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Show that ∇.(U×r)=r.(∇×U)where U is a vector function of x,y,zandr=xi+yj+zk .

Short Answer

Expert verified

It has been proved that ∇.(U×r)=r.(∇×U)

Step by step solution

01

Given Information.

It has been given that U is a vector function of x,y and z and r=xi+yj+zk

02

Definition of vector.

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines it magnitude.

03

Use the formula to evaluate the expression.

Use the identity mentioned below.

∇.U×V=V.∇×U-U.∇×V

Now letV→r,

∇.U×r=r.∇×U-U.∇×r

But r is a radial vector field, so it doesn't curl at all, hence its curl should vanish.

∇.U×r=r.∇×U

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Most popular questions from this chapter

Evaluate each of the integrals in Problems 3to 8as either a volume integral or a surface integral, whichever is easier.

∭(∇×F)»åÏ„over the region x2+y2+z2≤25, where localid="1657282505088" F=(x2+y2+z2)(xi+yj+zk)

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(c) Evaluate around the boundary of the rectangle and thus verify Stokes' theorem for this case.

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(Physically, is the rate of change of density with time as we follow the fluid along a streamline; ∂p/∂tis the corresponding rate at a fixed point.) For a steady state (that is, time-independent), ∂p/∂t=0, but ∂p/∂t is not necessarily zero. For an incompressible fluid, ∂p/∂t=0. Show that then role="math" localid="1657336080397" ∇×v=0. (Note that incompressible does not necessarily mean constant density since ∂p/∂t=0does not imply either time or space independence of ÒÏ; consider, for example, a flow of watermixed with blobs of oil.)

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