Chapter 6: Q. 11 (page 575)

The volume of the solid obtained by revolving the region between the graphs of $f (x) = \sqrt{x}$ and $g (x) = x^{3}$ on $[0, 1]$ around (a) the y-axis and (b) the line $x = 2$ .

Short Answer

Expert verified

(a) The volume is $\frac{2}{5} π$ cubic units.

(b) The volume is $\frac{19}{15} π$ cubic units.

Step by step solution

Volume around y-axis

The required region is shown below,

$y = \sqrt{x} \Rightarrow x = y^{2} y = x^{3} \Rightarrow x = y^{1 / 3}$

The volume is computed as,

$V = π \int_{0}^{1} [y^{2 / 3} - y^{4}] dy = π {[\frac{3}{5} y^{5 / 3} - \frac{y^{5}}{5}]}_{0}^{1} = π [\frac{3}{5} - \frac{1}{5}] = \frac{2}{5} π cubic units$

Volume around line x=2

The outer radius is $2 - y^{2}$ .

Inner radius is $2 - y^{1 / 3}$ .

The volume is computed as,

$V = π \int_{0}^{1} [{(2 - y^{2})}^{2} - {(2 - y^{1 / 3})}^{2}] dy = π \int_{0}^{1} [y^{4} - 4 y^{2} - y^{2 / 3} + 4 y^{1 / 3}] dy = π {[\frac{y^{5}}{5} - \frac{4}{3} y^{3} - \frac{3}{5} y^{5 / 3} + 3 y^{4 / 3}]}_{0}^{1} = π [\frac{1}{5} - \frac{4}{3} - \frac{3}{5} + 3] = \frac{19}{15} π cubic units .$

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The volume of the solid obtained by revolving the region between the graphs of $f (x) = \sqrt{x}$ and $g (x) = x^{3}$ on $[0, 1]$ around (a) the y-axis and (b) the line $x = 2$ .

Short Answer

Step by step solution

Volume around y-axis

Volume around line x=2

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The volume of the solid obtained by revolving the region between the graphs of f(x)=x and g(x)=x3 on 0,1 around (a) the y-axis and (b) the line x=2.

Short Answer

Step by step solution

Volume around y-axis

Volume around line x=2

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Statistics

Logic and Functions

Mechanics Maths

Decision Maths

Theoretical and Mathematical Physics

Pure Maths

Study anywhere. Anytime. Across all devices.

The volume of the solid obtained by revolving the region between the graphs of $f (x) = \sqrt{x}$ and $g (x) = x^{3}$ on $[0, 1]$ around (a) the y-axis and (b) the line $x = 2$ .