Two volume/integration questions (1 Viewer)

[davidbarnes]

I just don't know how to do these two questions. The textbook doesn't have examples like this which makes it hard, so would appreciate it if somone would be kind enouhg to show how to solve these.

parts b and c

a and b

Thanks again if anyone can help.

 

[kurt.physics]

Hello,

I can help you with some, and the other i can not because i dont have any year 12 text handy, only my year 10 text.

for question b, it is actually quite easy, because they have done "most" of the work! the equation to find the area of a function around the x-axis is

(the integral between; b and a) ∫ y² dx

they gave the y² so we plug it into the equation

It happens that i dont remember to do the rest of b, hope that gets you on the track


Question c

all you have to do is solve it as a quadratic simultaniously, first let the right side of the 1st equation equal the right side of the 2nd.

c ii) No idea

a & b) Cant remember formula

Hope i was any help at all lol

 

[undalay]

Sorry if there are any mistakes, its hard to check my working in a post; it should be correct thogh.
(b) There are a number ways of to tackle this problem. Heres one:

The area enclosed is the same as the area in x^2 = y + 1 and the x axis.
(We just turned the x to y and vice versa, this way it is a regular parabola and shouldn't look too daunting)

y=x^2-1
y = (x+1)(x-1)
Curve cuts x axis at -1, and 1.
Y = x^3/3 - x

Area = [x^3/3 - x](-1 -> 1)
Area = (1/3-1) - (-1/3+1)
Area = -4/3 units^2
Area = 4/3 units

(c)
(i)
Solve simultaneously.
y= -x^2 + 9x - 6 = 3x^2 - 3x - 6
0 = 4x^2 - 12x
0 = 4x(x-3)
Thus intersect at x = 0, x = 3

(ii) You take one curve away from the other. (It doesn't matter which from which, on will give a negative answer, simply turn it into a positive since area is always positive)

y = 4x^2 - 12x
Y = 4x^3/3 - 6x^2
area = [4x^3/3 - 6x^2](0->3)
area = (36 - 54) - (0)
area = 18 units^2

Need to sleep; type up the last 2 questions some other time.

 

[Forbidden.]

Finding the volume of a function using the Simpson's Rule is given by:

Image Below

 

[Poad]

Finding the volume of a function using the Simpson's Rule is given by:

Image Below

You shouldn't be trying to memorise that, by the way. Just plug it in a table and its super easy that way.

 

[flaming.robo]

y=x^2-1
y = (x+1)(x-1)
Curve cuts x axis at -1, and 1.
Y = x^3/3 - x

Area = [x^3/3 - x](-1 -> 1)
Area = (1/3-1) - (-1/3+1)
Area = -4/3 units^2
Area = 4/3 units

(c)
(i)
Solve simultaneously.
y= -x^2 + 9x - 6 = 3x^2 - 3x - 6
0 = 4x^2 - 12x
0 = 4x(x-3)
Thus intersect at x = 0, x = 3

(ii) You take one curve away from the other. (It doesn't matter which from which, on will give a negative answer, simply turn it into a positive since area is always positive)

y = 4x^2 - 12x
Y = 4x^3/3 - 6x^2
area = [4x^3/3 - 6x^2](0->3)
area = (36 - 54) - (0)
area = 18 units^2