Domanda del libro di testo
88. The region in Exercise 87 is revolved about the x-axis to generate a solid.
a. Find the volume of the solid.
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Ch. 8 - Techniques of Integration
Hass15th EditionThomas' CalculusISBN: 9780137616077
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TOC.tableOfContents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions490
- Ch. 8 - Techniques of Integration399
- Ch. 9 - First-Order Differential Equations66
- Ch. 10 - Infinite Sequences and Series132
- Ch. 11 - Parametric Equations and Polar Coordinates58
Capitolo 8, Problema 8.6.59a
Centroid:
Find the centroid of the region cut from the first quadrant by the curve
y = 1/√(x + 1) and the line x = 3.
Guida verificata passo dopo passo1
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. The region lies between \(x = 0\) and \(x = 3\), and above the \(x\)-axis.
Set up the integral for the area \(A\) of the region using the formula \(A = \int_0^3 y \, dx = \int_0^3 \frac{1}{\sqrt{x + 1}} \, dx\). This 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:
\(\bar{x} = \frac{1}{A} \int_0^3 x y \, dx = \frac{1}{A} \int_0^3 x \frac{1}{\sqrt{x + 1}} \, dx\)
and
\(\bar{y} = \frac{1}{2A} \int_0^3 y^2 \, dx = \frac{1}{2A} \int_0^3 \left(\frac{1}{\sqrt{x + 1}}\right)^2 \, dx\).
Evaluate each integral separately: the area integral, the \(x\)-moment integral \(\int_0^3 x y \, dx\), and the \(y\)-moment integral \(\int_0^3 y^2 \, dx\). Use substitution if necessary, for example, let \(u = x + 1\) to simplify the integrals.
After computing the integrals, substitute the results back into the centroid formulas to find \(\bar{x}\) and \(\bar{y}\). These values give the coordinates of the centroid of the region.
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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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Percorso guidato
07:45
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 x or y coordinates. These integrals are essential to determine the centroid coordinates.
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Percorso guidato
05:43Definition 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 the region is crucial for setting up the correct integrals to find area and moments.
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05:23Finding Area Between Curves on a Given Interval
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Domanda del libro di testo
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Domanda del libro di testo
Finding volume: Find the volume of the solid generated by revolving the region bounded by the x-axis and the curve y = x sin(x), 0 ≤ x ≤ π, about
a. The y-axis.
(See Exercise 57 for a graph.)
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Domanda del libro di testo
Evaluate ∫ sec θ dθ by:
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