Note that the equation of the plane is x/a + z/b = 1.
==> z = b - bx/a.
So, a normal vector to the plane with positive k component is
<-z_x, -z_y, 1> = <b/a, 0, 1>.
Hence the flux ∫∫s F · dS equals
∫∫ <1, -1, -5> · <b/a, 0, 1> dA
= (b/a - 5) * (ab, the area of the rectangular region of integration)
= b^2 - 5ab.
I hope this helps!
Copyright © 2024 1QUIZZ.COM - All rights reserved.
Answers & Comments
Verified answer
Note that the equation of the plane is x/a + z/b = 1.
==> z = b - bx/a.
So, a normal vector to the plane with positive k component is
<-z_x, -z_y, 1> = <b/a, 0, 1>.
Hence the flux ∫∫s F · dS equals
∫∫ <1, -1, -5> · <b/a, 0, 1> dA
= (b/a - 5) * (ab, the area of the rectangular region of integration)
= b^2 - 5ab.
I hope this helps!