Flux Over Parabolic Strip

Calculus Level 3

Consider the following surface:

− 1 ≤ x ≤ 1 y = x 2 0 ≤ z ≤ 1 -1 \leq x \leq 1 \\ y = x^2 \\ 0 \leq z \leq 1

Find the flux of the vector field F ⃗ = ( F x , F y , F z ) = ( 0 , y , 0 ) \vec{F} = (F_x,F_y,F_z) = (0,y,0) over the surface. Assume that the surface normal vector generally has a negative y y component.


The answer is -0.667.

Steven Chase by Steven Chase , 37, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Otto Bretscher
Dec 18, 2018

The standard normal vector is N ⃗ = ( 2 x , − 1 , 0 ) \vec{N}=(2x,-1,0) so

∫ ∫ S F ⃗ ⋅ d S ⃗ = ∫ ∫ D F ⃗ ⋅ N ⃗ d A = − ∫ − 1 1 ∫ 0 1 x 2 d z d x = − 2 3 ≈ − 0.667 \int\int_S \vec{F}\cdot d\vec{S}=\int\int_D \vec{F}\cdot \vec{N} \ dA=-\int_{-1}^1\int_0^1 x^2\ dz\ dx=-\frac{2}{3}\approx\boxed{-0.667}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...