^3 \frac{1}{2} \left( \frac{1}{\sqrt{x + 1}} \right)^2 \, dx\). Note that \(\bar{y}\) is found by integrating the square of the function divided by 2 because the region is bounded below by \(y=0\).
Evaluate each integral separately: the area integral, the \(x\)-moment integral, and the \(y\)-moment integral. Use substitution methods if necessary to simplify the integrals.
After computing the integrals, substitute the values back into the centroid formulas to find \(\bar{x}\) and \(\bar{y}\). These coordinates represent the centroid of the region.

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Key Concepts

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

Centroid of a Region

The centroid is the geometric center or 'balance point' of a plane region. It is found by calculating the average position of all points in the area, typically using integrals to find the coordinates (x̄, ȳ). For regions bounded by curves, the centroid coordinates are given by the moments divided by the area.
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Area of Polar Regions

Definite Integrals for Area and Moments

Definite integrals are used to compute the area under a curve and the moments about the axes. The area is found by integrating the function over the given interval, while moments involve integrating the product of the function and the coordinate variable. These integrals are essential for determining the centroid.
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Definition of the Definite Integral

Region Bounded by a Curve and a Vertical Line

The region is defined by the curve y = 1/√(x + 1), the vertical line x = 3, and the coordinate axes in the first quadrant. Understanding the limits of integration and the shape of this region is crucial for setting up the correct integrals to find area and moments.
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Finding Area Between Curves on a Given Interval
Related Practice
Textbook Question

In Exercises 11–22, estimate the minimum number of subintervals needed to approximate the integrals with an error of magnitude less than 10^-4 by (b) Simpson’s Rule. (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)

∫ from 0 to 2 of sin(x + 1) dx

Textbook Question

Consider the region bounded by the graphs of

y = arctan(x), y = 0, and x = 1.

b. Find the volume of the solid formed by revolving this region about the y-axis.

Textbook Question

In Exercises 11–22, estimate the minimum number of subintervals needed to approximate the integrals with an error of magnitude less than 10^-4 by (b) Simpson’s Rule. (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)

∫ from 1 to 2 of x dx

Textbook Question

Consider the region bounded by the graphs of y = sin⁻¹(x), y = 0, and x = 1/2.

b. Find the centroid of the region.

Textbook Question

Finding area

Find the area of the region enclosed by the curve y = x cos(x) and the x-axis (see the accompanying figure) for:

b. 3π/2 ≤ x ≤ 5π/2.

1
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Textbook Question

In Exercises 11–22, estimate the minimum number of subintervals needed to approximate the integrals with an error of magnitude less than 10^-4 by (b) Simpson’s Rule. (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)

∫ from 2 to 4 of 1/(s - 1)² ds