query 121

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course Mth 277

4/13 2

Question: `q001. Evaluate the double integral over R with respect to A of x/4 where R: 1 <= x <= 8, -2 <= y <= 0.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Int( Int(x/4 dy, -2,0) dx,1,8)

Int(x/4 dy, -2,0) = xy/4, evaluating at our limits on the integral

= (x*0/4) - (x*-2/4) = 2x/4 = x/2

Int(x/2 dx, 1,8) = x^2,1,8 = 8^2 - 1^2 = 63

R = 63

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Given Solution:

The region runs from x = 1 to x = 8. For each value of x, y runs from -2 to 0.

So the integral is

int( (int( x / 4 dy, -2, 0) dx, 1, 8) ).

The inner integral, then the complete double integral, are written

The inner integral is with respect to y, so x^2 / 4 is treated as a constant, and an antiderivative is just x / 4 * y (note that y is an antiderivative of 1). Evaluating between limits -2 and 0 we get (x/4 * 0) - (x/4 * (-2)) = x / 2.

The outer integral is then int(x/2 dx, 1, 8), written

This integral is easily evaluated. We get 63 / 4.

The integral could be interpreted as the mass of a rectangular of dimensions 7 units by 2 units, with a density x / 4 that varies linearly from 1/2 to 2 as we move along the 7 unit length. The average density would be 9/8, the area is 14, so the mass is 9/8 * 14 = 63/4.

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Self-critique (if necessary):

I don’t know what I was thinking, I saw 2x instead of x/2 but I understand

Int (x/2 dx,1,8) = (1/2)*(x^2/2) = (x^2/4)

Then evaluated at x=1 and x=8.

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Self-critique rating:3

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Question: `q002: Compute the double integral over R with respect to A of cos(x+y) where R: 0 <= x <= pi/2, 0 <= y <= pi/2.

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Your solution:

Int( Int(cos(x+y) dy, 0,pi/2) dx,0,`pi/2)

Int(cos(x+y) dy, 0,pi/2) = sin(x+y), evaluating at our limits on the integral

= sin(x+`pi/2) - sin( x), not a constant value rather a fn of x

Int(sin(x+`pi/2) - sin( x) dx, 0,`pi/2) = -cos(x + `pi/2) + cos(x) evaluating at our limits on the integral

= [cos(`pi/2) - cos(`pi/2 + `pi/2)] - [cos(0) - cos(`pi/2)] = [0 - (-1)] - [1-0] = 0

R = 0

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Given Solution:

Choosing to integrate first with respect to y, the inner integral is int(cos(x+y) dy, 0, pi/2), which is equal to sin(x+y) evaluated between y = 0 and pi/2.

The resulting inner integral is therefore sin(x + pi/2) - sin(x).

The entire double integral is

int ( int(cos(x+y) dy, 0, pi/2) dx, 0, pi/2) = int((sin(x+pi/2) - sin(x)) dx, 0, pi/2). Using -cos(x + pi/2) and - cos(x) as antiderivatives and evalutating between x = 0 and pi/2, we get -cos(pi) - (-cos(pi/2) ) - cos(pi/2) - cos(0) = -(-1) - 0 - 0 - 1 = 0

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Self-critique (if necessary):

Just to double check is it appropriate to say the resulting value of the double integral is R??

Ex) R = 0, for the last problem

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Self-critique rating:

@& R is the region. It's the double integral of the given function over the region R.*@

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Question: `q003. Find the volume of the solid bounded above by the graph of z = f(x,y) and below by the xy-plane when f(x,y) = ye^(xy) and R: 0 <= x <= ln2, 0 <= y <= 1.

We integrate the vertical distance from the plane to the graph, with respect to area, over the given region.

Choosing to integrate first with respect to x (we could as well have chosen y first), we get inner integral

int( y e^(x y) dy, 0, 1).

Then integrating this with respect to x we obtain our double integral

int(.int( y e^(x y) dy, 0, 1),x,0, ln(2))

For the inner integral x is treated as a constant.

We must integrate y e^(x y) with respect to y, for which we use integration by parts with u = y and dv = e^(x y) dy.

We get du = dy and v = 1/x e^(x y), so that our antiderivative is

u v - int(v du) = y / x e^(x y) - 1/x int ( e^(x y) dy) = y/x e^(x y) - 1/x^2 e^(x y).

Evaluating between y limits 0 and 1 we get 1/x e^x - 1/x^2 e^x - (0/x * e^0 - 1 / x^2 * e^0 ) = e^x ( 1/x - 1/x^2) + 1 / x^2.

Unfortunately e^x / x cannot be integrated in closed form.

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Your solution:

I believe the solution is included in the question but we would have something we can visualize as the volume of a given object that follows the equation

Int( Int(ye^(xy) dy, 0,1) dx,0,ln(2))

Int(ye^(xy) dy, 0,1) = int. by parts, u=y, du = 1, dv = e^(xy), v = 1/xe^(xy)

Int(ye^(xy) dy, 0,1) = y(e^(xy)/x) - Int(1*(e^(xy)/x) dy)

= (y*e^(xy))/x - (e^(xy))/x^2, evaluating at integral limits

[(1*e^(1*x))/x - (e^(1*x))/x^2] - [(0*e^(0*x))/x - (e^(0*x))/x^2]

= e^(x)/x - e^(x)/x^2 + 1/x^2, now evaluating our outer integral

Int(e^(x)/x - e^(x)/x^2 + 1/x^2 dx,0,ln(2)) = ????So the is no way to find the volume of this problem exactly, only an approximation can be made??????

@& That's right. The approximation can be as close as you wish to the actual value, whatever that is, but you can never get the actual value (e.g., you can know you're within .000000001 of the actual value, but you can't know exactly how close you are; if you did you'd know the actual value).*@

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Given Solution:

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Self-critique (if necessary):

What exactly does that mean in close form is that like saying by taking an exact integral? i remember reading about it before but don’t recall definition of closed form.

@& Closed form means you can write down a specific number, or a specific function in terms of sines, cosines, exponentials, polynomials, etc.. that exactly expresses the result.

You can integrate e^x / x just fine as a power series, but that's an infinite series, not a closed form.*@

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Question: `q004. Use a grid with 16 cells to approximate the volume under the surface of 4 - x^2 - y^2, above the rectangle R: 0 <= x <=1, 0 <= y <= 1.

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Your solution:

We will have a graph that has rectangle covering area centered within the vertices (0,0),(1,0),(1,1),(0,1)

or we can visualize drawing line starting at x axis from x=1 to y=1, then another line back to y axis to give us our rectangle. Now we would draw lines from x axis to our line at the top of our rectangle( y=0)

at x=1/4,1/2,3/4 and from y axis to other side of rectangle(x=1) at y=1/4,1/2,3/4. These lines give us our 16 cells. Each cell is a square of width(1/4)unit and length (1/4)unit and with an altitude of alt =4 - x^2 - y^2

So that each cell has volume = c.s area * ht = (x*y)*( 4 - x^2 - y^2), where x and y are given by the point at which we are looking at in our rectangle.

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Given Solution:

A grid with 16 cells could correspond to a partition of the x interval [0, 1] into subintervals [0, .25], [.25, .50], [.50, .75], [.75, 1.00], and the y interval [0, 1] also into subintervals [0, .25], [.25, .50], [.50, .75], [.75, 1.00].

We might number the x intervals with index i, which takes values 1, 2, 3, and 4, and the y intervals with index j, also taking values 1, 2, 3, 4.

With this numbering the region 0 <= x <= .25, .50 <= y <= .75 would be the i = 1, j = 3 interval. Its length in the x direction is `dx_1 = .25, its width in the y direction is `dy_3 = .25, so its area is `dA_1,3 = .25 * .25 = .0625. (more generally the area of the region defined by the ith x interval and the jth y interval would be denoted `dA_i,j = `dx_i * `dy_j).

For sample points we could use x coordinates .2, .3, .6 and .9, with y coordinates .1, .4, .7 and .8. Each sample coordinate is within the corresponding interval (e.g., using 'hat' to denote sample points, x_1_hat = .2 is in the i = 1 interval of the x partition, the interval [0, .25], and the j = 3 interval [.50, .75] contains the point y_3_hat = .7).

Using this scheme the point (x_1_hat, y_3_hat) = (.2, .7) would be the sample point of the i,j = 1, 3 region of the plane, which has area A_i,j = A_1,3 = .0625.

Our sample coordinates are not completely random within the various intervals, but they illustrate how any point in an interval can serve as a sample point.

Thus our partition divides the region up into 16 squares, indexed 1, 1 throug 1,4. Each square has dimension `dx = 1/4 by `dy = 1/4, and hence has total area 1/4 * 1/4 = 1/16.

The resulting sample points, values of z = 4 - x^2 - y^2, and contributions to the integral (calculated by value * area increment) are

sample point ( 0.2 , 0.1 ); evaluating function: z = 3.95 . Value * area increment = 0.246875

sample point ( 0.3 , 0.1 ); evaluating function: z = 3.9 . Value * area increment = 0.24375

sample point ( 0.6 , 0.1 ); evaluating function: z = 3.63 . Value * area increment = 0.226875

sample point ( 0.9 , 0.1 ); evaluating function: z = 3.18 . Value * area increment = 0.19875

sample point ( 0.2 , 0.4 ); evaluating function: z = 3.8 . Value * area increment = 0.2375

sample point ( 0.3 , 0.4 ); evaluating function: z = 3.75 . Value * area increment = 0.234375

sample point ( 0.6 , 0.4 ); evaluating function: z = 3.48 . Value * area increment = 0.2175

sample point ( 0.9 , 0.4 ); evaluating function: z = 3.03 . Value * area increment = 0.189375

sample point ( 0.2 , 0.7 ); evaluating function: z = 3.47 . Value * area increment = 0.216875

sample point ( 0.3 , 0.7 ); evaluating function: z = 3.42 . Value * area increment = 0.21375

sample point ( 0.6 , 0.7 ); evaluating function: z = 3.15 . Value * area increment = 0.196875

sample point ( 0.9 , 0.7 ); evaluating function: z = 2.7 . Value * area increment = 0.16875

sample point ( 0.2 , 0.9 ); evaluating function: z = 3.15 . Value * area increment = 0.196875

sample point ( 0.3 , 0.9 ); evaluating function: z = 3.1 . Value * area increment = 0.19375

sample point ( 0.6 , 0.9 ); evaluating function: z = 2.83 . Value * area increment = 0.176875

sample point ( 0.9 , 0.9 ); evaluating function: z = 2.38 . Value * area increment = 0.14875

We add the contributions of the 16 area increments to get our total approximation: 3.3075

Naturally a different selection of sample points would be expected to provide a different total. However there isn't that much room to move around in any square, so our sample point couldn't be too far from the sample point chosen here. No matter where the sample points are chosen, we would probably expect the final result to be reasonably close to the 3.3 obtained here. If the sample points are chosen at random (vs., for example, choosing a point very close to the upper right-hand corner of each interval, which is far from a random choice), we would expect to be closer to 3.3 than if we had made a completely random choice of the sample point.

Note: the exact value of the integral is 10/3 = 3.33, quite close to the value obtained here.

Note: The integral approximated here is

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Self-critique (if necessary):

I had a sort of right answer, but the problem was I still have some issues visualizing the surface or whatever. So the surface is not the altitude of the cell in the x,y plane

@& The surface f(x, y) = z = 4 - x^2 - y^2 is a paraboloid with circular cross-sections parallel to the x-y plane, vertex at (4, 0, 0). The intersections of this surface with the xz plane is the parabola z = 4 - x^2; the intersection with the y z plane is z = 4 - y^2.

The integral over this region will be the part of the surface lying above the square 0 <= x <= 1, 0 <= y <= 1.

The surface has an altitude above every point (x, y), obtained by plugging the values of x and y into the function.

There is an average altitude above every cell. If you multiply the average altitude by the area of the cell, you get the volume beneath the surface and above the cell.

If the cells are small then the altitude doesn't vary much from one point of the cell to another, and any point within the cell provides a reasonable sample point. Plugging in the coordinates of the sample point provides a good approximation to the average altitude of the surface above the cell.

As the dimensions of the cell approach zero, the difference between the altitude at the sample point and the average altitude approaches zero, and the sum of all the corresponding volumes approaches the exact volume.*@

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Question: `q005. Let R be a rectangular region within the boundary of a certain city defined by R: -2 <= x <= 3, -1 <= y <= 1. The units are in miles and (0,0) is the city center. Assume the population density is 13*e^(-0.08 * r) thousand people per square mile and r = sqrt(x^2 + y^2). Give the double integral which will model the total population of this region. Do not, however, solve the integral.

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Your solution:

R = Int(Int(13*e^(-0.08 * `sqrt(x^2+y^2)) dy,-1,1) dx,-2,3)

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Given Solution:

If we partition the x and y intervals in the usual manner, so that the typical region has dimensions `dx_i by `dy_i and contains sample point (x_i_hat, y_j_hat), then the region has area A_ij = `dx_i * `dy_j. The population density is given in terms of r, the distance from the point (0, 0). The point (x_i_hat, y_j_hat) lies at distance sqrt( (x_i_hat)^2 + (y_j_hat)^2 ) from the center, so the population density is 13 e^(-.08 r) = 13 e^(-.08 * sqrt((x_i_hat)^2 + (y_j_hat)^2 ) ).

If we sum over all such regions and take the limit as the `dx and `dy increments approach zero, our sum approaches the integral

int ( int(13 e^(-.08 sqrt(x^2 + y^2) ) dy, y from -1 to 1) dx, x from -2 to 3). In standard notation:

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Self-critique (if necessary):

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@& Good questions. Check my notes.*@