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Q. 40

Found in: Page 1120


Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

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Short Answer

F(x,y,z)=i+j+k, where S is the lower half of the unit sphere, with n pointing outwards.

The required flux of the vector field through the surface S is 2π.

See the step by step solution

Step by Step Solution

Step 1. Given information.

Consider the given question,


Step 2. Find the Flux of Fx,y,zthough an oriented surface S.

If a surface S is the graph of z=zx,y, then the Flux of Fx,y,z through S is given below,

SF(x,y,z)ndS=D(F(x,y,z)n)zx×zydA=D(F(x,y,z)n)zx2+zy2+1dA ... (i)

Here, the surface S is the lower half of the unit sphere, so its equation will be,


Now first find zx. The first partial derivates of z are given below,


Step 3. Find the value of ∂z∂x2+∂z∂y2+1.

On finding the value of zy,

role="math" localid="1650342662824" zy=y-1-x2-y2=y1-x2-y2



Step 4. Find the choice of n.

The choice of n should have pointing upwards. Then the following vector, v=zx,zy,1 is normal to the surface.

Now, for z=1x2y2, gives the following vector perpendicular to this surface,


Then the desired normal vector will be,


Step 5. Find the value of F(x,y,z)⋅n.

The value of F(x,y,z)n will be,


Step 6. Find the polar coordinates.

Substituting these values in equation (i),

SF(x,y,z)ndS=D(F(x,y,z)n)zx2+zy2+1dA=D1x2y21x+y1x2y211x2y2dA=D1x+y1x2y2dA .(ii)

The surface S is the lower half of the unit sphere, so the region of integration D is the unit disk in the xy-plane and centered at the origin.

Hence, in polar coordinates, the region of integration will be,


In this case,

localid="1650343222895" x=rcos θ,y=rsin θ

Step 7. Using equation (ii), write the flux of the vector field through the surface S.

Using equation (ii), the flux of the vector field through the surface S,

SF(x,y,z)ndS=D1x+y1x2y2dA=02π011rcos θ+rsin θ1r2rdrdθ=02π011dr(cos θ+sin θ)01r21r2drdθ=02π[10](cos θ+sin θ)121112+sin1 1120102+sin1 0dθ=02π1π4(cos θ+sin θ)dθ=θπ4(sin θcos θ)02π=2ππ4(sin 2πcos 2π)0π4(sin 0cos 0)=2π

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