32 open query

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course Mth 272

11/18 10:25 pm EST

032.

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Question: `qQuery problem 7.8.6 integrate (x^2+y^2) with respect to y from x^2 to `sqrt(x)

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Your solution:

Fy(x,y) = f(x,y) + C(x)

(x^2+y^2)

= x^2y + y^3/3

[x^2y + y^3/3] b = [sqrt]x ; a = x^2

[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)

Rewriting:

-x^6/3 – x^4 + x^5/2 – x^3/2 / 3

confidence rating #$&*: 3

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Given Solution:

`a 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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Self-critique (if necessary):

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self-critique rating #$&*: ‘OK’ -Clear

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Question: `qQuery 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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Your solution:

“`a The limits on the first integral are 0 and x.

The result of the first integral 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.”

confidence rating #$&*: 0

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Given Solution:

`a The limits on the first integral are 0 and x.

The result of the first integral 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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Self-critique (if necessary):

#### Is it possible for much more elaboration on this one? I see that we use with respect to y, would this be the inner integral, and x the outer. However, had some issues rewriting to solve for the integration here; I’m having major issues with moving forward on this particular problem.

The integral for x in the given solution is completely in error.

An antiderivative of x sqrt( 1 - x^2 ) is easily found by letting u = 1 - x^2, so that du = -2 x dx and x dx = -du / 2.

The integrand becomes -sqrt(u), with antiderivative -2/3 u^(3/2). Substituting 1 - x^2 for u we get antiderivative -2/3 (1 - x^2)^(3/2). Between x = 0 and x = 1 this expression changes from -2/3 to 0, a change of 2/3.

The definite integral is therefore 2/3.

In the solution as previously given the function of integration was accidentally confused with the boundary on the region. I do apologize for the confusion.

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self-critique rating #$&*: 2

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Question: `qQuery 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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Your solution:

[int.] 1 ; b = 4-y^2 ; a = 0

[int.] x ; b ; a

= 4-y^2 – 0

= 4 – y^2

[int.] 4 – y^2 ; b = 2; a = -2

[4y = y^3/3] ; b = 2; a = -2

= (8 – 8/3) – (-8 + 8/3)

= 16/3 + 16/3

= 32/3

confidence rating #$&*: 2

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Given Solution:

`a 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. An antiderivative 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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Self-critique (if necessary):

#### Feel pretty good with regard to this problem; however, I had some doubts relative to changing the order of integration. How was [sqrt] x formulated here, in order to present the equivalent result of the previous integral?

The given solution included an implicit change of variable, with 4 - x being replaced by x. This should have been explicitly stated.

However don't worry about that. I've rewritten the second part of the solution to avoid that confusion, and here's a copy:

The region of integration is -2 <= y <= 2, 0 <= x <= 4 - y^2. This region lies between the parabola x = 4 - y^2 and the y axis. The vertex is (4, 0), the parabola opens to the left, and it intercepts the y axis at the points (0, 2) and (0, -2).

For a given x value between 0 and 4, a vertical line through (x, 0) intersects this region at the points where x = 4 - y^2. The y coordinates of these points are easily found by solving x = 4 - y^2 for y, obtaining y = +-sqrt(4 - x). Thus the vertical line intersects the region between (x, -sqrt(4-x)) and (x, sqrt(4 - x)).

If x < 0, or if x > 4 the point (x, 0) lies outside the region.

Thus the region can be described by

0 <= x <= 4, -sqrt(4-x) <= y <= sqrt(4 - x)

and we integrate 1 first with respect to y, from limit -sqrt(4 - x) to sqrt( 4 - x), then from x = 0 to 4.

The 'inner integral' with respect to y yields antiderivative y, which evaluated between the limits -sqrt(4 - x) and sqrt( 4 - x) gives us 2 sqrt(4 - x).

This result is then integrated between x = 0 and x = 4. Our antiderivative is easily found to be -2 * (2/3 (4 - x)^(3/2)) = -4/3 (4 - x)^(3/2). Evaluating between the limits x = 0 and x = 4 we obtain -4/3 (4 - 4)^(3/2) - (-4/3 * (4 - 0)^(3/2)) = 32/3.

This result agrees with our first integral, as expected.

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self-critique rating #$&*: 2

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Question: `qQuery problem 7.8.40 Area beneath curve 1/`sqrt(x-1) for 2 <= x <= 5

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Your solution:

0 <= y <= 1/[sqrt] x-1

[int,] 5,2 ; [int.] 1/[sqrt]x-1, 0

1/[sqrt]x-1 – 0 = 1/[sqrt]x-1

2<= x <= 5

[int.] 5, 2 (1/[sqrt]x-1)

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

du = 1 dx

1 [int.] 5,2 ; u^-1/2 + 1 / -1/2 + 1

1 [int.] 5,2; u^1/2 / ½

[int.] 5,2; 2[sqrt]x-1

2[sqrt]4 – (2[sqrt]1) = 4-2

= 2

confidence rating #$&*: 3

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Given Solution:

`a 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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Self-critique (if necessary):

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self-critique rating #$&*: ‘OK’

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Question: `qQuery problem 7.8.44 area bounded by xy=9, y=x, y=0, x=9

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Your solution:

“`a 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 ). ”

confidence rating #$&*: 1

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Given Solution:

`a 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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Self-critique (if necessary):

#### I wasn’t sure how to start this one; therefore, I had to refer to the solution. However, after doing so, having some doubts as to how the values for the outer and inner integration were found exactly? A bit of complexity here, so working on breaking things down to understand double integration more confidently, and the text is assisting somewhat, but would like to have your feedback as it is much more clear. – Thank you.

Here's a revised explanation:

The description of the region:

0 <= x <= 3, 0 <= y <= x

plus

3 <= x <= 9, 0 <= y <= 9/x.

We are integrating area so for our first region:

Our 'inner' integral is the integral of 1 (we integrate 1 when we want to find area) with respect to y, between the limits y = 0 and y = x. Our antiderivative is therefore y, which changes between the limits from 0 to x, a change of x. So our 'inner' integral gives us x.

Integrating this result x with respect to x, between limits x = 0 and x = 3, our antiderivative is x^2 / 2. Between the limits x = 0 and x = 3 our antiderivative changes from 0^2 / 2 = 0 to 3^2 / 2 = 9/2, so the change in the antiderivative is 9/2.

For our second region:

Our 'inner' integral is the integral of 1 with respect to y, between the limits y = 0 and y = 9 / x. Our antiderivative is y, which changes between the limits from 0 to 9 / x, a change of 9/x. So our 'inner' integral gives us 9 / x.

Integrating this result 9 / x with respect to x, between limits x = 0 and x = 3, our antiderivative is 9 ln | x |. Between the limits x = 3 and x = 9 our antiderivative changes from 9 ln ( 3) to 9 ln ( 9 ), a change of 9 ln(9) - 9 ln(3) = 9 ln(9/3) = 9 ln(3).

The area of our region is therefore 9/2 + 9 ln(3), which we simplify to 9 ( ln(3) + 1/2 ).

Let me know if you have specific questions.

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self-critique rating #$&*: 2

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Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment.

This assignment pertaining to the relative unit is somewhat challenging. I am trying to grasp certain areas of this section more clearly. A couple of these problems have presented issues, and I look forward to further progression here. However, I enjoy the aspect of challenge here. Any added help where addressed is more than welcome.

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&#Your work looks good. See my notes. Let me know if you have any questions. &#

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