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HSC 2015 MX2 Marathon (archive) (1 Viewer)

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Green Yoda

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Re: HSC 2015 4U Marathon

Wow this stuff is looking difficult! Anyone know good yr 10-11 tutor? just post a link if already on bos!
 

InteGrand

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Re: HSC 2015 4U Marathon

so y=x^2, y=1, x=3 about x axis would be
?
Note quite.

For this one, your typical slice will be a (vertical) washer. At a point x, the outer radius of the washer will be . The inner radius will be (constant). The area of an annulus (http://en.wikipedia.org/wiki/Annulus_(mathematics) ) is .

Then the area of a typical washer is .

Then integrate this area function between the appropriate bounds (which you have, i.e. x = 1 to x = 3).
 

Ekman

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Re: HSC 2015 4U Marathon

Note quite.

For this one, your typical slice will be a (vertical) washer. At a point x, the outer radius of the washer will be . The inner radius will be (constant). The area of an annulus (http://en.wikipedia.org/wiki/Annulus_(mathematics) ) is .

Then the area of a typical washer is .

Then integrate this area function between the appropriate bounds (which you have, i.e. x = 1 to x = 3).
Shouldn't the inner radius be 1?
 

Drsoccerball

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Re: HSC 2015 4U Marathon

Note quite.

For this one, your typical slice will be a (vertical) washer. At a point x, the outer radius of the washer will be . The inner radius will be (constant). The area of an annulus (http://en.wikipedia.org/wiki/Annulus_(mathematics) ) is .

Then the area of a typical washer is .

Then integrate this area function between the appropriate bounds (which you have, i.e. x = 1 to x = 3).
Can you explain as if youre talking to an idiot i still dont get it
 

Ekman

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Re: HSC 2015 4U Marathon

Interesting Question:



Don't use cylindrical shells to solve this.
 
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InteGrand

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Re: HSC 2015 4U Marathon

Can you explain as if youre talking to an idiot i still dont get it
Draw a diagram and shade in the region that will be revolved. Now draw a thin vertical rectangle at a point x (where x is between 1 and 3). This thin vertical rectangle starts at the line y = 1 and goes up to the point where it touches the parabola (which will be at a place where the y-value is x2). (This isn't exactly a rectangle, since at the top, it's not straight, but rather, parabolic. But in the limit as you take very small strips, it becomes rectangular.)

Now, imagine rotating this rectangular strip, which is between y = 1 and y = x2, about the line y = 0 (i.e. the x-axis). What you'd get would be an annulus (basically a doughnut shape). You're not getting a full circle because the rectangular strip does not include any area between the x-axis and the line y = 1. Now, the radius of 'hole' of the annulus is the distance from the x-axis to the line y = 1, which is 1. So the inner radius of the annulus is r(x) = 1 (a constant function of x). The outer radius of the annulus is the distance of the furthest end of the strip from the x-axis to the x-axis, which is just x2. Hence R(x) = x2. (Using R to denote outer radius and r to denote inner radius.)

Then the area of a typical annulus is A(x) = π((R(x))2 – (r(x))2), where R(x) = x2, r(x) = 1. Now you can find the volume using .
 

VBN2470

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Re: HSC 2015 4U Marathon

VOLUMES QUESTION

 
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Re: HSC 2015 4U Marathon

Interesting Question:



Don't use cylindrical shells to solve this.
Would you jsut find the volume of the circle rotated around x = 4 subtracted by volume of vertical ellipse rotated around x = 4? Or letting slice have end point P(4cost,4sint) [lies on circle] and Q(2cost,4sint) [lies on ellipse] with height of delta y? Then you have an integral with 't' and 'delta y' so to eliminate delta y I use y = 4sint ==> dy=4cost dt ?
 
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