Chapter 13: Q.27 (page 991) URL copied to clipboard! Now share some education! Each of the integrals or integral expressions in Exercises 27 represents the area of a region in the plane. Use polar coordinates to sketch the region and evaluate the expressions.2∫-π/4π/2∫0(2/2)+sinθrdrdθ-2∫-π/2-π/4∫0(2/2)+sinθrdrdθ Short Answer Expert verified The value of integral is2∫-π/4π/2∫0(2/2)·sinθrdrdθ-2∫-π/2-π/4∫0(2/2)sinθrdrdθ=32 Step by step solution 01 Given information The expression is2∫-π/4π/2∫0(2/2)+sinθrdrdθ-2∫-π/2-π/4∫0(2/2)+sinθrdrdθ 02 Simplification Here, r=0,r=(2/2)+sinθand θ=-π/4,θ=-π/2,θ=π/2θr=(1/2)+cosθ01.5π/41.2071π/20.53π/4-0.2071π-0.55π/4-0.20714π/30To sketch the region use the above tablePlot of r=(2/2)+sinθI=2∫-π/4π/2∫0(2/2)+sinθrdrdθ-2∫-π/2-π/4∫0(2/2)+sinθrdrdθI=I1-I2Where,I1=2∫-π/4π/2∫0(2/2)+sisθrdrdθI1=2∫-π/4π/2r220(2/2)·sinθdθI1=2∫-π/4π/2{(2/2)+sinθ}2-02θI1=2∫-π/4π/212+2sinθ+sin2θ2dθI1=2∫-π/4π/212+2sinθ+12(1-cos2θ)2dθIntegrate with respect to θ.localid="1651553993945" I1=2θ2-2cosθ+12θ-12sin2θ-π/4π/221-θ-2cosθ-14sin2θ-π/4π/221=2π2-2cosπ2-14sin(π)--π4-2cos-π4-14sin-π22I1=2π2--π4-1+142I1=2[π+1)I1=34[π2now,I2=2∫-π/2-π/4∫0(2/2)+sinθrdrdθI2=2∫-π/2-π/4r220(2/2)+sinθdθI2=2∫-π/2-π/412+2sinθ+12(1-cos2θ)2dθI2=2θ-2cosθ-14sin2θ2-π/2-π/4I2=2-π4-2cos-π4-14sin-π2--π2-2cos-π2-14sin(-π)2I2=2-π4-1+14--π22I2=34(π-1)Therefore,I=34(π+1)-34(π-1)I=32Thus, the value of integral is2∫-π/4π/2∫0(2/2)·sinθrdrdθ-2∫-π/2-π/4∫0(2/2)·sinθrdrdθ=32 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!