Problems: Flux Through a Paraboloid - Edubirdie

University:
Massachusetts Institute of Technology
Course:
18.02SC | Multivariable Calculus
Academic year:

2019
Views:

193

Pages:

1
Author:

Myles Abbott

Problems: Flux Through a Paraboloid Consider the paraboloid z = x2 + y 2 . Let S be the portion of this surface that lies below the plane z = 1. Let F = xi + yj + (1 − 2z)k. Calculate the ﬂux of F across S using the outward normal (the normal pointing away from the z-axis). Answer: First, draw a picture: The surface S is a bowl centered on the z-axis. The outward normal n points away from the outside of the bowl and downward. The region R is the shadow of the bowl – the unit circle in the xy-plane. We know the z component of n is negative, so n dS = (zx , zy , −1) dx dy = (2x, 2y, −1) dx dy. Thus, F · n dS = (2x2 + 2y 2 + 2z − 1) dx dy = (4z − 1) dx dy = (4r2 − 1) dx dy. �� F · n dS = S � R 2π = 0 0 � = 0 = π. (4r2 − 1) dx dy � 1 (4r2 − 1)r dr dθ 2π 1 dθ 2

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