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/ap...s_ab_formb.pdf
and the scoring guidelines here
http://apcentral.collegeboard.com/ap...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
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
You can find the question here
http://apcentral.collegeboard.com/ap...s_ab_formb.pdf
and the scoring guidelines here
http://apcentral.collegeboard.com/ap...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
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