Math 214 Advanced Calculus II Midterm Exam II SOLUTIONS

April 21, 2001 Ali Sinan Sertöz

1) Suppose V is open in ℝ^{n} and f : V → ℝ is continuously differentiable with Δ_{f} ≠ 0 on V. Prove that lim_{r→0} \frac{Vol(f(B_{r}(x_{0})))}{Vol(B_{r}(x_{0}))} = |Δ_{f}(x_{0})| for every x_{0} ∈ V.

Solution:

Fix an arbitrary x_{0} ∈ V. Note that \overline{B_{r}(x_{0})} is compact, Δ_{f} attains its inf and sup there, and

Vol(f(B_{r}(x_{0}))) = ∫_{f(B_{r}(x_{0}))} du  (1)
= ∫_{B_{r}(x_{0})} |Δ_{f}(x)| dx  (2)
= c Vol(B_{r}(x_{0}))  (3)
= |Δ_{f}(x_{1})| Vol(B_{r}(x_{0}))  (4)

where

(1) follows from Cor. 7.5 (p349),

(2) follows from Thm. 7.14 (p375),

(3) holds for some c between the inf and sup of Δ_{f} on B_{r}(x_{0}), Thm. 7.9 (p354),

(4) holds for some x_{1} ∈ \overline{B_{r}(x_{0})} since |Δ_{f}(x)| is continuous there. Finally taking the limit as r → 0 gives the required result. Note that this is Exercise 7 on page 382. (The problem would be more meaningful if the range of f is ℝ^{n})...!

2) Suppose V is open in ℝ^{n} and f : V → ℝ is continuous. Prove that if ∫_{E} f(x)dx = 0 for all Jordan regions E ⊂ V, then f = 0 on V.

Solution: Assume that there is an x_{0} ∈ V where f(x_{0}) = 2ϵ > 0. Let U = {x ∈ V | f(x) > ϵ > 0 }. We know that U is not empty since x_{0} ∈ U. Since f is continuous, U is open. (U is the preimage of the open interval (ϵ, ∞)). By continuity of f there is an open ball B_{2r} of radius 2r around x_{0}, where f is strictly positive. The integral of f on B_{r} is zero by the assumption of the theorem. (Note that B_{r} is a Jordan region). By the mean value theorem this is equal to c V ol(B_{r}) for some c between the inf and sup of f on B_{r}. Since f is continuous on B_{2r}, there is a point x_{1} ∈ \overline{B_{r}} where f(x_{1}) = c, so c ≠ 0. But this forces the volume of B_{r} to be zero, a contradiction. So there is no point where f is strictly positive. Similarly f cannot take a strictly negative value. So f is identically zero on V. Note that this is Exercise 9 on page 357.

3) Let S be the surface defined by z = 7\sqrt{x^{2} + y^{2}}, 7 ≤ z ≤ 28. Evaluate the integral ∫∫_{S} F · n⃗ dσ where F (x, y, z) = (x^{2}, y(1 − 2x), −z) and n⃗ is the unit normal vector pointing outward.

Solution: Evaluating this integral directly requires a lot of tedious work, so we try using Stokes’ theorem which reduces this integral to an integral of the boundary. Since the boundary consists of two circles, we expect that the resulting integral will be easy. But to apply Stokes’ theorem we must express F as curlG. Since divF = 0, such a G exists, see Theorem 8.6 on page 450. Set G = (P, Q, R). Trial and error, with some luck, will give you a G with curlG = F. For example I set Q = 0 and tried to solve the remaining equalities: R_{y} = x^{2}, P_{z} − R_{x} = y − 2xy, −P_{y} = −z. This easily yields a solution, for example G = (yz, 0, yx^{2}). Now Stokes’ theorem says that the required integral is equal to ∫_{∂S} G · T ds. We now parametrize the boundary ∂S = C_{1} − C_{2}. C_{1}: ϕ(t) = (cos t, sin t, 7), t ∈ [0, 2π], C_{2}: ψ(t) = (4 cos t, 4 sin t, 28), t ∈ [0, 2π]. G(ϕ(t)) · ϕ′(t)dt = −7 sin^{2} t dt. G(ψ(t)) · ψ′(t)dt = −448 sin^{2} t dt. So we integrate 441 sin^{2} tdt from 0 to 2π to obtain 441π as the answer.

4) Find the surface area of the cap x^{2} + y^{2} + z^{2} = R^{2}, z ≥ \sqrt{R^{2} − A^{2}}, where R ≥ A ≥ 0.

Solution: We first parametrize the surface: ϕ(r, θ) = (r cos θ, r sin θ, \sqrt{R^{2} − r^{2}}), (r, θ) ∈ [0, A] × [0, 2π] = E. To find the surface area we need to evaluate ∫_{E} ∥ N_{ϕ_{r}×ϕ_{θ}} ∥ d(r, θ), see Definition 8.12 on page 424. A straightforward calculation gives ∥ N_{ϕ_{r}×ϕ_{θ}} ∥= r (R^{2}/(R^{2} − r^{2}))^{1/2}. Integrating this over E gives 2πR^{2} − 2πR\sqrt{R^{2} − A^{2}}.

5) (i) Construct a C^{∞} vector field F : ℝ^{3}\{(0, 0, 0)} → ℝ^{3} such that F is not defined at the origin and divF = 0. (ii) Let E_{1} be the surface given by x^{2} + y^{2} + z^{2} = 1 and E_{2} the surface given by x^{2}/25 + y^{2}/36 + z^{2}/49 = 1. Show that ∫∫_{E_{1}} F · n⃗_{1} dσ = ∫∫_{E_{2}} F · n⃗_{2} dσ where F is the vector field you found in the first part, and n⃗_{i} is the unit normal pointing outward of the surface E_{i}, i = 1, 2.

Solution: The easiest way to construct a vector field F with zero divergence is to start with a vector field G and set F to be curlG. Since we want F to be defined everywhere except at the origin we can start with an arbitrary C^{∞} vector field G which is defined everywhere except at the origin. One such easy example is G = (log(x^{2} + y^{2} + z^{2}), log(x^{2} + y^{2} + z^{2}), log(x^{2} + y^{2} + z^{2})). To prove the second part let the region between these surfaces be denoted by E. Then use Gauss’ theorem, Theorem 8.4 on page 441, and observe that the right hand side is zero since divF = 0. This gives, after observing correct orientations of the surfaces E_{1} and E_{2}, that the required equality holds. Note that this is Exercise 10.c on page 454. I solved parts (a) and (b) in class and promised to ask part (c) in the exam!