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MATH 263 Solutions: TEST III
Instructions: Answer any 8 of the 10 questions. You may answer more than 8 to earn extra credit.
Whoever despises the high wisdom of mathematics nourishes himself on delusion and will never still the sophistic sciences whose only product is an eternal uproar.
- Leonardo da Vinci (1452-1519)
1. (a) Let f and g be (scalar-valued) functions of x, y and z and let F and G be vector fields defined on R3. For each of the following expressions, determine if the result is a scalar, a vector, or meaningless. (No need to give reasons.)
(b) Prove that div(curl F) = 0.
Solution: Let F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k.
Then curl F = (Qz – Ry) i – (Rx – Pz) j + (Qx – Py) k.
Thus div(curl F) = (Qz – Ry)x – (Rx – Pz)y + (Qx – Py)z = Qzx – Ryx – Rxy + Pzy + Qxz – Pyz = 0 by Clairaut’s theorem.
2. [Stewart] Let R be the trapezoidal region with vertices
(1, 0), (2, 0), (0, -2), and (0, -1). Evaluate the integral
Re(x+y)/(x-y)dA
Hint: Make the change of variables u=x+y, v=x-y.
Note: Be certain to sketch R in the xy-plane and its pullback in the uv-plane.
(Give a numerical answer.)
u = x + y and v = x – y
by solving for x and y.
So x = ½ (u + v) and y = ½ (u – v).
3. Find the path integral of f(x, y, z) = x + y + z over the straight-line segment from
(1, 2, 3) to (0, -1, 1). (Give a numerical answer.)
Solution: Define a parametrization as follows:
s (t) = t(0, -1, 1) + (1 – t)(1, 2, 3) for 0 ≤ t ≤ 1 .
So x(t) = 1 – t, y(t) = -t + 2 (1 – t) = 2 – 3t, z(t) = t + 3(1 – t)= 3 – 2t
Now s¢(t) =(-1, -3, -4), so || s¢(t) || = (1 + 9 + 4)1/2 =(14)1/2.
Also f(s (t)) = 1 – t – t + 2 (1 – t) + t + 3(1 – t)= 6 – 6t = 6(1 – t)
Thus
4. If a circleCwith radius 1 rolls along the outside of the circlex2+y2=16, a fixed pointPonCtraces out a curve called anepicycloid, with parametric equations
x=5cos(t)−cos(5t),y=5sin(t)−sin(5t).
Below is a graph of the epicycloid. Find the area it encloses.
Solution: Let s(t) = (x(t), y(t)) = (5 cos t – cos 5t, 5 sin t – sin(5t), for 0 ≤ t ≤ 2p.
Let E be the region in question. Now, invoking Green’s theorem for area:
area of E =x dy-y dx
Now x dy = (5 cos t – cos 5t)(5cos t – 5 cos 5t) = 25 cos2 t – 25 cos t cos 5t + 5cos2 5t
and, similarly:
– y dx = 25 sin2 t – 25 sin t sin 5t + 5 sin2 5t
Now x dy – y dx = 30 – 25(cos t cos 5t – sin t sin 5t) = 30 – cos 6t.
Hence Area of E =
12x dy-y dx=12 02π(30-cos6t) dt= 12 02π30 dt= 30π
5. Let G(x, y) = 3x i + y2 j be given.
(a) Find div G.
Solution:
(b) Find curl G.
Solution:
(c) Find the circulation of G around the closed curve C that is the triangle joining the points (0, 0), (2, 0), and (0, 4). Assume that the curve is oriented in the counterclockwise direction and that the curve begins and ends at the origin. Show all work!
Solution: Let T denote the triangle including its interior.
Using Green’s theorem,
circulation= T3+2ydA=0204-2x3+2ydy dx=0234-2x+(4-2x)2 dx=132
6. Find a potential function for the vector field
Solution:
Assume that F = grad f = fx i + fy j + fz k
Then fxx,y,z=y2-1x+z; so fx, y, z=xy2-lnx+z+gy, z
Next, fyx,y,z=2xy+∂gx,z∂y and so
2xy+∂gx,z∂y=2xy+5 which implies that gx,z=5y+h(z)
Now fx, y, z=xy2-lnx+z+5y+h(z)
Next, fzx,y,z=-1x+z+dhdzand so 3z2-1x+z+dhdz= -1x+z which implies thatdhdz=0.
Finally, a potential function is
fx, y, z=xy2-lnx+z+5y
7. Find the work done by the vector field
F(x, y, z) = (x2 + y) i + (y2 + x) j + zez k
over the path given by the curve
s(t) = (2 cos t) i + (2 sin t) j + (2t/p) k, 0 ≤ t ≤ 2p.
Solution: Note that this vector field is conservative: F = g,
where g(x, y, z) = (x3 + y3)/3 + xy + zez – ez.
Invoking the Fundamental theorem of line integrals:
8. Let C be the helix given by s(t) = (3cos t, 3sin t, 4t) , defined for 0 £ t £ 2p and let
T(x, y, z) = ln (1 + x + y + 2z) be the temperature (in degrees Fahrenheit) at a point (x, y, z). Calculate the average temperature of the curve C. Express your answer as a Riemann integral. (No need to evaluate.) Solution: Computing the length of the curve: L=C1ds= 02π(dxdt)2+(dydt)2+(dzdt)2 dt= 02π(-3sint)2+(3cost)2+(4)2 dt = 5(2p) = 10p
Thus the average temperature is
110π CT ds= 110π02πTσt|σ't|dt= 110π 02πln(1+3cost+3sint+4t) dt ℉
9. (Stewart) Evaluate the following integral, where C is the circle x2 + y2 = 9.
C(2y- esinx)dx+5x+ y4+1 dy
(Give a numerical answer.)
Solution: First observe that the scalar curl of the vector field is 5 – 2 = 3.
Using Green’s theorem,
C(2y- esinx)dx+5x+ y4+1 dy= Bdiv G dA=3area of C=3π9=27π
10. Let F(x, y) = 2y i + x j, and let Ca be the circle of radius a centered at the origin, oriented in a counter-clockwise direction.
(a) Compute
Solution: Note that curl F = 1 – 2 = -1. Using Green’s theorem,
CaF∙ ds= Ddiv G dA=-1area of C=-πa2
(b) Show that
Solution: lima→0-πa2=0
Extra Extra Credit:
Let R be the region in the first-quadrant that lies inside x2 + y2 = 1 but outside x2 + y2 = 2y.
(a) Sketch this region.
Solution:
Here is the region of integration.
Water is fluid, soft and yielding. But water will wear away rock, which is rigid and cannot yield. As a rule, whatever is fluid, soft and yielding will overcome whatever is rigid and hard. This is another paradox: what is soft is strong.
- Lao-Tzu (600 B.C.)
Klein bottle (from virtual math museum)