Evaluate∫∫\intop\intop∫∫ x2+y2 +2
Where Ris bounded by x2+ y2 =1
Let us evaluate ∬R(x2+y2+2)dxdy.\iint\limits_R(x^2+y^2+2)dxdy.R∬(x2+y2+2)dxdy.
∬R(x2+y2+2)dxdy=\iint\limits_R(x^2+y^2+2)dxdy=R∬(x2+y2+2)dxdy=
|x=rcosφ, y=rsinφ, dxdy=rdrdφ,x2+y2=r2x=r\cos \varphi,\ y=r\sin\varphi,\ dxdy=rdrd\varphi, x^2+y^2=r^2x=rcosφ, y=rsinφ, dxdy=rdrdφ,x2+y2=r2 |
=∫02πdφ∫01(r2+2)rdr=2π12∫01(r2+2)d(r2+2)=π(r2+2)22∣01=π(92−2)=5π2.=\int\limits_0^{2\pi}d\varphi\int\limits_0^1(r^2+2)rdr=2\pi\frac{1}{2}\int\limits_0^1(r^2+2)d(r^2+2)= \pi\frac{(r^2+2)^2}{2}|_0^1=\pi(\frac{9}{2}-2)=\frac{5\pi}{2}.=0∫2πdφ0∫1(r2+2)rdr=2π210∫1(r2+2)d(r2+2)=π2(r2+2)2∣01=π(29−2)=25π.
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