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HSC 2015 MX2 Marathon (archive) (2 Viewers)

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Drsoccerball

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Re: HSC 2015 4U Marathon

screw this... i learn the normal way properly but still dont know how to do washers method and anything with annulie :(
 

Ekman

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Re: HSC 2015 4U Marathon

Just a simple question, for questions like:



Is the outer radius 2+x or 2-x? Some argue that it is 2-x as x is technically speaking negative in the second quadrant
 
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InteGrand

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Re: HSC 2015 4U Marathon

(A quick way to see it is to test when x = -2: The outer radius should clearly be 4 here, so we should be using 2 – x (2 – (-2) = 4).)
 

VBN2470

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Re: HSC 2015 4U Marathon

^ Do you have Cambridge 4U? That may help since they have pretty solid examples you can follow, if you're struggling, then go back to 2U and revise (and try to understand) the ideas behind basic volumes of solids of revolution. This can help form a basis on which you can strengthen your ability to solve 4U problems (even the basic ones).
 

InteGrand

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Re: HSC 2015 4U Marathon

(Or, R(x) = 2 + |x|, and since x < 0, |x| = -x, so R(x) = 2 – x.)
 

Drsoccerball

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Re: HSC 2015 4U Marathon

^ Do you have Cambridge 4U? That may help since they have pretty solid examples you can follow, if you're struggling, then go back to 2U and revise (and try to understand) the ideas behind basic volumes of solids of revolution. This can help form a basis on which you can strengthen your ability to solve 4U problems (even the basic ones).
I have new senior maths the examples are so good but i cant to the questions that involve two radii
 

Drsoccerball

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Re: HSC 2015 4U Marathon

thanks guys understand it now :))) thanks to integrand and VBN :)
 

Drsoccerball

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Re: HSC 2015 4U Marathon

now i need to know how you do a question where like this where theres two areas or 2 annuli:
 

InteGrand

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Re: HSC 2015 4U Marathon

now i need to know how you do a question where like this where theres two areas or 2 annuli:
For this, the inner radius is the distance from the y-axis to the nearer half of the parabola, and the outer radius is the distance from the y-axis to the further half of the parabola.

To find these for a given y, solve the equation for x using the quadratic formula (there will be two solutions). The x you get with the smaller value (the one with the negative square root) will be the inner radius r(y). The bigger solution will be the outer radius R(y). After finding these two functions, you can find as a function of y (using should make this less tedious), and then after simplifying, you integrate with respect to y between the range of y-values, which will be between y = 0 and y = ymax., the y-value of the vertex of the parabola (which you'll need to find).
 

Drsoccerball

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Re: HSC 2015 4U Marathon


Using the method of cylindrical shells to calculate the volume of this solid is the answer :
 
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