Chapter 6: Q6P (page 323)

Evaluate each of the integrals in Problems $3$ to $8$ as either a volume integral or a surface integral, whichever is easier.
$∭ \nabla \times Vdτ$ over the unit cube in the first octant, where $V = (x^{3} - x^{2}) yi + (y^{3} - 2 y^{2} + y) xj + (z^{2} - 1) k$

Short Answer

Expert verified

The solution of the integrals is $∭ \nabla \times Vdτ = 1$ .

Step by step solution

Given Information.

The given integral is $∭ \nabla \times V d τ$ .

Definition of Divergence’s Theorem.

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. According to this theorem, 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.

Solve the equation.

Solve the equation as shown below.

$\nabla \times V = (3 x^{2} - 2 x) y + (3 y^{2} - 4 y + 1) x + 2 z$

$= 3 x^{2} y + 3 x y^{2} - 6 x y + x + 2 z$

Then, the integrals become as shown below.

$\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} (3 x^{2} y + 3 x y^{2} - 6 x y + x + 2 z) d x d y d z$

$\int_{0}^{1} \int_{0}^{1} (y + \frac{3 y^{2}}{2} - 3 y + \frac{1}{2} + 2 z) d y d z$

$\int_{0}^{1} (\frac{1}{2} + \frac{1}{2} - \frac{3}{2} + \frac{1}{2} + 2 z) d z = 1$

$∭ \nabla \times V d τ = 1$

Hence, the solution of the integrals is $∭ \nabla \times V d τ = 1$ .

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Evaluate each of the integrals in Problems $3$ to $8$ as either a volume integral or a surface integral, whichever is easier.
$∭ \nabla \times Vdτ$ over the unit cube in the first octant, where $V = (x^{3} - x^{2}) yi + (y^{3} - 2 y^{2} + y) xj + (z^{2} - 1) k$

Short Answer

Step by step solution

Given Information.

Definition of Divergence’s Theorem.

Solve the equation.

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Most popular questions from this chapter

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Evaluate each of the integrals in Problems 3to 8as either a volume integral or a surface integral, whichever is easier.∭∇×Vdτ over the unit cube in the first octant, where V=(x3−x2)yi+(y3−2y2+y)xj+(z2−1)k

Short Answer

Step by step solution

Given Information.

Definition of Divergence’s Theorem.

Solve the equation.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Linear Momentum

Modern Physics

Engineering Physics

Magnetism and Electromagnetic Induction

Electrostatics

Study anywhere. Anytime. Across all devices.

Evaluate each of the integrals in Problems $3$ to $8$ as either a volume integral or a surface integral, whichever is easier.
$∭ \nabla \times Vdτ$ over the unit cube in the first octant, where $V = (x^{3} - x^{2}) yi + (y^{3} - 2 y^{2} + y) xj + (z^{2} - 1) k$