Chapter 13: Q. 40 (page 1027)

Each of the integrals or integral expressions in Exercises 39-46 represents the volume of a solid in $R^{3}$ . Use polar coordinates to describe the solid, and evaluate the expressions.
$2 \int_{0}^{2 π} \int_{0}^{R} r \sqrt{R^{2} - r^{2}} drdθ$

Short Answer

Expert verified

The value of integral is

$\int_{0}^{2 π} \int_{0}^{R} r \sqrt{R^{2} - r^{2}} d r d θ = \frac{4}{3} π R^{3}$

Step by step solution

Given information

The expression is

$I = 2 \int_{0}^{2 π} \int_{0}^{R} r \sqrt{R^{2} - r^{2}} d r d θ$

Calculation

Here, $r = 0, r = R$ and $θ = 0, θ = 2 π$

To integrate with respect to $r$ first,

Put

localid="1651390046153" role="math" $R^{2} - r^{2} = t^{2} - 2 r d r = 2 t d t r d r = - t d t r = 0 \Rightarrow t = R and r = R \Rightarrow t = 0$

So integral $I$ become

localid="1650634581886" $I = 2 \int_{0}^{2 π} [\int_{R}^{0} \sqrt{t^{2}} (- t d t)] d θ [\int_{0}^{b} f (x) d x = - \int_{0}^{a} f (x) d x] I = 2 \int_{0}^{2 π} [\int_{0}^{R} t^{2} d t] d θ I = 2 \int_{0}^{2 π} {[\frac{t^{3}}{3}]}_{0}^{R} d θ I = 2 \int_{0}^{2 π} [\frac{R^{3} - 0}{3}] d θ I = 2 \int_{0}^{2 x} \frac{R^{3}}{3} d θ I = \frac{2 R^{3}}{3} \int_{0}^{2 π} d θ I = \frac{2 R^{3}}{3} [θ]_{0}^{2 π} I = \frac{4}{3} π R^{3}$

Thus, the value of integral is

$2 \int_{0}^{2 r} \int_{0}^{R} r \sqrt{R^{2} - r^{2}} d r d θ = \frac{4}{3} π R^{3}$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Each of the integrals or integral expressions in Exercises 39-46 represents the volume of a solid in $R^{3}$ . Use polar coordinates to describe the solid, and evaluate the expressions.
$2 \int_{0}^{2 π} \int_{0}^{R} r \sqrt{R^{2} - r^{2}} drdθ$

Short Answer

Step by step solution

Given information

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Applied Mathematics

Discrete Mathematics

Probability and Statistics

Mechanics Maths

Logic and Functions

Study anywhere. Anytime. Across all devices.

Each of the integrals or integral expressions in Exercises 39-46 represents the volume of a solid in R3. Use polar coordinates to describe the solid, and evaluate the expressions.2∫02π ∫0R rR2−r2drdθ

Short Answer

Step by step solution

Given information

Calculation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Applied Mathematics

Discrete Mathematics

Probability and Statistics

Mechanics Maths

Logic and Functions

Study anywhere. Anytime. Across all devices.

Each of the integrals or integral expressions in Exercises 39-46 represents the volume of a solid in $R^{3}$ . Use polar coordinates to describe the solid, and evaluate the expressions.
$2 \int_{0}^{2 π} \int_{0}^{R} r \sqrt{R^{2} - r^{2}} drdθ$