Question

The value of ∬R sin⁡(x^2 + y^2) dx dy where R is the region bounded by circle centered at origin with radius r=2 is _____

(a) πcos 4

(b) π(1-cos 4)

(c) π

(d) π(1-sin 4)

I got this question in final exam.

I'd like to ask this question from Change of Variables In a Double Integral in chapter Multiple Integrals of Engineering Mathematics

Answer 1

Right option is (b) π(1-cos 4)

To explain: Using Polar variable transformation x = r cos θ & y=r sin θ, r varies from 0 to 2 & θ varies from 0 to 2π because radius of circle i.e r=2 & centered at origin

\(\int\int_P f(r cos θ,r sin θ )rdr \,dθ = \int\int_R sin⁡ (x^2+y^2) \,dx \,dy = \int_0^2π \int_0^2 r sin⁡ r^2 \,dr \,dθ\)

Using substitution t=r^2 integral changes to \(\int_0^{2π} \int_0^4 0.5 \,sin⁡t \,dt \,dθ\)

\(\int_0^{2π} 0.5\big[-cos⁡t\big]_0^4  dθ = \int_0^{2π} 0.5(1-cos⁡4) dθ = π(1-cos 4)\).