Chapter 6: Q11P (page 335)

$\iint V - ndσ$ over the entire surface of a cube in the first octant with edges of length $2$ along the coordinate axes, where $V = (x^{2} - y^{2}) i + 3 yj - 2 xzk$ .

Short Answer

Expert verified

The solution derived is $I = 24 .$

Step by step solution

Given Information.

The given equation is $V = (x^{2} - y^{2}) i + 3 yj - 2 xzk$

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.

Apply divergence theorem.

Apply the divergence theorem and the fact that $\nabla \times V = 2 x + 3 - 2 x = 3$ then the integral becomes as mentioned below.

$I = 3 ∭_{V} dτ$

$= 3 \times 8$

$= 24$

Hence, the solution derived is $I = 24 .$ .

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$\iint V - ndσ$ over the entire surface of a cube in the first octant with edges of length $2$ along the coordinate axes, where $V = (x^{2} - y^{2}) i + 3 yj - 2 xzk$ .

Short Answer

Step by step solution

Given Information.

Definition of vector.

Apply divergence theorem.

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∬V−ndσover the entire surface of a cube in the first octant with edges of length 2along the coordinate axes, whereV=(x2−y2)i+3yj−2xzk.

Short Answer

Step by step solution

Given Information.

Definition of vector.

Apply divergence theorem.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Radiation

Physics of Motion

Fundamentals of Physics

Linear Momentum

Electromagnetism

Work Energy and Power

Study anywhere. Anytime. Across all devices.

$\iint V - ndσ$ over the entire surface of a cube in the first octant with edges of length $2$ along the coordinate axes, where $V = (x^{2} - y^{2}) i + 3 yj - 2 xzk$ .