Assignment 32

course Mth 272

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032.

Applied Calculus II

05-12-2009

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15:07:54

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

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integrate with respect to y

x^2y + y^3/3 is the antideriv.

x^2(sqrt(x) + (sqrt(x))^3/3 - x^2(x^2) + (x^2)^3/3

x+x^2/3 - x^4 + x^6/3

confidence assessment: 3

I think you have errors on your fractional exponents. Procedures are right, just that one detail differs:

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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15:07:59
What is your result?

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answered
confidence assessment: 3

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15:08:07
What was your antiderivative?

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answered
confidence assessment: 3

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15:18:08
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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first the anti deriv
[2/3(1-x^2)^3/2]from 0 to x
2/3[(1-x^2)^3/2- (1-0^20^3/2]
(1-x^2)^3/2 - 1
2/3[-x^3]
=-2/3x^3 integ from 0 to 1
= -8/3x^4
-8/3(1)^4 - (-8/3(0)^4)
=-8/3
confidence assessment: 3

The first integration is with respect to y. Note also that 2/3 ( 1 - x^2)^(3/2) is not an antiderivative of sqrt(1 - x^2); if you take the derivative of 2/3 ( 1 - x^2)^(3/2) you have to use the chain rule, which gives you an extra factor of -2x.

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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15:18:14

What is your result?

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answered

confidence assessment: 3

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15:18:38

What is your result?

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15:19:37

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

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confidence assessment: 3

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15:19:45

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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answered

confidence assessment: 3

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15:19:49

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

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confidence assessment: 3

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15:23:53

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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integ -2 to 2 [x] from 0 to 4-y^2

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

=[4y-y^3/3] from -2 to 2

4(2) - (2)^3/3 - (4(-2) + (2)^3/3) = -16/3 = area

confidence assessment: 3

** 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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15:24:04

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

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15:24:39

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

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it is a rectangular peice, under a lareger curve

confidence assessment: 3

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15:26:55

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

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15:27:06

What did you get when you evaluated the original integral?

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answered

confidence assessment: 3

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15:27:31

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

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im not sure how to get the new limits

confidence assessment: 3

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15:28:13

Why should both integrals give you the same result?

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becuase they are describing the same area

confidence assessment: 3

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16:27:51

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

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the two limits, i think are 0 to y and 2 to 5

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

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

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

= 1.2679

confidence assessment: 3

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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16:27:59

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

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confidence assessment: 3

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16:28:01

What is the area?

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confidence assessment: 3

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16:28:33

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

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i didnt really understand how to do this one could you explain?

confidence assessment:

**It is recommended that you graph y = 9 / x, y = x, y = 0 and x = 9, then follow the details given below.

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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16:28:36

What is the area of the region?

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16:28:38

Describe the region in detail.

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16:28:40

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

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16:28:42

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

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16:29:26

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

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This one was pretty difficult. I didn't understand how to set up the last one. If you could explain a little that would be much appreciated. Thankyou.

confidence assessment: 3

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Assignment 32

You're doing pretty well here; see my notes to clarify a few things. Hopefully my notes on that last problem will also be helpful.