Students' answers to review problems for the second exam
in Math 291, fall 2002

#1 Ellway #2 Gradziadio #3 Grubin #4 Gurkovich #5 Hort
#6 Huang #7 Jackson #8 Johnson (etc.) #9 Kravtsov #10 Lunemann
#11 Odoi #12 Pergament #13 #14 Ryslik #15 Sullivan
#16 WHO? #17 Tokayer #18 Vander Valk #19 Walsh #20 Zhu

Students have so far (11/14) requested no formulas for the second exam. O.k.: I got (11/23) one request for one formula: less work for "the management"! Now (11/25) a draft formula sheet exists. And today (11/26) an updated version has been posted.

1. Use Lagrange multipliers to find the maximum and minimum values of f(x,y)=x+y2 subject to the constraint x2+2y2=4.

g(x,y)= x^2+2y^2=4

df/dx= 1
df/dy= 2y

dg/dx= 2x
dg/dy= 4y
Using Lagrange Multipliers: (Note L is used to represent lambda)
  1. L= 2x
  2. 2yL = 4y
  3. x^2+2y^2= 4
From equation 2, we can solve and find that L = 2 or y=0. If L= 2 then by plugging this value into equation 1, you get x=1. By putting x=1 into the last equation, y= (3/2)^(1/2) or -(3/2)^(1/2).
If y=0 then x= +2 or -2.
You must then plug these values into the function, x+y^2. The minimum is therefore at (-2,0) and equals -2. The maximum is therefore at both (0,sqrt(3/2)) and (0,-sqrt(3/2)) and equals 5/2.

Sent by Jordana Ellway on Thu, 14 Nov 2002 02:09:15

2. Evaluate the iterated integral int01int0vsqrt{1-v2} du dv.

Evaluate the inside integral first:
int sqrt(1-v^2) du = u*sqrt(1-v^2)|u=0 to u=v
= v*sqrt(1-v^2) - 0*sqrt(1-v^2)
= v*sqrt(1-v^2)

Evaluate the outside integral:
int v*(1-v^2)^(1/2) dv = -(1/2)*(2/3)*(1-v^2)^(3/2)|v=0 to v=1
= -(1/3)*(1-1^2)^(3/2) - -(1/3)*(1-0^2)^(3/2) = 0 - -(1/3) = 1/3

Comment by the management I am almost sure that the instructor who wrote this problem wanted to write the following question:

Evaluate the iterated integral int01intu1sqrt{1-v2} dv du.

Then students would need to recognize this iterated integral as a creature somewhat difficult to compute, and then would need to rewrite it as an iterated integral in the "other" order and compute the result, as was done above.

Sent by Chris Gradziadio on Thu, 21 Nov 2002 13:08:13

3. Change the integral int0aint0sqrt(a2-y2) (x2+y2)3/2 dx dy to polar coordinates.

The first step in changing this integral to polar coordinates is setting x equal to r cos(theta) and y equal to r sin(theta). Then, change (x^2+y^2)^(3/2) to polar using the equations above. This results in r^3. Then, change dx dy to polar, which including the Jacobian, is r dr d(theta). This leaves r^4 dr d(theta) to be integrated. Next, put the area over which this is to be integrated in terms of r and theta. The upper bound for x was (a^2-y^2)^(1/2). Plugging in r cos(theta) and r sin(theta) for x and y respectively, r ends up equaling a. Using the same procedure for the lower x, it is found that r equals zero. Now, put the y bounds in terms of theta. The upper bound becomes pi/2 and the lower bound is zero. Fully converted, this integral becomes: int_0^(pi/2) int_0^a r^4 dr d(theta).

Sent by Jeremy Grubin on Sat, 23 Nov 2002 16:46:00

4. Evaluate the integral int03inty29y cos(x2) dx dy by reversing the order of integration.

Reversing the order of integration results in the following limits of integration: x is integrated from 0 to 9 and y goes from 0 to sqrt(x). These limits can be found through analysis of a graph of the original limits.

The result of the inside integral is 1/2 y^2 cos(x^2) evaluated from y=0 to y=sqrt(x), which simplifies to (1/2)x cos(x^2). This can be integrated with respect to x by letting u=x^2 so du=2xdx. The new limits are u=0^2 to u=9^2 and the integrand becomes (1/4)cos(u) du . The result of this integral is (1/4)sin(u) evaluated from u=0 to u=81. The final answer is (1/4)sin(81) .

Sent by Corey Gurkovich on Sat, 23 Nov 2002 05:43:56

5. Set up a double integral in polar coordinates to find the volume of the solid that lies under the paraboloid z=x^2+y2, above the xy-plane, and inside the cylinder x2+y2=2x.
Do not evaluate the integral.

z=x^2+y^2. Above xy plane, which is z=0 And bounded by cylinder x^2+y^2=2x.

This is the double integral: int(int(x^2+y^2))dA over a region R

Converting to polar coordinates:
z=x^2+y^2 => r^2
The bounds are ordinarily 0 < theta < 2*pi (but see below!)
0 < r < (2cos(theta)).
This is found because the lower part of the integral is z=0 and the upper bound is this cylinder. When you convert to polar coordinates for the cylinder you get something like r^2=2 r cos(theta) and that means r goes from 0 to (cos(theta)). You go once around the entire circle if theta goes from -pi/2 to pi/2 (put in some values of theta, and see what happens!). theta of course goes from -pi/2 to pi/2 because it is going around an entire circle for each "slice" of the cylinder. The final integral is: int(int(r^3,r=0...sqrt(2cos(theta))),theta=-pi/2..pi/2)  dr dtheta.

Sent by Greg Hort on Fri, 22 Nov 2002 16:56:20

6. Set up a triple integral in spherical coordinates to find int int intEx2 dV, where E lies between the spheres rho=1 and rho=3 and above the cone phi=pi/4.
Do not evaluate the integral.

The function in the integral is x^2, thus it is (rho*sin(phi)*cos(theta))^2 in spherical, and dV becomes rho^2*sin(phi) d(rho) d(theta) d(phi).

The problem tells us that rho goes from 1 to 3, and above the cone phi=pi/4, so the limit for this triple integral would be E = {(rho,theta,phi); 1<=rho<=3, 0<=theta<=2pi, 0<=phi<=pi/4} and the integrand would be rho^4*(sin(phi))^3*cos(theta))^2.

Sent by James Huang on Sat, 23 Nov 2002 19:50:19

7. Evaluate intC(x-2y2) dy where C is the arc of the parabola y=x2 from (-2,4) to (1,1).

To evaluate the line integral int_C (x-2y^2)dy over the arc of y=x^2 from (-2,1) to (1,1) the formula for the dy portion of a line integral has to be applied:
int_C f(x,y) dy=int_a^b f(x(t),y(t)) y'(t) dt
x(t) = t; y(t) = t^2; y'(t) = 2t.

Then: x-2y^2 = (t)-2(t^2)^2 = t-2t^4
So: int_C (x-2y^2)dy = int_(-2)^1 (t-2t^4)(2t)dt =int_(-2)^1 (2t^2-4t^5)dt =2/3t^3-2/3t^6]_(-2)^1 =(2/3(1)^3-2/3(1)^6)-(2/3(-2)^3-2/3(-2)^6)=48.

Sent by David Jackson on Sun, 24 Nov 2002 19:24:09

8. (Lagrange multipliers) Let f(x,y,z)=xy+xz+yz.
(a) Find the maximum value of $f$ on the sphere x^2+y^2+z^2=1. (b) Find the minimum value of $f$ on the sphere x^2+y^2+z^2=1.

The Management together with Mr. Johnson contributed to the solution of this problem. Mr. Hort's persistence was also useful.

We formulate the Lagrange multiplier equations, and L will also be used here for the traditional lambda.

Add the first three equations to get 2(x+y+z)=2L(x+y+z). Now EITHER x+y+z=0 OR L=1.

If L=1, the first equation becomes y+z=2x, so x=(1/2)y+(1/2)z. Then the second equation x+z=2Ly becomes (substituting for L and x):
and this is (3/2)z=(3/2)y which implies z=y. Going back to x=(1/2)y+(1/2)z we see that x=y also: all the variables are equal. then the constraint equation x^2+y^2+z^2=1 implies that x=y=z={+/-}(1/sqrt(3)). Substitute into f's formula, so at these two points, f is 1.

Now what if x+y+z=0? Square this equation. The result is x^2+y^2+z^2+2(xy+xz+zy)=0. Use the constraint equation. This becomes 1+2(the value of f)=0. So the values of f when x+y+z=0 are just -1/2.

Apparently the maximum value of f on the sphere is 1 and the minimum value of f on the sphere is -1/2.

Initially there is a distraction: if x+y+z=0 then x+y=-z so L=-(1/2) but that doesn't seem to help in finding values of f.

Some pictures
The geometry of the level surfaces of f together with the constraint surface are interesting and a bit complicated.

This is the level set f=-1/2 together with the plane x+y+z=1 and the unit sphere. The level set is tangent to the unit sphere exactly where the plane cuts the sphere. There is a circle of Lagrange multiplier candidates. This is the level set f=1 together with the unit sphere. Note that there are 2 Lagrange multiplier candidates. Here is the level surface f=0 together with the unit sphere. This level surface is not tangent to the unit sphere but cuts through it.
Sent by Brent Johnson on Sun, 24 Nov 2002 14:59:24 (with some additions made by the management 11/26/2002.)

9. Evaluate the integral intCF· dr, where F(x,y)=x2y3i-y*sqrt(x)j and r(t)=t2i-t3j, 0=<t<=1.

Using the formula of line integrals of vector fields [int(F,r) along C = int(F(r(t)) dot r'(t),t=a..b) where a<=b<=b (<= is less or equal and dot represents the dot product)] we obtain int((t^4*(-t)^9*i+t^4*j)dot(2*t*i - 3*t^2*j), t=0..1) which, after applying dot product rule becomes int(-2*t^14-3*t^6,t=0..1)=-59/105.

Sent by Andrey Kravtsov on Mon, 25 Nov 2002 09:41:50

10. Evaluate int01int01 emax(x2,y2) dx dy where max(x2,y2) means the larger of the numbers x2 and y2.

To begin this problem, I realized that this problem appeared to have two different paths, one for where x^2 is the max and the other for when y^2 was the max. In order to do away with this problem, I turned the iterated integral into a double integral over the region of the unit square. Then, I said that by splitting the unit square in half by the line x=y, one can then integrate the integral of e^(w^2), where w is either x or y. Then, I converted this double integral of the triangle into a new iterated integral, multiplied by 2 to make it equal to the region of the square. Therefore, the new iterated integral was 2e^(x^2) dy dx from y=0 to x and x=0 to 1. When one takes the case of y^2 as the max, the result is the same with x's and y's reversed. After taking the integral with respect to y, one is left with the integral of 2xe^(x^2) dx from x=0 to 1. By setting u=x^2, one is left with the integral of e^u du. By evaluating this integral, one receives e-1. The same is the case for where y^2 is the maximum since dA can be made dx&dy in that case thereby leading to the same result of e-1. So the total answer is 2(e-1).

Sent by Matthew Lunemann on Sat, 23 Nov 2002 18:40:01

11. I) Set up the triple integral int int intR f(x,y,z) dV where R is the region inside the sphere x2+y2+z2=2 and inside the cone z2=x2+y2 (inside the cone means z2>=x2+y2) in the following systems of coordinates:
a) Rectangular;    b) Cylindrical;    c) Spherical.

II) Evaluate the volume of the region R described in part I.

a) We set up the triple integral over the region R where R is inside the sphere x^2+y^2+z^2=2 and inside the cone z^2=x^2+y^2 in the rectangular system of coordinates.

First, we divide the solid into four parts, the cone z=(x^2+y^2)^(1/2) above z=0 and below z=1(We denote this R1) and the solid above z=1 and below x^2+y^2+z^2=2 (We denote this R2). Similarly, the cone z=(x^2+y^2)^(1/2) below z=0 and above z=-1 (We denote this R1') and the solid below z=-1 and above x^2+y^2+z^2=2 (We denote this R2').

For R1 and R1'; If we cut the cone along xy-plane first, we get a circle (x^2+y^2)^(1/2)=+-z on the xy-plane. The boundary of +z is (x^2-y^2)^(1/2)<=z<=1 where the lower boundary moves as x and y change, and the boundary of -z is -1<=z<=-(x^2-y^2)^(1/2) where the upper boundary moves as x and y change.

So we compute the iterated integral of the circle (x^2+y^2)^(1/2)=z with assuming z is constant. The boundary of y, x should be where x^2+y^2=1, so -(1-x^2)^(1/2)<=y<=(1-x^2)^(1/2), and -1<=x<=1.

Thus we can express R1and R1' as the following; R1={(x,y,z)|-1<=x<=1, -(1-x^2)^(1/2)<=y<=(1-x^2)^(1/2),(x^2+y^2)^(1/2)<=z<=1 }
R1'={(x,y,z)|-1<=x<=1, -(1-x^2)^(1/2)<=y<=(1-x^2)^(1/2), -1<=z<=-(x^2+y^2)^( 1/2)}

Therefore, we have int(-1to1){int(-(1-x^2)^(1/2))to(1-x^2)^(1/2))[int((x^2+y^2)^(1/2)to1)f(x,y, z)dz]dy}dx and int(-1to1){int(-(1-x^2)^(1/2))to(1-x^2)^(1/2))[int(-1 to -(x^2+y^2)^(1/2))f(x,y,z)dz]dy}dx.

For R2 and R2': Similarly if we cut the cone along xy-plane, we get a circle (x^2+y^2)^(1/2)=+-z on the xy-plane. Then the boundary of +z is 1<=z<=(2-x^2-y^2)^(1/2) where the upper boundary moves as x and y change, and the boundary of -z is -(2-x^2-y^2)^(1/2)<=z<=-1 where the lower boundary moves as x and y change.

We compute the iterated integral of the circle (x^2+y^2)^(1/2)=z in the same way as R1, so we get -(1-x^2)^(1/2)<=y<=(1-x^2)^(1/2), and -1<=x<=1.

We can express R2 as the following; R2={(x,y,z)|-1<=x<=1, -(1-x^2)^(1/2)<=y<=(1-x^2)^(1/2),1<=z<=(2-x^2-y^2)^(1/ 2)}
R2'={(x,y,z)|-1<=x<=1, -(1-x^2)^(1/2)<=y<=(1-x^2)^(1/2), -(2-x^2-y^2)^(1/2)< =z<=-1}.

Therefore, we have

int(-1to1){int(-(1-x^2)^(1/2))to(1-x^2)^(1/2))[int(1to(2-x^2-y^2)^(1/2))f(x, y,z)dz]dy}dx and int(-1to1){int(-(1-x^2)^(1/2))to(1-x^2)^(1/2))[int(-(2-x^2-y^2)^(1/2)to -1)f (x,y,z)dz]dy}dx

Therefore, R=R1+R2+R1'+R2', so the triple integral over R is; int(-1to1){int(-(1-x^2)^(1/2)) to(1-x^2)^(1/2))[int((x^2+y^2)^(1/2)to1)f(x,y,z)dz]dy}dx + int(-1to1){int(-(1-x^2)^(1/2))to(1-x^2)^(1/2))[int(1 to(2-x^2-y^2)^(1/2))f(x,y,z)dz]dy}dx

+int(-1to1){int(-(1-x^2)^(1/2))to(1-x^2)^(1/2))[int(-1to -(x^2+y^2)^(1/2))f( x,y,z)dz]dy}dx + int(-1to1){int(-(1-x^2)^(1/2))to(1-x^2)^(1/2))[int(-(2-x^2-y^2)^(1/2)to -1)f (x,y,z)dz]dy}dx

b)We set up the triple integral over the region R where R is inside the sphere x^2+y^2+z^2=2 and inside the cone z^2=x^2+y^2 in the cylindrical system of coordinates.

Let x=r*sin(theta), y=r*cos(theta), z=z.

Similar to a), we divide the solid into four parts, the cone z=(x^2+y^2)^(1/2) above z=0 and below z=1(We denote this R1), the solid above z=1 and below x^2+y^2+z^2=2 (We denote this R2), the cone z=(x^2+y^2)^(1/2) below z=0 and above z=-1 (We denote this R1') and the solid below z=-1 and above x^2+y^2+z^2=2 (We denote this R2').

For R1 and R1', when we cut the solid parallel to xy-plane, we get the circle x^2+y^2=r^2=z^2, therefore, the boundary of r is 0<=r<=z, 0<=r<=-z, respectively.

For R2 and R2', we get the circle x^2+y^2==r^2=2-z^2 where z is fixed. therefore,the boundary of r is 1<=r<=(2-z^2)^(1/2) for both regions.

Therefore these regions can be expressed by R1={(r,theta,z)|0<=theta<=2*pi, 0<=r<=z, 0<=z<=1} R1'={(r,theta,z)|0<=theta<=2*pi, 1<=r<=(2-z^2)^(1/2), 1<=z<=sqrt(2)} R2={(r,theta,z)|0<=theta<=2*pi, 0<=r<=z, -1<=z<=0} R2'={(r,theta,z)|0<=theta<=2*pi, 1<=r<=(2-z^2)^(1/2), -sqrt(2)<=z<=-1}

Thus we get int(0to1){int(0to2*pi)[int(0toz)f(r*cos(theta),r*sin(theta),z)rdr]d(theta)}d z + int(1to sqrt(2)){int(0to2*pi)[int(1to (2-z^2)^(1/2))f(r*cos(theta),r*sin(theta),z)rdr]d(theta)}dz +int(-1to0){int(0to2*pi)[int(0toz)f(r*cos(theta),r*sin(theta),z)rdr]d(theta) }dz + int(-sqrt(2)to-1){int(0to2*pi)[int(1to(2-z^2)^(1/2))f(r*cos(theta),r*sin(the ta),z)rdr]d(theta)}dz

c) We set up the triple integral over the region R where R is inside the sphere x^2+y^2+z^2=2 and inside the cone z^2=x^2+y^2 in the spherical system of coordinates.

Let x=rho*sin(phi)*cos(theta), y=rho*sin(phi)*sin(theta), z=rho*cos(phi).

We divide the solid into two parts, the solid above z=0 (R1) and the solid below z=0 (R2).

Notice that rho^2=x^2+y^2+z^2, so the boundary of rho is 0<=rho<=sqrt(2).

Then we get R1={(rho,theta,phi)|0<=rho Thus we get int((0 to 2*pi){int(0 to pi/4) [int(0 to sqrt(2)) f(rho*sin(phi)*cos(theta), rho*sin(phi)*sin(theta),rho*cos(phi))(rho)^2 sin(phi) d(rho)]d(phi)}d(theta) + int((0 to 2*pi){int(3*4/pi to pi) [int(0 to sqrt(2) f(rho*sin(phi)*cos(theta), rho*sin(phi)*sin(theta),rho*cos(phi))(rho)^2 sin(phi) d(rho)]d(phi)}d(theta).

II) We choose the system of coordinates of the simplest computation, c) and evaluate the volume of R.

We take f(rho*sin(phi)*cos(theta), rho*sin(phi)*sin(theta),rho*cos(phi))=1 so that we can get the volume of R. Also, the volume of R1=the volume of R2, so we compute only R1 and double it.

Then int(0 to 2*pi){int(0 to pi/4) [int(0 to sqrt(2) (rho)^2 sin(phi) d(rho)]d(phi)}d(theta) =int(0 to 2*pi){int(0 to pi/4) [(1/3) (rho)^3 sin(phi)](0 to sqrt(2))d(phi)}d(theta) =int(0 to 2*pi){int(0 to pi/4) [(2*sqrt(2)/3*sin(phi)]d(phi)}d(theta) =int(0 to 2*pi)[-(2*sqrt(2))/3*sin(phi)](0 to pi/4)d(theta) =int(0 to 2*pi)[((2*sqrt(2)+2)/3]d(theta) =[(4*sqrt(2)-4)*pi]/3

Therefore the volume of R is [(8*sqrt(2)-8)*pi]/3.

Comment by the management Yes, this problem was given by one of my colleagues in a recent Math 291 exam. The problem and the solution are long. I can't offer any serious "shortening" comments to the solution given above.

Sent by Atsuko Odoi on Thu, 21 Nov 2002 14:23:14

12. The constraint x4+x2y2+2y4+z4=1 defines a closed and bounded set in R3, and thus the continuous function f(x,y,z)= xyz attains its maximum value on that set. What is the maximum value of xyz subject to this constraint? Be sure to analyze carefully and completely any system of equations you solve.

First, we use Lagrange Multipliers, by taking the gradient of the constraint, multiplying it by L (which I will use for lambda here), and setting it equal to the gradient of the function we want to maximize. This yields 3 equations, the constraint equation, and four unknowns. I will label each equation as:

  1. (4x^3 + 2xy^2)L = yz
  2. (2yx^2 + 8y^3)L = xz
  3. (4z^3)L = xy
  4. x^4 + (x^2)(y^2) + 2y^4 + z^4 = 1
I will multiply both sides of equation (1) by x, both sides of equation (2) by y, and both sides of equation (3) by z. If L=0, then at least one of the variables is 0 and then the function xyz is zero, but it has positive values, so such a point can't be a max. Then, we divide both sides by L. This will give me:
  1. 4x^4 + 2(x^2)(y^2) = xyz/L
  2. 2(y^2)(x^2) + 8y^4 = xyz/L
  3. (4z^4) = xyz/L
We can set equation (5) = equation (6), and we get either both x and y = 0, or
  1. x = +/- Wy
where W is 2^(1/4) (approximately 1.18921). If the first case is true, we have all three variables equal to zero. This can't happen because of the constraint equation (4). which is the first candidate for a min or a max (but we know its neither, because we know we can have both positive and negative values of the function). If the second case is true, we move on. We then set (5) = (7), and plug in Wy for x.

From this, we get

  1. z= +/- Vy
where V is (8+2sqrt(2))^(1/4) (approximately 1.28270). Now, since we have all the variables in terms of y, we can plug in for x and z into the constraint equation (4). When we plug in, we get 8.12132y^4 = 1, and y= +/- .59237. We plug this back into (eq. 8) to get x = +/- .704451. Then we plug in y into (9) and get z = +/- .759836. We have eight candidates for max and min, but 4 of them will give the same value, and the other four will give the negative of that value. We plug all these candidates into the original f(x,y,z)=xyz equation. Since these 8, and (0,0,0) are the only candidates for min and max, we know that the min is -.317076 where we have (+,+,-), (+,-,+), (-,+,+), or (-,-,-). And the max is .317076 where we have (+,+,+), (+,-,-), (-,+,+), or (-,-,+).

Sent by Alexander Pergament on Sun, 17 Nov 2002 21:30:22

14. Find the line integral intC dr where F=2xyi+x2jand C is an arc of a circle centered at (0,1), of radius 1 joining the points P1(0,0) and P2(0,2).

Here F= 2xyi+x^2j. First we notice that the field is conservative. Since P = 2xy and Q = x^2, dP/dy = 2x and dQ/dx = 2x. Since dP/dy and dQ/dx are both 2x, we suspect that the field is conservative. Therefore, we now must get an f so that F = grad(f).

Since f_x = 2xy, if we get the antiderivative with respect to x, f(x,y) = x^2  y + C_1(y). Since f_y = x^2, if we get the antiderivative with respect to y, f(x,y) = x^2 y+ C_2(x). Logically, we can now deduce that C_1(y) and C_2(x) are not necessary (since we are computing a line integral and the difference of two values of f will just cause the resulting constants to cancel) and therefore equal 0 and f(x,y) = (x^2)y.

Now, because the function is conservative and F = grad(f), we can say that the integral(over the path C) of F dot dr equals the integral(over the path C) of grad(f) dot dr. Then we can apply the fundamental theorem for line integrals and say that the above integral is equal to f(r(b)) - f(r(a)) where r is the position vector and a<=t<=b. But, since the beginning and final points are provided in the problem, we can simply rewrite this to be f(0,2) - f(0,0).

Therefore the answer is (0^2)(2) - (0^2)(0) which equals 0.

Note from the management: you can also "do" this problem by a direct parameterization using x=cos(t) and y=1+sin(t), where t goes from -Pi/2 to Pi/2. The parameterized integral can be done, and is really not too bad. The answer will again be 0. The computation above is much shorter, though!

Sent by Greg Ryslik on Mon, 18 Nov 2002 21:45:04

15. Suppose R is the trapezoid with vertices (1,0), (2,0), (0,-2), and (0,-1). Use the change of variables u=x+y and v=x-y to compute the double integral over R of e^(x+y)/(x-y) dA.

First, solve for x and y in terms of u and v. You get:

By sketching R, you'll notice that the trapezoid is made up of four lines (x=0, y=0, y=x-1, and y=x-2), which can be written in terms of u and v:
x=0 => (u+v)/2=0 => u=-v
y=0 => (u-v)/2=0 => u=v
y=x-1 => (u-v)/2=(u+v)/2+1 => u-v=u+v-2 => v=1
y=x-2 => (u-v)/2=(u+v)/2+2 => u-v=u+v-4 => v=2

The Jacobian is |x_u*y_v-x_v*y_u|. So the Jacobian in this case is 1/2. Since it is constant, we can just pull it out of the integrand, and multiply through by 1/2 at the end.

So, the double integral becomes this iterated integral:
(1/2) int(1 to 2) [int(-v to v) e^(u/v) du] dv.

First, we can solve the inner integral by using the substitution w=u/v, therefore making dw=1/v du, and the limits of integration change from u=-v to w=-1 and u=v to w=1. So the inner integral is:
int(-1 to 1) e^w v dw=v int(-1 to 1) e^w dw = v(e-(1/e)).

Now use this to compute the outer integral. We can factor out the constant e-(1/e) and multiply through at the end to make it easier:
int(1 to 2) v dv = 4/2-1/2 = 3/2.

So, double integral over R of e^(u+v)/(u-v) dA, (by multiplying by the Jacobian and (e-1/e), which were factored out in the above steps) turns out to be:
(1/2)(e-(1/e))(3/2) = 3e/4-3/(4e)

Sent by Jason Sullivan on Mon, 18 Nov 2002 19:08:02

16. Use Green's Theorem to evaluate the line integral int_C (y^2-arctan x)dx+(3x + sin y)dy, where C is the boundary of the region enclosed by y=x^2 and y=4.

Here P is y^2-arctan x and Q is 3x+sin y. So the integrand over the region is Q_x-P_y (note that the more horrible things [they aren't SO horrible!] like arctan x and sin y disappear). I get Q_x-P_y=3-2y.

We need to integrate this over the region bounded by y=x^2 and y=4. I don't think it matters too much which order this is done in. x for me will be on the outside and y, on the inside. When y=4 then x=+/-2. So we seem to have int_{-2}^2 int_{x^2}^4 (3-2y) dy dx.

The inner integral: 3y-y^2]_{y={x^2}}^{y=4}=(12-16)-(3*x^2-x^4)=-4-3*x^2+x^4.

Now int_{-2}^2 -4-3*x^2+x^4 dx=-4x-x^3+(x^5)/5]_{x=-2}^{x=2}.

When x=2 this is -8-8+(32)/5=(-80+32)/5=-48/5. When x=-2, the value is 48/5, but we subtract it. The total answer seems to be -96/5.

P.S.: I "cheated". after I did it by hand and I had Maple check. In fact, I had Maple do it both dy dx AND dx dy. I got the same answer, which is above.

Done by the management after prompting by Mr. Hort on Mon, 25 Nov 2002 20:17:00

17. Show that (2xy2 -1)i + (6y+2x2y)j is a gradient vector field, and evaluate intGamma (2xy2 -1) dx + (6y+2x2y) dy where Gamma is the graph of $y= (cos x)10 from x=0 to x=pi/2.

In order to show that this is a gradient (conservative) vector field, we can find a potential=>the logic that leads to this conclusion is as follows: if grad F=P(X,Y)i+Q(X,Y)j, then the line integral of Pdx+Qdy=F(end)-F(start). By definition, F is a potential for P(X,Y)i+Q(X,Y)j(which is a gradient vector field). So if i can find a potential, i can verify that this is a gradient vector field.

a) By definition, grad F=dF/dX(X,Y)i+dF/dY(X,Y)j. Applying the definition to this problem gives dF/dX=(2*x*y^2-1) and dF/dY=(6*y+2*x^2*y). I can now integrate each of these individually. The first one then becomes F=x^2*y^2-x+C(y), where C(y) is a constant with respect to x, and can have a y in it. The second equation becomes F=3*y^2+x^2*y^2+C(x). Fortunately, I can easily combine these two equations: so i have found a potential: F(X,Y)=x^2*y^2-x+3*y^2. I have thus verified that (2*x*y^2-1)i+(6*y+2*x^2*y)j is a gradient vector field. [I can check this by taking the first partials of F]

b) Since F is a conservative vector field, the work around any closed curve on F with be 0--if the work done around any closed curve of F is zero, then the work done on any curve on F will depend only on the endpoints. Therefore, i can simply evaluate the function at the given points to solve the problem. At x=0, y=(cos(0)^10), y=1. At x=pi/2, y=0. Plugging these endpoints into the potential equation gives: F(end)= ((pi/2)^2*0^2-(pi/2)+3*0^2)= -pi/2. F(start)=(0^2*1^2-0+3*1^2)=3. Therefore, the work done around this curve(the line integral) is -(pi/2)-3.

Sent by Jason Tokayer on Sun, 24 Nov 2002 18:53:17

18. Use Green's Theorem to evaluate intT(x2y) dx - (xy2 dy, where T is the circle x2+y2=4.

Green's Theorem states that intCP dx + Q dy] = int intRdQ/dx - dP/dy) dA where C is the boundary of R.

For this problem P=x^2*y and Q=x*y^2. So dQ/dx=-y^2 and dP/dx=x^2.

By Green's Theorem the integral can be written as int[intR -1(x^2+y^2) dA where R is the inside of the circle of radius 2 centered at (0,0).

We use polar coordinates for this double integral. If we sub in the values x=r cos(w) and y=r sin(w), the integrand becomes -r^2. Now if we plug in the limits, we get inttheta=0theta=2*piintr=02-r^2 r dr d(theta). This is then evaluated as follows:

inttheta=0theta=2*pi(-1/4)*r^4 |r=0r=2 d(theta) = inttheta=0theta=2*pi(-4) d(theta)=-4*theta |theta=0theta=2*pi=-8*pi.

Sent by Nicholas Vander Valk on Mon, 18 Nov 2002 22:25:37

19. The plane curve C1 is a rectangle whose corners are (0,0), (5,0), (5,4), and (0,4), oriented in the usual (counterclockwise) fashion. The plane curve C2 is the circle of radius 1 whose center is (2,2), oriented in the usual (counterclockwise) fashion. Compute intC1 M dx + N dy if the following information is known:

  1. M and N are continuously differentiable functions of x and y in the region between the two curves. In that region, Mx = arctan(x3), My = 3y, Nx = 5, and Ny = cos(sqrt(y)).
  2. intC2M dx + N dy = 8. .

To find this line integral, we can use Green's Theorem on the area between to two curves, and then add the line integral along C_2 to the result. In order to use Green's Theorem, first picture a curve that travels counter-clockwise around C_1, at some point stops and travels inward to C_2, travels clockwise around C_2, then travels back along the same line joining the two original curves, and then completes the original path of C_1. (I know this is incredibly hard to see without an actual picture, but bear with me.) The resulting enclosed area is the region between the two original curves. According to Green's Theorem, since M and N are continuous and differentiable on this region the line integral of Mdx + Ndy along the strange curve described above is equal to int int_R N_x - M_y dA.

int int_R N_x - M_y dA = int int_R 5-3y dA.

Since 5-3y is continuous everywhere in the xy-plane, it doesn't matter what M and N are: we can integrate over this region by integrating over the entire rectangle followed by subtracting the integral over the cutout circle region. The rectangle integral becomes int(0,5) int(0,4) 5-3y dy dx = -20. (Boring parts of calculations omitted. See me if you have questions.) The equation for the boundary of the circle area is (x-2)^2 + (y-2)^2 = 1. I changed the variables to (u,v) such that (x,y) = (u+2, v+2). The integrand changed from 5-3y to -1-3v, and the boundary for the circle is now u^2+v^2 = 1. The Jacobian for this change is 1. Another change of variables (to polar coordinates) transforms (u,v) to (r*cos(theta), r*sin(theta)). The Jacobian is r. The new integral is int(0,2PI) int(0,1) -r-3(r^2)sin(theta) dr d(theta) = -PI (Once again, boring computations omitted.) So the double integral over the original region is the integral over the rectangle minus the integral over the circle, or (-20)-(-PI) = PI-20. By Green's Theorem, this is the value of the line integral over the weird-looking curve I tried to describe earlier. Adding this to the value of the line integral over C_2, (value = 8) we find that the line integral over C_1 of M dx + N dy = PI-12.

Sent by Joe Walsh on Thu, 21 Nov 2002 15:48:46

20. a) Compute int01int0xint0y2xy2z3  dz dy dx.

b) Write this iterated integral in ``dx dy dz'' order. You may want to begin by sketching the volume over which the triple integral is evaluated. You are not asked to evaluate the ``dx dy dz'' result.


 int_0^1 int_0^x int_0^(y^2) xy^2z^3 dz dy dx
=int_0^1 int_0^x (xy^2/4)z^4]z=0 to z=y^2) dy) dx
=int_0^1 int_0^x) (xy^10/4) dy) dx
=int_0^1 (x/44)y^11]y=0 to y=x) dx
=int_0^1 (1/44)x^12) dx

By staring at the picture of the integral limits, we see that

int_0^1 (int_0^x (int_0^(y^2) F(x,y,z) dz) dy) dx=int_0^1 ( int_sqrt(z)^1 (int_x^1 F(x,y,z) dx) dy) dz

Note from the management: This problem was done with the help of several somewhat confusing pictures at the review session. I regret that I will probably not have time today to create pictures to put here. At the review session, we evaluated the answer in b) and got something resembling the answer to a).

Sent by Siwei Zhu on Mon, 25 Nov 2002 04:18:35