Integration by parts (Gauss’ Formula) Recall the Product

Chapter 14, Problem 48AE

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QUESTION:

Integration by parts (Gauss' formula) Recall the Product Rule or Theorem 14.1 l:\(\nabla \cdot(u \mathbf{F})=u \nabla \cdot \mathbf{F}+\mathbf{F} \cdot \nabla u\)..

a. Integrate both sides of this identity over a solid region D with a closed boundary S and use the Divergence Theorem to prove an integration by parts rule:

\(\iiint_{D} u \nabla \cdot \mathbf{F} d V=\iint_{S} u \mathbf{F} \cdot \mathbf{n} d S-\iiint_{D} \mathbf{F} \cdot \nabla u d V\)

b. Explain the correspondence between this rule and the integration by parts rule for single-variable functions.

c. Use integration by parts to evaluate \(\iiint_{D}\left(x^{2} y+y^{2} z+z^{2} x\right) d V\), where D is the cube in the first octant cut by the planes x = 1, y = 1, and z = 1..

Text Transcription:

Nabla cdot (uF) = u nabla cdot F + F cdot nabla u

iiint_D u nabla cdot F dV = iint_S uF cdot n dS - iiint_D F cdot nabla u dV

iiint_D(x^2y+ y^2z + z^2x) dV

Questions & Answers

QUESTION:

Integration by parts (Gauss' formula) Recall the Product Rule or Theorem 14.1 l:\(\nabla \cdot(u \mathbf{F})=u \nabla \cdot \mathbf{F}+\mathbf{F} \cdot \nabla u\)..

a. Integrate both sides of this identity over a solid region D with a closed boundary S and use the Divergence Theorem to prove an integration by parts rule:

\(\iiint_{D} u \nabla \cdot \mathbf{F} d V=\iint_{S} u \mathbf{F} \cdot \mathbf{n} d S-\iiint_{D} \mathbf{F} \cdot \nabla u d V\)

b. Explain the correspondence between this rule and the integration by parts rule for single-variable functions.

c. Use integration by parts to evaluate \(\iiint_{D}\left(x^{2} y+y^{2} z+z^{2} x\right) d V\), where D is the cube in the first octant cut by the planes x = 1, y = 1, and z = 1..

Text Transcription:

Nabla cdot (uF) = u nabla cdot F + F cdot nabla u

iiint_D u nabla cdot F dV = iint_S uF cdot n dS - iiint_D F cdot nabla u dV

iiint_D(x^2y+ y^2z + z^2x) dV

ANSWER:

Solution 48AE1. F= F- F2. ()() - ()()Which is the usual ru

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