Chapter 6: Q23MP (page 338)

$\iint F \times ndσ$ where $F = (y^{2} - x^{2}) i + (2 xy - y) j + 3 zk$ and $σ$ is the entire surface of the tin can bounded by the cylinder
role="math" localid="1657353627256" $x^{2} + y^{2} = 16$
role="math" localid="1657353639412" $z = 3$
role="math" localid="1657353647648" $z = - 3$

Short Answer

Expert verified

The solution is $\iint_{\partial T} F \times ndσ = 192 π$ .

Step by step solution

Given Information.

$F = (y^{2} - x^{2}) i + (2 xy - y) j + 3 zk$

Definition of Divergence 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.

Find the solution.

Use the divergence theorem $∭_{T} \nabla \times VdT = \iint_{\partial T} V \times ndσ$ , where $\partial T$ is the surface area that encloses the volume $T$ .

$\nabla \times F = \frac{\partial F_{x}}{\partial x} + \frac{\partial F_{y}}{\partial y} + \frac{\partial F_{z}}{\partial z}$

$= - 2 x + (2 x - 1) + 3$

$= 2$ $= 2$

Here, $\partial σ$ is the entire surface of the tin, and can be bounded by the cylinder.

$x^{2} + y^{2} = 16$

$z = 3$

$z = - 3$

The tin’s radius and height are $4$ and $6$ respectively.

$\iint_{\partial T} F \times ndσ = ∭_{T} \nabla \times FdT$

$= 2 \iint_{T} d T$

$= (2 π) (4)^{2} (6)$

$= 192 π$

Hence, the solution is $\iint_{\partial T} F \times ndσ = 192 π$ .

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

$\iint F \times ndσ$ where $F = (y^{2} - x^{2}) i + (2 xy - y) j + 3 zk$ and $σ$ is the entire surface of the tin can bounded by the cylinder
role="math" localid="1657353627256" $x^{2} + y^{2} = 16$
role="math" localid="1657353639412" $z = 3$
role="math" localid="1657353647648" $z = - 3$

Short Answer

Step by step solution

Given Information.

Definition of Divergence Theorem.

Find the solution.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Mechanics and Materials

Physical Quantities And Units

Classical Mechanics

Waves Physics

Quantum Physics

Oscillations

Study anywhere. Anytime. Across all devices.

∬F×ndσwhere F=(y2−x2)i+(2xy−y)j+3zkand σis the entire surface of the tin can bounded by the cylinderrole="math" localid="1657353627256" x2+y2=16role="math" localid="1657353639412" z=3role="math" localid="1657353647648" z=-3

Short Answer

Step by step solution

Given Information.

Definition of Divergence Theorem.

Find the solution.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Mechanics and Materials

Physical Quantities And Units

Classical Mechanics

Waves Physics

Quantum Physics

Oscillations

Study anywhere. Anytime. Across all devices.

$\iint F \times ndσ$ where $F = (y^{2} - x^{2}) i + (2 xy - y) j + 3 zk$ and $σ$ is the entire surface of the tin can bounded by the cylinder
role="math" localid="1657353627256" $x^{2} + y^{2} = 16$
role="math" localid="1657353639412" $z = 3$
role="math" localid="1657353647648" $z = - 3$