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 int₀¹int_u¹sqrt{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

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

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

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

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

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

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.

y+z=2Lx
x+z=2Ly
x+y=2Lz
x^2+y^2+z^2=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):
(1/2)y+(1/2)z+z=2y
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.

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

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

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

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

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

1. M and N are continuously differentiable functions of x and y in the region between the two curves. In that region, M_x = arctan(x³), M_y = 3y, N_x = 5, and N_y = cos(sqrt(y)).
2. int_C2M 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

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.

a)

 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

=1/(44*13).

b)
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