Math. 311 (Fulling) 19 April 2000

Test C – Solutions (corrected)

Calculators may be used for simple arithmetic operations only!

1. (30 pts.) Let F⃗ = (x − y)î+ zĵ+ (z − y)k⃗ . (a) Calculate ∇ · F⃗ .

\frac{∂(x − y)}{∂x} + \frac{∂z}{∂y} + \frac{∂(z − y)}{∂z} = 1 + 0 + 1 = 2.

(b) Calculate ∇ × F⃗ .

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

(c) Calculate ∫∫R F⃗ · dS⃗ when R is the portion of the plane z = 2 − 5x that lies in the

quadrant z > 0 , x > 0 and between the planes y = 0 and y = 1 .

Note first that the equation of the surface can also be written as

x = \frac{2 − z}{5} ,

and that when x = 0 , z = 2 , and when z = 0 , x = \frac{2}{5} . Thus the integration will be from 0 to 2 in z or from 0 to \frac{2}{5} in x . Now let’s write the integral as

∫ ∫ [F_{x} dy dz + F_{y} dz dx + F_{z} dx dy]

and ponder how best to integrate each term.

Method 1: Integrate each term over its own plane. The projection onto the x – z plane has zero area, so the F_{y} term is zero. The others are

∫_{0}^{1} dy ∫_{0}^{2} dz (x−y)+∫_{0}^{1} dy ∫_{0}^{2/5} dx (z−y) = ∫_{0}^{1} dy ∫_{0}^{2} dz (\frac{2 − z}{5} − y)+∫_{0}^{1} dy ∫_{0}^{2/5} dx (2−5x−y)

= ∫_{0}^{2} dz [\frac{2 − z}{5} y − \frac{y^{2}}{2} ]_{0}^{1} +∫_{0}^{1} dy [2x − \frac{5x^{2}}{2} − yx]0^{\frac{2}{5}} = ∫_{0}^{2} dz [\frac{2 − z}{5} − \frac{1}{2}]+∫_{0}^{1} dy [\frac{4}{5} − \frac{2}{5} − \frac{2y}{5} ]

= [− \frac{z}{10} − \frac{z^{2}}{10}]_{0}^{2} + [\frac{2y}{5} − \frac{y^{2}}{5} ]_{0}^{1} = − \frac{1}{5} − \frac{2}{5} + \frac{2}{5} − \frac{1}{5} = − \frac{2}{5} .

Method 2: Integrate everything over x and y . We have dz = −5 dx + 0 dy . Therefore, the integral is

∫ ∫_{D} [(x − y) dy (−5 dx) + F_{y} (dx)^{2} + (z − y) dx dy] = ∫_{0}^{1} dy ∫_{0}^{2/5} dx [+5(x − y) + (2 − 5x − y)]

= ∫_{0}^{1} dy ∫_{0}^{2/5} dx [5x − 5y + 2 − 5x − y] = ∫_{0}^{1} dy ∫_{0}^{2/5} dx [−6y + 2]

= ∫_{0}^{2/5} dx [−3y^{2} + 2y]_{0}^{1} = \frac{2}{5} (−3 + 2) = − \frac{2}{5} .

311C-S00 Page 2

(d) Calculate ∫∫S F⃗ · dS⃗ when S is the sphere (x − 1)2 + y2 + (z + 2)2 = 25 .

By Gauss’s theorem, this is the integral of ∇·F⃗ over the ball whose boundary is S . Since ∇·F⃗ = 2 , this is just twice the volume of the ball. Since the radius is 5 , this is

2 · \frac{4π}{3} · 5^{3} = \frac{8π}{3} · 125 = \frac{1000π}{3} .

2. (10 pts.) Find the volume of the parallelepiped generated by the edges v⃗1 = (1, 2, 1), v⃗2 = (2, 0, 2), v⃗3 = (1, 2, 3).

| \begin{matrix}1 & 2 & 1 \\ 2 & 0 & 2 \\ 1 & 2 & 3 \end{matrix} | = −2 | \begin{matrix}2 & 2 \\ 1 & 3 \end{matrix} | − 2 | \begin{matrix}1 & 1 \\ 2 & 2 \end{matrix} | = −2(6 − 2) − 0 = −8.

So the volume is +8 .

3. (30 pts.) Define curvilinear coordinates (u, t) by \{ \begin{matrix}x & = eu cosh t, \\ y & = eu sinh t. \end{matrix}

(a) Find the formulas for the tangent vectors to the coordinate curves (at a generic point

(u, t) ).

\frac{∂r⃗}{∂u} = ( \begin{matrix}e^{u} cosh t \\ e^{u} sinh t \end{matrix} ) , \frac{∂r⃗}{∂t} = ( \begin{matrix}e^{u} sinh t \\ e^{u} cosh t \end{matrix} ) .

For future use, let us put these together (as columns) into the Jacobian matrix,

J = ( \begin{matrix}e^{u} cosh t & e^{u} sinh t \\ e^{u} sinh t & e^{u} cosh t \end{matrix} ) .

(b) Find the formulas for the normal vectors to the coordinate “surfaces” (which are ac-

tually curves in this two-dimensional case).

These are the rows of J^{−1} . So we start with

det J = | \begin{matrix}e^{u} cosh t & e^{u} sinh t \\ e^{u} sinh t & e^{u} cosh t \end{matrix} | = e^{2u}(cosh^{2} t − sinh^{2} t) = e^{2u}.

Therefore, by Cramer’s rule,

J^{−1} = e^{−2u} | \begin{matrix}e^{u} cosh t & −e^{u} sinh t \\ −e^{u} sinh t & e^{u} cosh t \end{matrix} | .

Thus

∇u = (e^{−u} cosh t, −e^{−u} sinh t) , ∇t = (−e^{−u} sinh t, e^{−u} cosh t) .

It is easy to check that these have the reciprocal orthonormality properties that they ought to have:

⟨\frac{∂r⃗}{∂u} , ∇u⟩ = 1, ⟨\frac{∂r⃗}{∂u} , ∇t⟩ = 0, ⟨\frac{∂r⃗}{∂t} , ∇u⟩ = 0, ⟨\frac{∂r⃗}{∂t} , ∇t⟩ = 1.

311C-S00 Page 3

(c) Calculate ∫∫D xy2 dx dy when D is the region bounded by the curves u = 0 , u = 2 , t = 0 , t = 1 .

∫_{0}^{2} du ∫_{0}^{1} dt xy^{2}J = ∫_{0}^{2} du ∫_{0}^{1} dt e^{5u} cosh t sinh^{2} t

= \frac{1}{5} e^{5u}|_{0}^{2} \frac{1}{3} sinh^{3} t|_{0}^{1} = \frac{1}{15} (e^{10} − 1) sinh^{3} 1.

4. (15 pts.) Tell whether each of these formulas defines an inner product on the space C(0, 5) (the real-valued continuous functions of t , where 0 < t < 5 ). If not, briefly explain why not.

(a) ⟨f, g⟩ = ∫{5 \atop 0} f(t)2g(t)2 dt

NO — not bilinear.

(b) ⟨f, g⟩ = ∫{5 \atop 0} \frac{f(t)g(t)}{1 + t2} dt

YES.

(c) ⟨f, g⟩ = ∫{π/2 \atop 0} f(t)g(t) dt

NO — not positive definite: If f(t) = 0 for t < \frac{π}{2} (but f is not zero everywhere in the interval from \frac{π}{2} to 5 ), then ⟨f, f⟩ = 0 although f is not the zero vector!

5. (15 pts.) Find an orthonormal basis for R3 whose first element is û1 = \frac{1}{\sqrt{6}}(1, 1, 2) .

Note that û_{1} has norm one, so we can put it into the basis unchanged. We continue by the Gram–Schmidt procedure. Choose any vector linearly independent of û_{1} to be v⃗_{2} ; for example, v⃗_{2} = (1, 0, 0) . Its projection onto û_{1} is

v⃗_{∥} = (û_{1} · v⃗_{2})û_{1} = \frac{1}{6} (1 + 0 + 0)(1, 1, 2) = \frac{1}{6} (1, 1, 2).

So the perpendicular part is

v⃗_{⊥} = (1, 0, 0) − \frac{1}{6} (1, 1, 2) = \frac{1}{6} (5, −1, −2).

Since \sqrt{25 + 1 + 4} = \sqrt{30} , the normalized vector in this direction is

û_{2} = \frac{1}{\sqrt{30}} (5, −1, −2).

Now we need to find a unit vector perpendicular to the two we’ve found so far.

Method 1: û_{3} ≡ û_{1} × û_{2} =

\frac{1}{\sqrt{6 · 30}} | \begin{matrix}î & ĵ & k̂ \\ 1 & 1 & 2 \\ 5 & −1 & −2 \end{matrix} | = \frac{1}{6\sqrt{5}} (0 î + 12 ĵ − 6 k̂) = \frac{1}{\sqrt{5}} (0, 2, −1).

311C-S00 Page 4

Method 2: Let v⃗_{3} = (0, 1, 0) . Its projection onto the plane of the first two vectors is

v⃗_{∥} = (û_{1} · v⃗_{3})û_{1} + (û_{2} · v⃗_{3})û_{2} = \frac{1}{6} (1, 1, 2) + \frac{−1}{30} (5, −1, −2).

So

v⃗_{⊥} = (− \frac{1}{6} + \frac{5}{30} , 1 − \frac{1}{6} − \frac{1}{30} , − \frac{2}{6} − \frac{2}{30}) = (0, \frac{4}{5} , − \frac{2}{5}) ,

which normalizes to the same û_{3} as before.