Question Number 100320 by bemath last updated on 26/Jun/20

Evaluate ∫∫_s  F^→ .n^�  dS where F^→ =4xi^�  −2y^2 j^�  +z^2 k^�    and S is the surface of the cylinder  bounded by x^2 +y^2 =4 ,z = 0 and z=3 .

Answered by smridha last updated on 26/Jun/20

∫∫_s F^→ .n^� ds=∫∫_R F^→ .n^� ((dxdz)/(∣n^� .j^� ∣)) [ take a projection on x−z plane]  our scaler point f^n  𝚽=x^2 +y^2 −4  so the unit normal of this surface  n^� =((▽𝚽)/(∣▽𝚽∣))=((xi^� +yj^� )/2) so ∣n^� .j^� ∣=(y/2)  and F^→ .n^� =((4x^2 −2y^3 )/2).  now our integral looks like  ∫_0 ^3 ∫_0 ^2 [((4x^2 )/y)−2y^2 ]dxdz  =∫_0 ^3 dz[∫_0 ^2 (((4x^2 )/(√(4−x^2 )))+(2x^2 −8))dx]  =3[−4.∫_0 ^2 {(√(4−x^2 ))  −(4/(√(4−x^2 )))}dx+((2/3)x^3 −8x)_0 ^2 ]  =3[−4(((x(√(4−x^2 )))/2)−2sin^(−1) ((x/2)))_0 ^2 +((16)/3)−16]  =3[4𝛑−((32)/3)]=[12𝛑−32]

Commented bybemath last updated on 26/Jun/20

great