MATH 263                          TEST III                  answers & brief solutions

PART I   (Answer any 9 of the ten problems.  You may answer more for extra credit.)

1.    Let F and G be two 2-dimensional fields,

F = 3x i + 5y j         and          G = 3y i + 5x j.

Let C1 be the circle with center (2, 2) and radius 1 oriented counterclockwise.  Let C2 be the path consisting of the straight line segments from (0, 4) to (0, 1) and from (0, 1) to

(3, 1).   Find each of the following line integrals.  Explain your reasoning!

(A)         Since F is conservative, the circulation is zero.

(B)          A potential function for F is f(x,y) = .  Hence = g(3,1) – g(0,1) = 16 – 40 = -24.

(C)         Using Green’s Theorem, the circulation around the boundary of the circle is the double integral of the scalar curl over the circle.  Since the scalar curl is just 2, the desired result is 2 times the area of the circle of radius 1, or 2p.

(D)             Here we must integrate the line integral over each of two curves.  The final result is 9.

2.     What is the flux of the vector field F = 9 i + 13 j + 5 k through a circle in the xy-plane of radius 5 oriented upward with the center at the point (5, 4, 0).

The area vector is A = 25p k.  The flux is constant throughout the circle.  Thus the total flux equals the dot product of A and F, or  (25 p)(5) = 125 p.

3.  Let  H = 2x i – 3xy j + xz2 k be a vector field defined in three-space.

(A)   Find div H(x,y,z).

div H(x,y,z) = 2 – 3x + 2xz.

(B)    Compute    where S is the (hollow) cube with vertices (0,0,0), (1,0,0), (0,1,0), (1,1,0), (0,0,1), (1,0,1), (0,1,1) and (1,1,1)?

=  the triple integral of div H over the solid cube.   This evaluates to 1.

4.    Let v(x,y) be the velocity vector field of a two-dimensional fluid flow defined as follows:
v(x,y) = (x2+y2)ax i + (x2+y2)ay j,  where a is a constant.

(Note that a may be negative, but, in this case, v is not defined at the origin.)

(A)      Is fluid flowing away from the origin, toward the origin, or neither?

Away from the origin, since v(x,y) is a multiple of the position vector, xi + yj.

div v(x,y) simplifies to 2(a+1)(x2+y2)a.

(C)       For which values of a is div v positive?  Zero?  Negative?

Using part (B), div v is positive when a>-1, zero when a = -1, and negative when a <-1.

5.    Let   G = 2x i – 3xy j + xz2 k.   Calculate curl G.

A standard calculation shows that curl G = -z2 j – 3y k.

6.   The two-dimensional vector field F satisfies the condition  ||F(x,y)|| < 13 everywhere and C is the circle of radius 4 centered at the point (5, 7).   What is the largest possible value of  ?    The smallest possible value?    Explain!

The work done is maximized when F is always tangent to the circle and points in the direction of motion.  In this case, the work performed is 13 times the circumference of the circle.  This is 13(2p)(4) = 104p.

## The work done is smallest when F opposes the motion to the maximal degree.  This will yield work of – 104p.

7.    Let F(x,y,z) = .

Compute  where C is the curve parameterized by s(t) = (t, t2, t3), for 0 < t < 1.

A potential function for F is g(x,y,z) = exp(x2y2z).  Hence  equals g(1,1,1) – g(0,0,0) = e – 1.

8.   Let S denote the sphere of radius 2 centered at the point Q = (3,2,5).  Let F denote the 3-dimensional vector field given by F(x,y,z) = (x+yz) i +(x-2y+z2) j + (xy+3z) k.  Using the divergence theorem, compute the flux integral  .

div F equals 2.   Thus, using the Divergence Theorem, equals the triple integral of 2 over the solid ball of radius 2.   This is just 2(4/3)p(23) = 64 p / 3.

9.    Let  F = ax i + by j + cz k,  where a, b, and c are constants.  Suppose that the flux of F through a surface of area 5 lying in the plane y = 3, oriented in the positive y-direction, is 14.  Find the flux of F through a surface of area 7 lying in the plane y = 9, oriented in the negative y-direction.

In the plane y = 3, F = ax i + 3b j + cz k.  The area vector, A1, of the surface of area 5 lying in the plane y=3, is 5j.   Thus the total flux passing through this region is the dot product of F and A1 that equals 15b.  Thus b equals 14/15.

In the plane y=9, F = ax i + 9b j + cz k.  The area vector A2 of the surface of area 7 lying in the plane y=9 is given by –7j.   Thus the flux passing through this region is given by the dot product of F and A2 which equals 9b(-7) = 9(14/15)(-7) = -294/5, or approximately 58.8.

10.   Find a potential function for the vector field

F(x,y) = (3x2 sin y + 1 – y sin x) i + (x3 cos y + cos x + 2y ) j.

Using the standard technique of setting up two partial differential equations, we obtain the potential function g(x,y) = x3 sin y + x + y cos x + y2.

PART II   (Answer any 8 of the 9 problems.   You may answer more than 9 for extra credit.)

1.    Let F = xy i + ln(x+1) j denote the velocity vector field of a fluid in the xy-plane.  Suppose that, at time t = 0 minutes, a cork is dropped into the fluid at the position (1,2).  Use Euler’s method with ∆t = 0.2 minutes, to estimate the position of the cork at time t = 0.6 minutes.

Using the standard technique that the next location is given by the current location plus Δt  F(current location), we obtain s(0.2) = (1.4, 2.14),

s (0.4) = (1.99, 2.31), s (0.6) = (2.92, 2.53).

2.    Find a potential function for the vector field

Using the standard technique of setting up three partial differential equations, we obtain the potential function g(x,y,z) = xy2 – ln(x+z) + 5y + z3.

3.     Let  G(x,y) = x2 i + y2 j  be a vector field on the plane.  Find the equation of the flow line s(t) that passes through the point  at time 0.

Let s (t)= (x(t), y(t)) be the unknown function.

Begin with the requirement that G(s (t))  = s’(t).  Thus (x2, y2) = (dx/dt, dy/dt).

Now dx/dt = x2 and dy/dt = y2.

Separating variables:   x -2 dx = dt.  Integrating yields –x -1 = t + C.  So x = -1/(t+C).  Since x = ½ when t = 0 , we find that C = -2.   Thus x = -1/(t-2) = 1/(2-t).

Similarly separating variables and solving for y yields:  y = 1/(1-t).

Hence s(t) = (1/(2-t), 1/(1-t).

4.    Using Green’s Theorem, compute , where C is the boundary of the square with vertices (0,0), (1,0), (1,1), and (0,1).

Using Green’s Theorem, the circulation equals the double integral of the scalar curl of the vector field over the square.  The scalar curl is 3x.   Hence we must integrate 3x over the unit square to obtain the answer 3/2.

5.   Consider the cardioid pictured below:

This curve is parameterized by the equation

r(t) =  3(1+cos t)(cos t) i +  3(1+cos t)(sin t) j,   0< t <2p.

Using Green’s area formula, express the area of the cardioid as a Riemann integral.

Note:  Do not evaluate the integral that you obtain.

We may choose any of several vector fields (each with scalar curl 1) to obtain the required area.  A popular choice was  F(x,y) = xj.

To use Green’s area theorem we must find the circulation of F over the closed curve r(t).

This is just the line integral of  the dot product of F(r(t)) with r’(t).  Since F(r(t)) equals 3(1+cos t)(sin t) j,  we need find only the j component of r’(t).   The j component equals the derivative of 3(1+cos t)sin t.   Using the product rule we obtain

3( (1+cos t)cos t +  (-sin t) sin t) = 3 (1 + cos2 t – sin2 t).

Thus the integral that represents the area of the region is

.  This could be simplified to:

.

6.  Let T be the triangle in the xy-plane with vertices (0,0), (2,0), and (0,4).  Let C denote the boundary of T endowed with the positive orientation.

Let F(x,y) = xy2 i + (x2y + x2) j.  Using Green’s Theorem, evaluate .

The triangle is bounded by the lines y = 4 – 2x, x = 0, and y = 0.  Using Green’s Theorem, the circulation around the boundary of the triangle equals the double integral of the scalar curl of F over the triangle.  The scalar curl of F equals 2x.

Thus, integrating 2x over the triangle yields the answer of 40/3.

7.  Using Green’s theorem, find the work done by the vector field  F = 2xy3 i + 4x2y2 j  in moving a particle counterclockwise around the boundary of the “crescent” in the first quadrant enclosed by the curves y = x4 and y = x5.

Note that the points of intersection occur at (0,0) and (1,1).

The scalar curl of F is 2xy2.    Using Green’s Theorem, the circulation around the crescent equals the double integral of the scalar curl of F over the crescent.  Expressing this as an iterated integral yields:

=  1/119.

Here it is important to note that x5 < x4 when 0 < x < 1.

8.  Let F = (x + sin z) i + (x - y) j +  (4z – 3xy) k be a velocity vector field.  Using the divergence theorem, find the total flux crossing the sphere of radius 4 centered at the origin.

The divergence of F is 1 –1 + 4 = 4.  Using the Divergence Theorem, the total flux crossing the sphere equals the triple integral of div F over the solid ball.

Since the divergence is constant, this triple integral reduces to 4 (volume of the ball) =

4 (4/3)p(43) = 1024p/3.

9.    Let  F(x,y)  be a conservative vector field with potential function g(x,y) satisfying g(0,0) = 3.   Let C1 be the line segment from (0,0) to (4,1),  C2 the path parameterized by

r(t) = (4 + 2 sin (t/2)) i + (cos t) j,  for 0 < t < p, and C3 the path parameterized by

s(t) = (t + 6) i + (t2 - t - 1) j, for 0 < t < 2.

Suppose that .

Evaluate g(4,1), g(6,-1), and g(8,1).

Using the Fundamental Theorem of Calculus for Line Integrals yields:

Hence g(4,1) = 4+3 = 7.

Similarly:

Hence g(6,-1) = 7-5 = 2.

and:

Hence g(8,1) = 11 + 2 = 13.