Chapter 14: Problem 2
Explain the meaning of the integral \(\iint_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d S\) in Stokes' Theorem.
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Chapter 14: Problem 2
Explain the meaning of the integral \(\iint_{S}(\nabla \times \mathbf{F}) \cdot \mathbf{n} d S\) in Stokes' Theorem.
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Let \(\mathbf{F}=\langle z, 0,0\rangle\) and let \(\mathbf{n}\) be a unit vector aligned with the axis of a paddle wheel located on the \(x\) -axis (see figure). a. If the paddle wheel is oriented with \(\mathbf{n}=\langle 1,0,0\rangle,\) in what direction (if any) does the wheel spin? b. If the paddle wheel is oriented with \(\mathbf{n}=\langle 0,1,0\rangle,\) in what direction (if any) does the wheel spin? c. If the paddle wheel is oriented with \(\mathbf{n}=\langle 0,0,1\rangle,\) in what direction (if any) does the wheel spin?
a. Show that a torus with radii \(R>r\) (see figure) may be described parametrically by \(r(u, v)=\langle(R+r \cos u) \cos v,(R+r \cos u) \sin v, r \sin u\rangle\) for \(0 \leq u \leq 2 \pi, 0 \leq v \leq 2 \pi\) b. Show that the surface area of the torus is \(4 \pi^{2} R r\).
Find the general formula for the surface area of a cone with height \(h\) and base radius \(a\) (excluding the base).
Use the procedure in Exercise 57 to construct potential functions for the following fields. $$\mathbf{F}=\langle x, y\rangle$$
Recall the Product Rule of Theorem \(14.11: \nabla \cdot(u \mathbf{F})=\nabla u \cdot \mathbf{F}+u(\nabla \cdot \mathbf{F})\) 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: $$\iiint_{D} u(\nabla \cdot \mathbf{F}) d V=\iint_{S} u \mathbf{F} \cdot \mathbf{n} d S-\iiint_{D} \nabla u \cdot \mathbf{F} d V$$ 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 \(z=1\)
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