Evaluate the surface integral ∬_s〖f (x,y,z)dS〗for the following equation:?
∬_S〖(4x+y+3z)dS,where S is the portion of the plane 4x+y+3z=12 inside x^2+y^2=1 〗
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∬_S〖(4x+y+3z)dS,where S is the portion of the plane 4x+y+3z=12 inside x^2+y^2=1 〗
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Verified answer
Starting with Cartesian coordinates,
∫∫s (4x + y + 3z) dS
= ∫∫ (4x + y + 3z) * √(1 + (z_x)^2 + (z_y)^2) dA
= ∫∫ (4x + y + 3(4 - 4x/3 - y/3)) * √(1 + (-4/3)^2 + (-1/3)^2) dA, since z = 4 - 4x/3 - y/3
= ∫∫ 12 * (1/3)√26 dA
= 4√26 * (Area inside unit circle x^2 + y^2 = 1)
= 4π√26.
I hope this helps!