Volumes by Slicing Question - Please Help (1 Viewer)

[Neon-Frog]

Arghhh

I have an idea of how this question works .. im just not sure which values to put in for the integration

anyway here it is

From an extremity A of the latus rectum AB of the parabola x^2=4ay interval AK is drawn perpendicular to the x axis. SHow that the volume formed by the rotation of the region OAK about the line AK is 2/3.pi.a^3

 

[wogboy]

Firstly, find the equation of the line AK, this happens to be x=2a. This is our axis of rotation.

Now draw up the sketch of the parabola and the region required to rotate, then imagine that this volume is composed of many thin cylindrical discs of varying radii, all stacked onto the line AK.

Now find a formula relating the radius (call it r) of each disc (which is actually the horizontal distance from any point on the line AK to the parabola itself, and more importantly this horizontal direction is and must be PERPENDICULAR to the axis of rotation), and the x-coordinate where that disc touches the parabola. r is given by:

r = 2a - x

the volume (dV) of each little disc (which has thickness dy), is:

dV = pi*(r^2)*dy

so,

dV = pi*(2a - x)^2 * dy

dV/dy = pi*(4a^2 - 4ax + x^2)

but x^2=4ay, so x = 2*sqrt(ay) (ignore the negative root of x since it is the positive x section of the parabola we're interested in)

therefore,

dV/dy = pi*(4*a^2 - 4a*sqrt(4ay) + 4ay)
= pi*(4a^2 - 8a^(3/2)*y^(1/2) + 4ay)

To find V, we just integrate the RHS with respect to y, and the limits which we integrate it with are actually where the volume starts and where it finishes (y-values, since we're integrating with respect to y).

so,

V = integral (0 -> a) of pi*(4a^2 - 8a^(3/2)*y^(1/2) + 4ay) dy

(note that "0" is the LOWER limit, while "a" is the upper limit)

solve this integral, and you should hopefully get

V = (2/3)*pi*a^3

 

[Neon-Frog]

Thankyou

 

[Harimau]

How did you know the length of the inverval AK was 2a? We just started volumes as well at school..

 

[ND]

Latus rectum has eqn: y = a, solve with x^2=4ay.

edit: it's not the length of AK that is 2a, the line AK has eqn: x = 2a.

 

[freaking_out]

hey where did u get the question from????

i had the exact same question that i had troubles with.

 

[freaking_out]

Originally posted by wogboy
...solve this integral, and you should hopefully get

well thats the bit i have trouble wif. i went down all the way to the integral but can't show the desired result!!!:mad1: so how do i do it?

 

[McLake]

I = pi*[4a^2y - (2/3)*8a^(3/2)*y^(3/2) + 2ay^2] between a and 0

so pi*[4a^3 - (16/3)a^3 + 2a^3] - 0
= 2/3*pi*a^3

 

[Chloe]

i do general u r lucky u hav a good brain,

 

[freaking_out]

Originally posted by Chloe
i do general u r lucky u hav a good brain,

don't worry, we (or atleast me:mad1: ) have crappy brains when it comes to english.