Chapter 13: Q. 31 (page 1083)

Evaluating triple integrals: Each of the triple integrals that follows represents the volume of a solid. Sketch the solid and evaluate the integral.
$\int_{0}^{2 π} \int_{0}^{3} \int_{0}^{\sqrt{9 - r^{2}}} r d z d r d θ$

Short Answer

Expert verified

$\int_{0}^{2 π} \int_{0}^{3} \int_{0}^{\sqrt{9 - r^{2}}} r d z d r d θ = 18 π$

Step by step solution

Given

We have to find the volume by using the triple integral.

\int_{0}^{2 π} \int_{0}^{3} \int_{0}^{\sqrt{9 - r^{2}}} r d z d r d θ

Evaluate the integral

$V = \int_{0}^{2 π} \int_{0}^{3} [\int_{0}^{\sqrt{9 - r^{2}}} r d z] d r d θ V = \int_{0}^{2 π} [\int_{0}^{3} r \sqrt{9 - r^{2}} d r] d θ$

Substitute,

$9 - r^{2} = t^{2} - 2 r d r = 2 t d t d r = - \frac{t}{r} d t When r = 0, t = \sqrt{9} = 3 When r = 3, t = 9 - 9 = 0$

Therefore,

$V = - \int_{0}^{2 π} [\int_{3}^{0} t^{2} d t] d θ V = - \int_{0}^{2 π} {[\frac{t^{3}}{3}]}_{3}^{0} d θ V = 9 \int_{0}^{2 π} d θ V = 9 (2 π - 0) V = 18 π cubic units$

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Evaluating triple integrals: Each of the triple integrals that follows represents the volume of a solid. Sketch the solid and evaluate the integral.
$\int_{0}^{2 π} \int_{0}^{3} \int_{0}^{\sqrt{9 - r^{2}}} r d z d r d θ$

Short Answer

Step by step solution

Given

Evaluate the integral

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Evaluating triple integrals: Each of the triple integrals that follows represents the volume of a solid. Sketch the solid and evaluate the integral.∫02π∫03∫09-r2rdzdrdθ

Short Answer

Step by step solution

Given

Evaluate the integral

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Most popular questions from this chapter

Recommended explanations on Math Textbooks

Geometry

Mechanics Maths

Logic and Functions

Probability and Statistics

Theoretical and Mathematical Physics

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Study anywhere. Anytime. Across all devices.

Evaluating triple integrals: Each of the triple integrals that follows represents the volume of a solid. Sketch the solid and evaluate the integral.
$\int_{0}^{2 π} \int_{0}^{3} \int_{0}^{\sqrt{9 - r^{2}}} r d z d r d θ$