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Chapter 6: Q3P (page 323)

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

∬r.ndσOver the whole surface of the cylinder bounded byx2+y2=,z=0,andz=3;rmeansix+jy+kz

Short Answer

Expert verified

The solution of the integrals is found to be as mentioned below.

∬r.ndσ=9π

Step by step solution

01

Given Information.

The given integrals is ∬r.ndσ.

02

Definition of Divergence’s Theorem.

The divergence theorem, often known as Gauss' theorem or Ostrogradsky’s theorem, is a theorem that connects the flow of a vector field across a closed surface to the field's divergence in the volume enclosed. The surface integral of a vector field over a closed surface, also known as the flux through the surface, equals the volume integral of the divergence over the region inside the surface, according to this theorem.

03

Step 3:Apply Gauss’ Theorem.

Apply Gauss' theorem and use the fact that∇.r=3

In cylindrical coordinatesdτ=rdθdrdz, then the solution is mentioned below

∬r.ndσ=∫03∫02π∫013rdrdθdz=3122π3=9π

Hence, the solution of the integrals is found to be as mentioned below.

∬r.ndσ=9π

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Most popular questions from this chapter

Hint:Integrate(g)Derive the following vector integral theorems

(a) ∫volumeτ∇ϕdτ=∮surfaceinclosingτϕndσ

Hint: In the divergence theorem (10.17), substitute V=ϕCwhere is an arbitrary constant vector, to obtain C⋅∫∇ϕdτ=C⋅∮ϕndσSince C is arbitrary, let C=i to show that the x components of the two integrals are equal; similarly, let C=j and C=k to show that the y components are equal and the z components are equal.

(b) ∫volumeτ∇×Vdτ=∮surfaceinclosingτn×Vdσ

Hint: Replace in the divergence theorem by where is an arbitrary constant vector. Follow the last part of the hint in (a).

(c) localid="1659323284980" ∫curveboundingσϕdr=∮surfaceσ(n×∇ϕ)dσ.

(d) ∫curveboundingσϕdr×V=∮surface(n×∇)×Vdσ

Hints for (c) and (d): Use the substitutions suggested in (a) and (b) but in Stokes' theorem (11.9) instead of the divergence theorem.

(e) ∫volumeτ∇ϕdτ=∮surfaceinclosingτϕV·ndσ-∮surfaceinclosingτϕV·∇ϕndτ.

Hint: Integrate (7.6) over volume and use the divergence theorem.

(f) localid="1659324199695" ∫volumeτV·(∇×∪)dτ=∫volumeτV·(∇×∪)dτ+∮surfaceinclosingτ(∪×V)·ndσ

Hint: Integrate (h) in the Table of Vector Identities (page 339) and use the divergence theorem.

(g) ∫surfaceofσϕ(∇×V)⋅ndσ=∫surfaceofσ(∇×∇ϕ)⋅ndσ+∮curveboundingϕV⋅dr

Hint:Integrate(g)in the Table of Vector Identities (page 339) and use Stokes' Theorem.

∬V−ndσover the curved part of the hemisphere in Problem 24, if role="math" localid="1657355269158" V=curl(yi−xj).

Evaluate∬V⋅ndσover the curved surface of the hemispherex2+y2+z2=9,z≥0, ifV=yi+xzj+(2z-1)k.Careful: See Problem 9.

Use Green’s theorem (Section 9) to do Problem 8.2.

Draw a figure similar to figurebut with q outside the surface. A vector (like rin the figure) fromq to the surface now intersects it twice, and for each solid angle dΩthere are two dΩwhere renters and where it leaves the surface. Show that D⋅ndσis given by (10.21) for the dσwhereleaves r the surface and the negative of(10.21)for thedσwhere renters the surface. Hence show that the totaldσ2over the closed surface is zero.
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