Evaluate the surface integral ∫∫S(x2z+y2z)ds where S is the hemisphere x2+y2+z2=9,z≥0

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Evaluate the surface integral  ∫∫S(x2z+y2z)ds  where S is the hemisphere x2+y2+z2=9,z≥0

irwchh · Accepted Answer

Answer:  The surface integral ∫∫S(x2z+y2z)ds (for the surface represented by the hemisphere x2+y2+z2=9,z≥0) is  ∫∫S(x2z+y2z)ds=18π  Explanation:  The surface S has parametric equations  x=rsin⁡(θ)cos⁡(ϕ)  y=rsin⁡(θ)sin⁡(ϕ)  z=rcos⁡(θ)  where 0≤θ≤π2 and 0≤ϕ≤2π and S  is the hemisphere x2+y2+z2=9 in the form x2+y2+z2=r2  where r is the radius of the sphere, thus  r=3  And the parametric equations becomes  x=3sin⁡(θ)cos⁡(ϕ)  y=3sin⁡(θ)sin⁡(ϕ)  z=3cos⁡(θ)  Thus, we get  x2z+y2z=(3sin⁡θcos⁡ϕ)2(3cos⁡θ)+(3sin⁡θsin⁡ϕ)2(3cos⁡θ)  =(9sin2θcos2ϕ)(3cos⁡θ)+(9sin2θsin2ϕ)(3cos⁡θ)  =27sin2θcos⁡θ(cos2ϕ+sin2ϕ)  =27sin2θcos⁡θ(1)=27sin2θcos⁡θ  Thus, the surface integral ∫∫S(x2z+y2z)ds becomes  ∫∫S(x2z+y2z)ds=∫θ=0π2[∫ϕ=02π(27sin2θcos⁡θ)dϕ]dθ

Evaluate the surface integral \int\int_S(x^2z+y^2z)ds where S is t

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