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course Mth 277
November 18 around 11:45am. I had trouble understanding section 11.7, so I just started working on chapter 12 instead and maybe we can talk about 11.7 next week. I find chapter 12 to be pretty understandable.
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 (x/4 dx) from 1 to 8
x^2/8 from 1 to 8
8-(1/8)=63/8
Int (63/8) from -2 to 0
63/8x from -2 to 0
0+63/4
=63/4
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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):
OK!
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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 (cos (x+y)) dx from 0 to pi/2
=sin (x+y) from 0 to pi/2
=sin (pi/2+y)-sin(y)
Int (sin (pi/2+y)-sin(y)) dy from 0 to pi/2
=cos (y) + sin (y) from 0 to pi/2
= approx. 0.03
I integrated first with respect to x, will this result in different answers? Or should it come out to be the same?
The region is the same either way, as is the function being integrated, so the integral should come out the same both ways.
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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):
Ended up with a slightly different answer. What did I do wrong?
Your integration was right on.
Do be careful about the numbers. Your solution looks like a calculator was involved, and for this problem it shouldn't have been.
You might well have used an approximation of pi rather than an exact value.
cos(0) = 1, cos( pi/2) = 0, sin(0) = 0, sin(pi/2) = 1. These are exact values.
You should never rely on the calculator when you can express exact values.
If you have to use the calculator, then when you use pi, use pi and not, say, 3.14. There is a pi button on almost any calculator.
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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.
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Your solution:
Int (ye^(xy) dx) from 0 to ln 2
e^(xy) from 0 to ln 2
=e^(ln 2*y)-e^0
=2^(y)-1
Int (2^(y)-1 dy) from 0 to 1
=[2^(y)/ln 2]-y from 0 to 1
= approx. 1.885
This is asking for the volume of the solid, did I even need to do what I did or do I just have another step to go?
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Given Solution:
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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Self-critique (if necessary):
OK, I understand now.
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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:
Int (4-x^2-y^2 dx) from 0 to 1
=(-x^3/3)+x(4-y^2) from 0 to 1
=(11/3)-y^2
Int (11/3-y^2 dy) from 0 to 1
=(11y/3)-(y^3-3) from 0 to 1
=10/3
What does the 16 cells have to do with this problem?
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Given Solution:
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Self-critique (if necessary):
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
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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:
I’m not sure about this but I’m going to attempt it:
Int Int (13*e^(-0.08*sqrt (x^2+y^2))dx dy)
dx goes from -2 to 3
dy goes from -1 to 1
However, I feel like dR should be in it somewhere…
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
If we partition the x and y intervals in the usual manner, so that the typical region has dimensions `dx_i by `dy_i and contains sample point (x_i_hat, y_j_hat), then the region has area A_ij = `dx_i * `dy_j. The population density is given in terms of r, the distance from the point (0, 0). The point (x_i_hat, y_j_hat) lies at distance sqrt( (x_i_hat)^2 + (y_j_hat)^2 ) from the center, so the population density is 13 e^(-.08 r) = 13 e^(-.08 * sqrt((x_i_hat)^2 + (y_j_hat)^2 ) ).
If we sum over all such regions and take the limit as the `dx and `dy increments approach zero, our sum approaches the integral
int ( int(13 e^(-.08 sqrt(x^2 + y^2) ) dy, y from -1 to 1) dx, x from -2 to 3).
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self-critique rating #$&*:
Good start. I've inserted solutions to the last two into this document. Check the solutions as now posted in the original document, where the formatting is probably better.
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