Chapter 6: Q. 10 (page 575)

The volume of the solid obtained by revolving the region between the graph of $f (x) = 9 - x^{2}$ and the x-axis on $[- 3, 3]$ around (a) the x-axis and (b) the line $y = - 3$ .

Short Answer

Expert verified

(a) The volume is $\frac{1296}{5} π$ cubic units.

(b) The volume is $\frac{216}{5} π$ cubic units.

Step by step solution

Volume around x-axis

The required region is shown below,

The volume is computed as,

$V = π \int_{- 3}^{3} y^{2} dx = π \int_{- 3}^{3} {(9 - x^{2})}^{2} dx = 2 π \int_{0}^{3} (81 + x^{4} - 18 x^{2}) dx = 2 π {[81 x + \frac{x^{5}}{5} - 6 x^{3}]}_{0}^{3} = \frac{1296}{5} π cubic units$

Volume around y=-3

The outer radius of the cross-section is,

$y = 6 - x^{2}$

The inner radius is, $y = - 3$

The volume is computed as,

$V = π \int_{- 3}^{3} [{(6 - x^{2})}^{2} - {(- 3)}^{2}] dx = 2 π \int_{0}^{3} (36 + x^{4} - 12 x^{2} - 9) dx = 2 π {[27 x + \frac{x^{5}}{5} - 4 x^{3}]}_{0}^{3} = \frac{216}{5} π cubic units$

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The volume of the solid obtained by revolving the region between the graph of $f (x) = 9 - x^{2}$ and the x-axis on $[- 3, 3]$ around (a) the x-axis and (b) the line $y = - 3$ .

Short Answer

Step by step solution

Volume around x-axis

Volume around y=-3

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The volume of the solid obtained by revolving the region between the graph of f(x)=9-x2 and the x-axis on -3,3 around (a) the x-axis and (b) the line y=−3.

Short Answer

Step by step solution

Volume around x-axis

Volume around y=-3

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Pure Maths

Theoretical and Mathematical Physics

Calculus

Probability and Statistics

Statistics

Study anywhere. Anytime. Across all devices.

The volume of the solid obtained by revolving the region between the graph of $f (x) = 9 - x^{2}$ and the x-axis on $[- 3, 3]$ around (a) the x-axis and (b) the line $y = - 3$ .