Answers to additional exercises for Colley, Chapter 7Section 7.1S1. x − y − 2z = 0S2. x(u, v) = (sin u cos v, sin u sin v, u), (0 ≤ u ≤ π, 0 ≤ v ≤ 2π)S3. (2 −√2)πS4. The tangent plane is z = x + y (which is also an equation satsified by the surface,although the latter only consists of points whose coordinates are nonnegative).S5. The parametrization is regular for all (u, v).S6. The parametrization is regular for all (u, v) 6= (1, −1).Section 7.2S1.√14S2.(5√5 − 1)π6S3.2π3·17√17 − 1S4.16kπ√53S5. −43S6.243π2S7. 32πSection 7.3S1. 1S2. −9S3.a24· (2a2+ a − 4)S4. 0S5. −16πS6. We shall use Exercise 24 on page 222 of the text (formula for the curl of a crossproduct). This identity yields∇ × (f ∇g) = f · ∇ × (∇g) + (∇f) ×(∇g)1and since ∇ × (∇g) = 0 by Theorem 3.4.3 on page 218, it follows that the first term on the righthand side drops out. This leaves us with just the second term on the right hand side, and hencewe have the identity we wanted to verify.S7. πS8.2π5S9. For the region under consideration, ∇ × F = 0 implies F = ∇f for some f. But then∇2f = ∇·(∇f) = ∇ · Fwhich is assumed to be zero. Therefore ∇2f = 0 as required.S10. The flux integral for F is equal to the scalar surface integral of the dot productF ·N, where N is the normal to the surface. If F is always tangent to the surface, then it is alwaysperpendicular to N and hence F ·N = 0, which means that the surface integral must also be equalto zero.Section 7.4S1. Since λ is an integrating factor for F, we have λ F = ∇g for some function g. Sincethe curl of a gradient is zero, this means that0 = ∇ × (∇g) = ∇ × (λF) = (∇λ) × F + λ (∇ × F) .Since λ is never zero we may rewrite this as∇ × F =−1λ(∇λ) × F .The right hand side of this equation is orthogonal to F, so the left hand side is also orthogonal toF.S2. Direct calculation implies that ∇ × F = (−1, −1, −1), and therefore (∇ × F) · F =−y − z −x. If all three of x, y, z are positive, then this dot product is negative. By the precedingexercise, it follows that one cannot find an integrating factor for F on the specified region. In fact,one can show that there is no integrating factor for F on the any nonempty region of 3-dimensionalcoordinate space.S3. Here is the formula from Exercise 15(a):∇ × (∇ × F) = ∇(∇ · F) − ∇2FIf ∇ × F = 0 and ∇ · F = 0, then the displayed formula reduces to0 = ∇ × 0 = ∇0 − ∇2F = 0 − ∇2Fwhich in turn means that ∇2F =