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∬(2xi−2yj+5k)ײԻåσover the surface of a sphere of radius 2and center at the origin.

Short Answer

Expert verified

The solution derived is I=0.

Step by step solution

01

Given Information.

The given expression is∬(2xi−2yj+5k)ײԻåσ.

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 its magnitude.

03

Apply divergence theorem.

Apply the divergence theorem and the fact that ∇×V=2−2=0then the integral becomes as shown below.

I=0×∭V »åÏ„

=0

Hence, the solution derived is I=0.

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Most popular questions from this chapter

∬FײԻåσwhere F=(y2−x2)i+(2xy−y)j+3zkand σis the entire surface of the tin can bounded by the cylinder

role="math" localid="1657353627256" x2+y2=16

role="math" localid="1657353639412" z=3

role="math" localid="1657353647648" z=-3

(a) Given Φ=x2-y2, sketch on one graph the curvesΦ=4,Φ=1,Φ=0,Φ=-1,Φ=-4. IfΦis the electrostatic potential, the curvesΦ=const. are equipotential, and the electric field is given byE=-∇Φ. IfΦis temperature, the curvesΦ= const. are isothermals and∇Φis the temperature gradient; heat flows in the direction-∇Φ.

(b) Find and draw on your sketch the vectors-∇Φat the points(x,y)=(+1,+1),(0,+2),(+2,0),. Then, remembering that∇Φis perpendicular toΦ= const., sketch, without computation, several curves along which heat would flow [see (a)].

The force on a charge moving with velocity v=dr/dtin a magnetic field B isF=q(v×B)we can write B asB=∇×Awhere A (called the vector potential) is a vector function of x,y,z,t . If the position vectorr=ix+jy+kzof the charge is a function of time, show that

dAdt=∂A∂t+v.∇A

Thus show that

F=qv×(∇×A)=q[∇(v.A)-dAdt+∂A∂t]

∬VײԻåσover the entire surface of the volume in the first octant bounded byx2+y2+z2=16and the coordinate planes, whereV=x+x2−y2i+(2xyz−2xy)j−xz2k

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

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