Math. 311 (Fulling) 19 April 2002

Test C – Solutions

Name:

Calculators may be used for simple arithmetic operations only!

1. (24 pts.) Let F⃗ (r⃗) = (ex + y)î+ y3ĵ+ (z2 − x)k̂.

(a) Calculate ∇ · F⃗ .

e^{x} + 3y^{2} + 2z.

(b) Calculate ∇ × F⃗ .

| \begin{matrix}î & ĵ & k̂ \\ ∂_{x} & ∂_{y} & ∂_{z} \\ e^{x} + y & y^{3} & z^{2} − x \end{matrix} | = 0 î + (−1)(−ĵ) + (−1)k̂ = ĵ − k̂.

(c) By the method of your choice, calculate ∮ F⃗ · dr⃗ around the rectangular path from the origin, to (1, 0, 1) , to (1, 1, 1) , to ( 0, 1, 0) , and back to (0, 0, 0) . (This path lies in the plane where z = x with y arbitrary.)

Method 1: Apply the Stokes theorem to the result of (b).

∫ ∫ 0 dy dz + ∫ ∫ 1 dz dx + ∫ ∫ (−1)dx dy = 0 + 0 + ∫_{0}^{1} dx ∫_{0}^{1} dy(−1) = −1.

Variant of Method 1: The upward unit normal to the plane is n̂ = \frac{1}{\sqrt{2}} (−î+k̂) . Thus n̂·∇×F⃗ = − \frac{1}{\sqrt{2}} . Multiply by the area of the rectangle, \sqrt{2} , to get −1 . Method 2: Note that y is constant on the first and third sides of the rectangle, while only y varies on the other two sides. So the line integral can be written out as (∫_{0}^{1}(e^{x} + 0) dx + ∫_{0}^{1}(z^{2} − z) dz) + ∫_{0}^{1} y^{3} dy + (∫_{1}^{0}(e^{x} + 1) dx + ∫_{1}^{0}(z^{2} − z) dz) + ∫_{1}^{0} y^{3} dy. Almost everything cancels! We are left with

− ∫_{0}^{1} dx = −1.

311C-S02 Page 2

2. (18 pts.) Find an orthonormal basis for the subspace of R4 spanned by

{v⃗1 = (2, 0, 0, 0), v⃗2 = (1, 1, 0, 1), v⃗3 = (−2, 1, 1, 0)}.

û_{1} = \frac{v⃗_{1}}{∥v⃗_{1}∥} = (1, 0, 0, 0).

v_{2∥} = (û_{1} · v⃗_{2})û_{1} = û_{1} .

v_{2⊥} = v⃗_{2} − û_{1} = (0, 1, 0, 1).

û_{2} = \frac{v⃗_{2⊥}}{∥v⃗_{2⊥∥}} = \frac{1}{\sqrt{2}} (0, 1, 0, 1).

v_{3∥} = (û_{1} · v⃗_{3})û_{1} + (û_{2} · v⃗_{3})û_{2} = −2û_{1} + \frac{1}{2} (1)(0, 1, 0, 1) = (−2, \frac{1}{2} , 0, \frac{1}{2} ).

v_{3⊥} = (−2, 1, 1, 0) − (−2, \frac{1}{2} , 0, \frac{1}{2} ) = (0, \frac{1}{2} , 1, − \frac{1}{2} ).

∥v_{3⊥}∥^{2} = \frac{1}{4} + 1 + \frac{1}{4} = \frac{3}{2} .

û_{3} = \sqrt{\frac{2}{3}} v_{3⊥} = \frac{1}{\sqrt{6}} (0, 1, 2, −1).

3. (15 pts.) Let A⃗(r⃗) = x3î + y3ĵ+ z3k̂, and let S be the hemispherical surface r = 2 , 0 ≤ θ < π, 0 ≤ ϕ < π (i.e., the part of the sphere of radius 2 that lies in the y > 0 half-space). Calculate ∫∫S A⃗ · dS⃗ by the method of your choice.

Note first that ∇ · A⃗ = 3x^{2} + 3y^{2} + 3z^{2} = 3r^{2} . Since the divergence is a rather simple function, it would be nice to be able to use Gauss’s theorem. We could close off the hemisphere by the y = 0 plane. The flux through that plane is 0 (since n̂ = ĵ and y^{3} = 0 ), so the integral we want is just the integral of 3r^{2} over the half ball (solid hemisphere). [Alternative argument: By symmetry, the flux through the other hemispherical surface is equal to the the flux we want, so we can integrate 3r^{2} over the entire ball and divide by 2 .]

∫_{0}^{2} r^{2} dr ∫_{0}^{π} sin θ dθ ∫_{0}^{π} dϕ 3r^{2} = 2π ∫_{0}^{2} 3r^{4} dr = \frac{6π}{5} r^{5}|_{0}^{2} = \frac{192π}{5} .

Brief look at direct methods: One must resist the temptation to write A⃗ = r^{3}r̂ — that is false! In polar coordinates,

n̂ = r̂ = sin θ cos ϕ î + sin θ sin ϕ ĵ + cos θ ẑ.

So

A⃗ · n̂ = r^{3} sin^{4} θ cos^{4} ϕ + r^{3} sin^{4} θ sin^{4} ϕ + r^{3} cos^{4} θ, dS = r^{2} sin θ dθ dϕ, r = 2.

Thus the integral equals I_{x} + I_{y} + I_{z} , where

I_{x} = 2^{5} ∫_{0}^{π} dϕ ∫_{0}^{π} dθ sin^{5} θ cos^{4} ϕ, I_{y} = 2^{5} ∫_{0}^{π} dϕ ∫_{0}^{π} dθ sin^{5} θ sin^{4} ϕ,

I_{z} = 2^{5} ∫_{0}^{π} dϕ ∫_{0}^{π} dθ cos^{4} θ sin θ.

(The parametric method of pp. 385–386 will lead to the same integrals.) These integrals are standard but a bit unpleasant, except for I_{z} = 64π/5 . The easiest way to do the other two is to note that by geometrical symmetry I_{x} = I_{z} and by translational symmetry of the trig functions I_{y} = I_{x} . Thus all three are equal and we get the Gauss result again.

4. (28 pts.) Define a new coordinate system in a region of the x–y plane by

x = u + \frac{1}{3}v3, y = ev.

(a) Find the formulas for the tangent vectors to the coordinate curves.

\frac{∂r⃗}{∂u} = ( \begin{matrix}1 \\ 0 \end{matrix} ) , \frac{∂r⃗}{∂v} = ( \begin{matrix}v^{2} \\ e^{v} \end{matrix} ) .

For future reference, let’s put these together into a Jacobian matrix,

J = ( \begin{matrix}1 & v^{2} \\ 0 & e^{v} \end{matrix} ) .

(b) Find the area of the region R bounded by the curves u = 0 , u = 3 , v = −2 , and

v = 1 .

We have

det J = e^{v},

so the area is

∫_{0}^{3} du ∫_{−2}^{1} e^{v} dv = 3(e − e^{−2}) .

(c) Find the formulas (in terms of u and v) for the normal vectors to the coordinate

curves.

We find

J^{−1} = e^{−v} ( \begin{matrix}e^{v} & −v^{2} \\ 0 & 1 \end{matrix} ) .

The normal vectors are the rows of this matrix, ∇u = (1, −v^{2}e^{−v}), ∇v = (0, e^{−v}).

(d) Do ONE of these (10 pts. extra credit for both): [Continue on next page if necessary.]

A. In the x–y plane sketch the region R, the basis of tangent vectors at the point where (u, v) = (0, 1) , and the basis of normal vectors at the point where (u, v) = (3, 1) . (Graphing calculators are allowed. Note that the curve u = 0 has an inflection and vertical tangent at (u, v) = (0, 0) .)

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B. Discuss the “global” properties of this coordinate transformation: Does it cover the whole x–y plane? What is the largest natural range of the variables (u, v) ? Is the mapping from (u, v) to (x, y) one-to-one? Does it have a smooth (differentiable) inverse?

Since det J is never 0 , the transformation has a smooth local inverse at every point. Since y = e^{v} is always positive, the coordinate transformation is defined only in the upper half plane. However, as v ranges from −∞ to +∞, y ranges from 0 to ∞ injectively. Furthermore, for a fixed v , u is uniquely determined by x and vice versa. Thus the natural range of (u, v) is all of R^{2} , and the coordinate transformation maps this plane bijectively onto the upper half plane −∞ < x < ∞, 0 < y < ∞. (Thus the “local inverse” is actually a global inverse.)

5. (15 pts.) Calculate the determinant | \begin{matrix}0 & 1 & 0 & 0 & 0 \\ 1 & π & 2 & 3 & 4 \\ 4 & π2 & 3 & 2 & 2 \\ 1 & π3 & 1 & 1 & 1 \\ π & π4 & π & 2π & π\sqrt{7} \end{matrix} |.

Expand in minors of the top row, and extract a factor from the bottom row:

(−π) | \begin{matrix}1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 2 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 2 & \sqrt{7} \end{matrix} | .

Add the first row to the second row:

(−π) | \begin{matrix}1 & 2 & 3 & 4 \\ 5 & 5 & 5 & 6 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 2 & \sqrt{7} \end{matrix} | .

Subtract 5 times the third row from the second row:

(−π) | \begin{matrix}1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 2 & \sqrt{7} \end{matrix} | .

Expand in minors of the second row:

(−π) | \begin{matrix}1 & 2 & 3 \\ 1 & 1 & 1 \\ 1 & 1 & 2 \end{matrix} | = (+π) | \begin{matrix}1 & 1 & 1 \\ 1 & 2 & 3 \\ 0 & 0 & 1 \end{matrix} | = π | \begin{matrix}1 & 1 \\ 1 & 2 \end{matrix} | = π.