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Chapter 8: Problem 6

The area enclosed between the \(x\) axis, the curve \(y=x(2-x)\) and the ordinates \(x=1\) and \(x=2\) is rotated through \(2 \pi\) radians about the \(x\) axis. Calculate (a) the rotating area and the coordinates of its centroid; (b) the volume of the solid of revolution formed and the coordinates of its centre of gravity.

Short Answer

Expert verified
(a) Rotated area \(\frac{5}{3}\), centroid \(\left( \frac{11}{7}, \frac{9}{10} \right)\); (b) Volume \(\frac{9\pi}{10}\), center of gravity \(\left( \frac{11}{7}, 0 \right)\).

Step by step solution

01

Define the region's area

The function is given as \(y = x(2-x)\). The ordinates are \(x = 1\) and \(x = 2\). First, we find the area under the curve by integrating \(y\) from \(x=1\) to \(x=2\). This area can be calculated using the integral: \[ A = \int_{1}^{2} x(2-x) \, dx \].
02

Compute the integral for the area

Integrate \(x(2-x)\):\[ \int x(2-x) \, dx = \int (2x - x^2) \, dx \].Calculate the integral:\[ \int (2x - x^2) \, dx = \left[ x^2 - \frac{x^3}{3} \right] \].Evaluate from 1 to 2:\[ A = \left[ (2^2 - \frac{2^3}{3}) - (1^2 - \frac{1^3}{3}) \right] = [4 - \frac{8}{3}] - [1 - \frac{1}{3}] \].This simplifies to:\[ A = \frac{5}{3} \].
03

Find the coordinates of the centroid

The formula for the centroid \((\bar{x}, \bar{y})\) of the region is:\( \bar{x} = \frac{1}{A} \int_{1}^{2} x \, (x(2-x)) \, dx \)\( \bar{y} = \frac{1}{2A} \int_{1}^{2} (x(2-x))^2 \, dx \).Use \(A = \frac{5}{3}\) to calculate these values.
04

Calculate the x-coordinate of the centroid (\(\bar{x}\))

Compute \(\bar{x}\):\[ \bar{x} = \frac{3}{5} \int_{1}^{2} x^2(2-x) \, dx \].This expands to:\[ \int_{1}^{2} (2x^2 - x^3) \, dx \].The integral evaluates to:\[ (\frac{2}{3}x^3 - \frac{x^4}{4}) \] from 1 to 2.Calculate this:\[ \bar{x} = \frac{3}{5}([\frac{16}{3} - 4] - [\frac{2}{3} - \frac{1}{4}]) \].This results in:\(\bar{x} = \frac{11}{7}\).
05

Calculate the y-coordinate of the centroid (\(\bar{y}\))

Compute \(\bar{y}\) using:\[ (x(2-x))^2 = (2x - x^2)^2 = 4x^2 - 4x^3 + x^4 \].So, \( \int_{1}^{2} (4x^2 - 4x^3 + x^4) \, dx \) results in:\([ \frac{4}{3} x^3 - x^4 + \frac{x^5}{5} ]\) from 1 to 2.Calculate this value:\(\bar{y} = \frac{3}{10} ([ (\frac{32}{3} - 16 + \frac{32}{5}) - (\frac{4}{3} - 1 + \frac{1}{5}) ])\).\(\bar{y} = \frac{9}{10}\).
06

Find the volume of solid of revolution

The volume of the solid formed by rotating the area about the x-axis is given by the formula:\[ V = \pi \int_{1}^{2} (x(2-x))^2 \, dx \].The integral has been calculated in Step 5. Thus,\[ V = \pi \cdot \frac{9}{10} = \frac{9\pi}{10} \].
07

Determine the center of gravity of the solid

The center of gravity for solids of revolution involves calculating the same \(\bar{x}\) as in Step 4 for the x-coordinate, and dividing the volume between x and y surfaces. For y-coordinate: rotate symmetry applies to involve the center is on x-axis so y=0.Therefore,\([\bar{x}_{cg}, \bar{y}_{cg}] = \left(\frac{11}{7}, 0 \right)\).

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Key Concepts

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

Integration
Integration is a powerful mathematical tool used to calculate the area under curves. Think of it as a way to sum up parts to find the whole. In the context of this exercise, we use integration to determine the area beneath the curve of the function given by \( y = x(2-x) \) between the limits \( x=1 \) and \( x=2 \).

We set up the integral \( \int_{1}^{2} x(2-x) \, dx \) to find the area enclosed by the curve and the x-axis. By breaking down the function to \( \int (2x - x^2) \, dx \), and calculating its antiderivative, we get \( \left[ x^2 - \frac{x^3}{3} \right] \).
The definite integral is then evaluated from 1 to 2, leading to the area \( A = \frac{5}{3} \).

Integration is key for finding the exact area and is particularly useful when this region is rotated around the x-axis.
Centroid Calculation
The centroid of a shape is its "center of mass." For geometrical shapes, it represents a balancing point. For our application, calculating the centroid of the area under the curve is crucial before it forms a solid of revolution.

To find the x-coordinate of the centroid, \( \bar{x} \), we set up the integral \( \bar{x} = \frac{1}{A} \int_{1}^{2} x \cdot (x(2-x)) \, dx \). By solving this, we arrive at \( \bar{x} = \frac{11}{7} \).

The calculation for the y-coordinate of the centroid makes use of \( \bar{y} = \frac{1}{2A} \int_{1}^{2} (x(2-x))^2 \, dx \). Once evaluated, it results in \( \bar{y} = \frac{9}{10} \).

This centroid serves as a pivotal axis around which the area is rotated, leading to a solid of revolution.
Volume Calculation
Volume calculation is essential in determining the size of the solid of revolution formed when an area is rotated around an axis. In this problem, our region is rotated around the x-axis to form the solid.

The volume can be calculated using the disk method. This involves finding the integral of the area of circular disks along the axis of rotation. The key formula here is \( V = \pi \int_{1}^{2} (x(2-x))^2 \, dx \). We have already computed such an integral earlier in the centroid calculation.

With the value evaluated in previous steps being \( \frac{9}{10} \), multiplying by \( \pi \), we find the volume of the solid to be \( \frac{9\pi}{10} \). Understanding the quantity of space this solid occupies is crucial, especially when relating it to its physical applications.

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