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Chapter 6: Q. 9 (page 574)

Finding geometric quantities with definite integrals: Set up and solve definite integrals to find each volume, surface area, or arc length that follows. Solve each volume problem both with disks/washers and with shells, if possible.
The volume of the solid obtained by revolving the region between the graph of $f (x) = x^{2}$ and the $y -$ axis for $0 \leq x \leq 2$ around (a) the $x$ -axis and (b) the $y$ -axis .

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Step by step solution

01

Step 1. Given Information.

The volume of the solid obtained by revolving the region between the graph of $f (x) = x^{2}$ and the $y -$ axis for $0 \leq x \leq 2$ .

02

(a) Step 2. Calculation.

The volume is given by :

$V = 2 π \int_{a}^{b} yg (y) dy$

As $y = x^{2}$

$V = 2 π \int_{a}^{b} y f (x) d y V = 2 π \int_{0}^{\sqrt{2}} y \cdot y^{2} d y V = 2 π \int_{0}^{\sqrt{2}} y^{3} d y V = 2 π {[\frac{y^{4}}{4}]}_{0}^{\sqrt{2}} V = \frac{2 π}{4} \times 4 V = 2 π$

03

(b) Step 3. Around the y-axis.

The volume of solid revolving about the $y -$ axis is

role="math" localid="1652262266017" $V = 2 π \int_{a}^{b} xf (x) dx V = 2 π \int_{0}^{2} x \cdot x^{2} dx V = 2 π \int_{0}^{2} x^{3} dx V = 2 π {[\frac{x^{4}}{4}]}_{0}^{2} V = \frac{2 π}{4} \times 16 V = 8 π$

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Most popular questions from this chapter

Approximate the arc length of f (x) on [a, b], using n line segments and the distance formula. Include a sketch of f (x) and the line segments .

$f (x) = x^{3}, [a, b] = [- 2, 2], n = 4$

Consider the region between the graph of $f (x) = x - 2$ and the x-axis on [2,5]. For each line of rotation given in Exercises 35– 40, use definite integrals to find the volume of the resulting solid.

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52

$\frac{d y}{d x} = \frac{x}{1 + x^{2}}$

Use antidifferentiation and/or separation of variables to solve each of the initial-value problems in Exercises 29–52

$\frac{d y}{d x} = e^{x + y}, y (0) = 2$

Find the exact value of the arc length of each function f(x) on [a, b] by writing the arc length as a definite integral and then solving that integral.

$f (x) = \frac{1}{3} x^{3 / 2} - x^{1 / 2}$ , $[a, b] = [0, 1]$

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