Assignment 32

course MTH 272

......!!!!!!!!...................................Applied Calculus II

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Asst # 32

07-12-2006

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12:03:08

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**** Query problem 7.8.6 integrate (x^2+y^2) with respect to y from x^2 to `sqrt(x)

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12:11:37

integral of (x^2+y^2) with respect to y from x^2 to `sqrt(x) =

= (x^2)y + (1/3)y^3 from [sqrt(x), x^2]

= [ (x^2)(x^(1/2)) + (1/3)(x^(1/2))^3) ] - [(x^2)(x^2) + (1/3)(x^2)^3 ]

= x^(5/2) + (1/3)(x^(3/2)) - x^4 - (1/3)(x^6)

= x^(3/2) [ x + 1/3 - x^(5/2) - (1/3)x^(9/2) ]

Very good.

You factored x^(3/2) out of the final expression, which is equivalent to the final expression in the following, which I've included just for comparison:

** The y derivative of x^2 is 0, so the y derivative of this expression is just y^2, not x^2 + y^2. **

** An antiderivative would be x^2 y + y^3 / 3.

Evaluating this antiderivative at endpoints y = x^2 and y = `sqrt(x) we get

[ x^2 * `sqrt(x) + (`sqrt(x) ) ^ 3 / 3 ] - [ x^2 * x^2 + (x^2)^3 / 3 ] =

x^(5/2) + x^(3/2) / 3 - ( x^4 + x^6 / 3) =

x^(5/2) + x^(3/2) / 3 - x^4 - x^6 / 3.

This can be simplified in various ways, but the most standard form is just decreasing powers of x:

- x^6/3 - x^4 + x^(5/2) + x^(3/2)/3. **

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12:11:38

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**** What is your result?

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12:11:59

x^(5/2) + (1/3)(x^(3/2)) - x^4 - (1/3)(x^6)

or, factoring out x^(3/2):

= x^(3/2) [ x + 1/3 - x^(5/2) - (1/3)x^(9/2) ]

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12:12:00

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**** What was your antiderivative?

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12:12:18

(x^2)y + (1/3)y^3 from [sqrt(x), x^2]

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12:12:19

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**** Query problem 7.8.14 integrate `sqrt(1-x^2) from 0 to x wrt y then from 0 to 1 wrt x

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12:21:44

Integral of `sqrt(1-x^2) dy from 0 to x :

You're taking the derivative with respect to y, but there is no y term in the integral. x is just a constant, so the whole term in the integrand is a constant, making the derivative wrt y :

(1 - x^2) ^(1/2) y

From 0 to x:

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

= x(1-x^2)^(1/2)

integral of (x(1-x^2)^(1/2)) dx from 0 to 1:

u = 1 - x^2

du = -2x dx

-(1/2) integral of (u^(1/2)) =

= -(1/2)*(2/3)u^(3/2)

= (-1/3)u^(3/2)

= (-1/3) (1-x^2)^(3/2) from (0 to 1)

= [(-1/3)(1-1)^(3/2)] - [(-1/3)(1-0)^(3/2)]

= 0 - (-1/3 (1))

= 1/3

The limits on this integral are 0 and x.

The result is then to be integrated with respect to x.

The solution:

** Since the function has no y dependence the whole expression is treated as a constant and an antiderivative with respect to y is just `sqrt(1 - x^2) * y. The definite integral from 0 to x is therefore `sqrt(1 - x^2) * x - `sqrt(1 - x^2) * 0 = x `sqrt(1-x^2).

Using a table you will find that an antiderivative with respect to x is arcsin(x)/2 + x `sqrt(1 - x^2) / 2. I don't think your text has dealt with the arcsin function, but arcsin(1) = `pi/2 and arcsin(0) = 0. So substituting limits we get

arcsin(1)/2 + 1•`sqrt(1 - 1^2)/2 - [arcsin(0)/2 + 0•`sqrt(1 - 0^2)/2 ] = `pi/4

(all terms except the first give you zero).

Note that y = `sqrt(1-x^2) from x = -1 to 1 is the graph of a circle of radius 1. The integral from x = 0 to x = 1 gives a quarter-circle of radius 1, which has area 1/4 * `pi r^2 = 1/4 * `pi * 1^1 = `pi/4, in agreement with the second integral. **

Note that the region is just 1/4 of the circle of radius 1 centered at the origin. This circle has area pi r^2 = pi * 1^2 = pi, and 1/4 of the circle has area pi/4, in agreement with the integral.

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12:21:45

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**** What is your result?

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12:21:53

1/3

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12:21:54

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**** With respect to which variable of integration did you first integrate?

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12:22:08

I first integrated with respect to y

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12:22:09

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**** What was the antiderivative of your first integration and what was the definite integral you obtained when you substituted the limits?

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12:22:47

derivative wrt y :

(1 - x^2) ^(1/2) y

From 0 to x:

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

= x(1-x^2)^(1/2)

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12:22:48

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**** What did you get when you integrated this expression with respect to the remaining variable of integration?

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12:23:23

integral of (x(1-x^2)^(1/2)) dx from 0 to 1:

u = 1 - x^2

du = -2x dx

-(1/2) integral of (u^(1/2)) =

= -(1/2)*(2/3)u^(3/2)

= (-1/3)u^(3/2)

= (-1/3) (1-x^2)^(3/2) from (0 to 1)

= [(-1/3)(1-1)^(3/2)] - [(-1/3)(1-0)^(3/2)]

= 0 - (-1/3 (1))

= 1/3

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12:23:24

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**** Query problem 7.8.32 sketch region and reverse order of integration for integral of 1 from 0 to 4-y^2, then from -2 to 2.

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12:25:26

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12:25:26

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**** What are the limits, inner limit first, when the order of integration was changed?

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12:33:54

The integral of dx from [0, 4-y^2]

= x from [ 0, (4-y^2)]

(4-y^2) - 0

= 4-(y^2)

Integral of (4-y^2)dy from (-2,2) =

= 4y - (1/3)y^3

= [4(2) - 1/3(8)] - [(4)(-2) - 1/3(-8)]

= 8 - (8/3) + 8 - (8/3)

= 16 - (16/3)

= 32/3

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12:33:54

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**** Describe the region in the xy plane whose area is given by the given integral.

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12:38:30

I'm really not sure how to do this.

I'm guessing the region in the plane is the area that is bounded by the equations y= 4-y^2 and y=0, but as for what this area looks like, I'm not sure.

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12:38:31

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**** Explain how you obtained the limits for the integral when the order of integration was changed.

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12:42:57

I first found the integral of dx from [0, 4-y^2]

= x from [ 0, (4-y^2)]

(4-y^2) - 0

= 4-(y^2)

I then found the integral of (4-y^2)dy from (-2,2) =

= 4y - (1/3)y^3

= [4(2) - 1/3(8)] - [(4)(-2) - 1/3(-8)]

= 8 - (8/3) + 8 - (8/3)

= 16 - (16/3)

= 32/3

I don't think this is right because I didn't change the order of integration from the order that was given in the book. When I did try to change the order, I got a different answer.

I did: integral from [0, 4-y^2] of the integral from [-2,2] dy dx:

integral of dy = x

from [-2.2]=

2 - -2

= 4

integral of 4dx

= 4x

from [0, 4-y^2]:

= 4(4-y^2) - 4(0)

= 16 - 4y^2

This is not the same as my first answer, so I'm not sure where I'm going wrong.

** We first integrate 1 from 0 to 4 - y^2, with respect to x. An antiderivative of 1 with respect to x is x; substituting the limits 0 and 4 - y^2 we get [ 4 - y^2 ] - 0, or just 4 - y^2.

Then integrating from y = -2 to 2, with respect to y, the antiderivative of 4 - y^2 is 4 y - y^3 / 3; substituting the limits -2 and 2 we get [ 4 * 2 - 2^3 / 3 ] - [ 4 * (-2) - (-2)^3 / 3 ] = 8 - 8/3 - [ -8 - (-8 / 3) ] = 16 - 16/3 = 32 /3.

Reversing the order of integration we integrate 1 from -`sqrt(x) to `sqrt(x), with respect to y. Anantiderivative of 1 with respect to y is y; substituting the limits we get `sqrt(x) - (-`sqrt(x) ) = 2 `sqrt(x).

We next integrate with respect to x, from 0 to 4. Antiderivative of 2 `sqrt(x) with respect to x is 2 * [ 2/3 x^(3/2) ] = 4/3 x^(3/2). Evaluating between limits 0 and 4 we get [ 4/3 * 4^(3/2) ] - [4/3 * 0^(3/2) ] = 32/3.

The two integrals are equal, as must be the case since they both represent the area of the same region. **

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12:42:57

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**** What did you get when you evaluated the original integral?

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12:43:10

With the original integral, I got 32/3

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12:43:10

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**** What did you get when you evaluated the integral with the order of integration reversed?

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12:43:24

With the order of integration reversed, I got 16-4y^2

The specifics of the region and the reversal of the integration:

** If x = 4 - y^2, then:

If y = -2 or 2 we have x = 0. So the points (0, 2) and (0, -2) lie on the graph.

If y = 0 we have x = 4, so the point (4, 0) lies on the graph.

Somehow, therefore, the graph goes from (0, 2) to (4, 0) to (0, -2).

The graph is quadratic in y, linear in x. So the graph is a parabola opening in the y direction. This implies that (4, 0) is the vertex, with the parabola opening to the left.

For each y, the value of x goes from x = 0 to x = 4 - y^2. So the region includes the region from the y axis (x = 0) to the parabola x = 4 - y^2.

If we specify a value of x, thinking of a point on the x axis, the value of y in the region ranges from the value on the lower part of the parabola to the value on the higher part. The corresponding values of y are found by solving x = 4 - y^2 for y, obtaining y = +- `sqrt(4 - x). The leftmost value of x for the region is 0; the rightmost possible value is 4. So the region would be described by

0 < x < 4

-`sqrt(x) < y < `sqrt(x),

where the < signs really should be less than or equal signs. **

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12:43:24

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**** Why should both integrals give you the same result?

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12:53:06

Oh, hold on, I found out what I did wrong in the last part, but I'm still not really sure how to fix it.

The bounds after switching the order should be completely different than the bounds before switching-- I was just switching the order, but not the actual values of the bounds themselves.

I think the second bounds should be [0, sqrt(x-4)] because if you set x = 4 - y^2 and solve for y, that is what you get. However, I'm not sure about the first bounds. I thought it may be [0,4], but this still doesn't give me the correct answer.

I still don't know why both integrals should give the same result, because when I try it, they don't.

I think you'll see it from my note. Let me know if not.

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12:53:06

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**** Query problem 7.8.40 Area beneath curve 1/`sqrt(x-1) for 2 <= x <= 5

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12:56:04

This is the same as the integral of (x-1)^(-1/2)dx, which equals 2(x-1)^(1/2) from [2,5]:

2(5-1)^(1/2) - 2(2-1)^(1/2)

= 2 sqrt(4) - 2 sqrt(1)

= 2(2) - 2

= 4-2

= 2

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12:56:04

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**** What are the limits and the order of integration for the double integral you used to find the area?

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12:58:13

I think I forgot to carry out the problem with a second derivative in my last response. I got:

2(5-1)^(1/2) - 2(2-1)^(1/2)

= 2 sqrt(4) - 2 sqrt(1)

= 2(2) - 2

= 4-2

= 2

Then, I should take the integral of 2 wrt y over the interval [0,1], because this is the y-bounds of the area.

= 2y

= 2(1) - 2(0)

= 2 - 0

= 2

So 2 is still the answer.

The order of integration I used was wrt x and then wrt y, and the limits were [2,5] wrt x and [0,1] wrt y.

Good.

This can also be regarded as a double integral, which is a more general form and can be extended to more complex situations (e.g., calculating moments, etc.):

** The region goes from x = 2 to x = 5, and from the x axis to the curve y = 1 / `sqrt(x-1).

To find the area you integrate 1 over the region.

The integral would be from y = 0 to y = 1/`sqrt(x-1) with respect to y (this is the inside integral), then from x = 2 to x = 5 with respect to x (the outside integral).

Inside integral: antiderivative of 1 is y; substituting limits we have [1/`sqrt(x-1)] - 0 = 1 / `sqrt(x-1).

Outside integral is now of 1/`sqrt(x-1) with respect to x, from x = 2 to x = 5. Antiderivative is 2 `sqrt(x-1) which can be found thru the u substitution u = x - 1. Evaluating at limits we have 2 `sqrt(5-1) - 2 `sqrt(2-1) = 2 `sqrt(4) - 2 `sqrt(1) = 2. **

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12:58:14

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**** What is the area?

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12:58:17

2

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12:58:18

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**** Query problem 7.8.44 area bounded by xy=9, y=x, y=0, x=9

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13:00:13

I have absolutely no idea how to solve this problem. I'm not sure what the graph even looks like, much less how to solve for the area.

xy=9 is the same as y = 9/x, so now I think I can graph it, but I'm not sure where the x=9 comes in, or what integral I'm supposed to be solving for, or what the boundaries are.

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13:00:13

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**** What is the area of the region?

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13:00:29

I really don't know how to solve this.

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13:00:29

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**** Describe the region in detail.

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13:06:37

I graphed y= 9/x, y=x, and y=0. The graphs y=9/x and y=x intersect at (3,3)

9/x = x --> x^2 = 9 --> x=3.

So maybe the x-bounds are [0,3], and the y-bound might also be [0,3].

I at least want to attempt this problem, so I'll set both bounds equal to [0,3] and take the integral of 9 dx dy.

9x over [0,3]

= 9(3) - 9(0)

= 27

integral of 27 dy

= 27y from [0,3]

= 27(3) - 27(0)

= 81 - 0

= 81

I got the area to be 81.

Good attempt. However we need to integrate over more than one region.

xy = 9 means y = 9/x; graph decreases at decreasing rate from asymptote at y axis toward asymptote at x axis.

y = x is straight line at 45 deg to x axis.

y = x intersects xy = 9 when x^2 = 9 or x = 3 (x^2 = 9 found by substituting y = x into xy = 9).

Graph is bounded by y = 0, which is the x axis. Also by x = 9. So the region lies above the x axis, below the y = x curve from x = 0 to x = 3 ( where y = x is lower than y = 9 / x) and below the xy = 9 curve from x = 3 to x = 9.

So we first integrate 1 from x = 0 to x = 3 (outer integral) and from y = 0 to y = x (inner integral). Inner integral gives antiderivative y; substituting limits we get x - 0 = x. Integrating x from x = 0 to x = 3 we get antiderivative x^2 / 2; substituting limits yields 3^2 / 2 - 0^2 / 2 = 9/2 (which is the area of the triangle formed between x = 0 and x = 3).

Then we integrate 1 from x = 3 to x = 9 (outer integral) and from y = 0 to y = 9 / x (inner integral). Inner integral gives antiderivative y; substituting limits we get 9/x - 0 = 9/x. Integrating x from x = 3 to x = 9 we get antiderivative 9 ln | x | ; substituting limits yields 9 ln | 9 | - 9 ln | 3 | = 9 ln | 9/3 | = 9 ln | 3 |.

Total area is therefore 9 ln | 3 | + 9/2 = 9 ( ln | 3 | + 1/2 ).

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13:06:38

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**** Did you have to use one double integral or two to find the area?

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13:07:06

I used two double integrals, first 9 wrt x and the 27 wrt y, and both with the limits [0,3].

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13:07:07

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**** What was the order of integration and what were the limits you used for each integral?

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13:07:50

The limits I used for both integrals are [0,3] , and I first took the integral of 9dx, and by taking the definite integral of that, I got 27, which I then took the integral of wrt y.

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13:07:51

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**** Query Add comments on any surprises or insights you experienced as a result of this assignment.

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13:08:57

I had a very hard time understanding this assignment. I really don't feel strong regarding the material in this section, and really, in this whole chapter.

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You made a good attempt on this assignment. See my notes and be sure to let me know if you have additional questions.