Volume generated by revolving a region

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Homework Statement

Hi, I'm trying to solve the 2010 AP Calculus AB Free-Response Question 1 (Form B) and can't figure out why my answer doesn't match up with the scoring guidelines for the life of me.

You can find the question here
http://apcentral.collegeboard.com/apc/public/repository/ap10_frq_calculus_ab_formb.pdf

and the scoring guidelines here
http://apcentral.collegeboard.com/apc/public/repository/ap10_calculus_ab_form_b_sgs.pdf

I'm trying to do part (b) and keep on getting the wrong answer

In the figure above, R is the shaded region in the first quadrant bounded by the
graph of y = 4ln(3-x) , the horizontal line y = 6, and the vertical line x = 2

(a) Find the area of R.

(b) Find the volume of the solid generated when R is revolved about the
horizontal line y = 8.

(c) The region R is the base of a solid. For this solid, each cross section
perpendicular to the x-axis is a square. Find the volume of the solid.

The scoring guidelines say that the answer is 168.179 or 168.180 and I keep on getting 143.676 and don't see what's wrong with what I am doing

Homework Equations

The Attempt at a Solution

the function that represents the y values of the region that are being rotated is simply
f(x) = 6 - 4 ln(3-x)

put in the original function f(x) = 4 ln (3) and y = 6 and you'll see that f(x) = 6 - 4 ln(3-x) is the region "rotated" so it's now flush with the x-axis and y-axis and one should be able to simply rotate this function about the x-axis using shells to get the same answer the college board did, yes I know this isn't how they did it but this is how I did it and thought about it before checking the answers, I reasoned I should be able to do this sense were only interested in rotating the region R and that it makes no difference if we were rotating the function around y = 8 or y = 6 we should still get the same answer right because were not interested in the volume between 8 and 6 but only the volume of the region...

f(0) = 6 - 4 ln(3)
f(2) = 6
solved for x
x = 3 - e^(3/2 - y/4)

2 pi integral [lower limit=6- 4 ln(3), upper limit = 6] y(3 - e^(3/2 - y/4))dy = 143.676

note that [lower limit=6- 4 ln(3), upper limit = 6] were the limits that were being integrated

 

[Rudeboy37]

I reasoned I should be able to do this sense were only interested in rotating the region R and that it makes no difference if we were rotating the function around y = 8 or y = 6 we should still get the same answer right because were not interested in the volume between 8 and 6 but only the volume of the region...

The axis you are rotating around DOES matter! When you rotate around an axis that's far away from your region, the shape and volume of the object is changing compared to an axis that's flush with your region.

The basic formula for shells is integral of 2pi*radius*height. When your axis is moving, your radius is changing. In this case it is no longer y but instead 8-y.

You can still do shells (if you want to), but you must split it into 2 separate integrals because the height is 2 from 4 ln(3) to 6, but the height is 2-(3 - e^(y/4)

2 pi integral [lower limit=4 ln(3), upper limit = 6] (8-y)(2)dy

+

2 pi integral [lower limit=0, upper limit = 4 ln(3)] (8-y)(2-(3 - e^(y/4)))dy

which gets the same answer the scoring guide