Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Each of the integral expressions that follow represents the area of a region in the plane bounded by a function expressed in polar coordinates. Use the ideas from this section and from Chapter 9 to sketch the regions, and then evaluate each integral

02π312+cosθ2dθ-π4π312+cosθ2dθ

Short Answer

Expert verified

The value of integral isπ4

Step by step solution

01

.Given information

Integral;

02π312+cosθ2dθ-π4π312+cosθ2dθ

02

Step 2. Plot the region:

In the given integral we can see that there is a subtraction occurs between integral.

By comparing the given integral with the formula of area of region of the polar curve r=f(θ):

12abr2dθ

We get r=212+cosθ

Now plot this curve in both given interval of θ

03

Step 3. Simplify integral

02π312+cosθ2dθ-π4π312+cosθ2dθ=02π314+cos2θ+cosθdθ-π4π314+cos2θ+cosθdθ=02π314+1+cos2θ2+cosθdθ-π4π314+1+cos2θ2+cosθdθ=02π314+12+cos2θ2+cosθdθ-π4π314+12+cos2θ2+cosθdθ=02π334+cos2θ2+cosθdθ-π4π334+cos2θ2+cosθdθ

04

Step 4. Solve integral

Now integrall can be solved as

34θ+12sin2θ2+sinθ02π3-34θ+12sin2θ2+sinθπ4π3=3×2π34-14sin4π3+sin2π3-0-3×4π34-14sin8π3+sin4π3-34π+14sin2π+sinπ=π2-38+32-π-38+32-34π-0-0=π2-38+32-π+38-32+34π=3π-2π4=π4

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.

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

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free