Question

f(x)= x^3 and g(x)= 8x-x^2. Rotate region A around x-axis and find volume of resulting body.

Answer 1

Well this is an easy problem
\[(f-g)(x)=f(x)-g(x)=8x-2-(2x-3)6x+1\]

Answer 2

i still don't understand what you did.. i found the intersection points already: x=0, x= (-1 +or- sqrt(33))/2. i am not sure how to find the A

Answer 3

First solve x and y to get the upper and lower limit
http://www.wolframalpha.com/input/?i=y%3D+x^3%2C+y%3D+8x-x^2
\[ \int_{a}^{b} \pi (f^2 - g^2) dx\]

where a and b are lower and upper limits.

Answer 4

\[\int\limits_{(-1-\sqrt{33})/2}^{(-1+\sqrt{33})/2}(x^6-(64x^2-8x^3-8x^2+x^4))dx \]

yes?

Answer 5

pi is missing.

Answer 6

oh! right.

Answer 7

Area = pi * r^2 <---- y is your r here,


integrate pi * r^2 * dh <--- should give volume of solid of rotation.

Answer 8

also if I look at the graph properly, now I think ... it was actually g - f not f - g
http://www.wolframalpha.com/input/?i=y%3D+x^3%2C+y%3D+8x-x^2
if you get negative volume, you can remove minus later

Answer 9

this is my work so far. am i even on the right track?

Answer 10

i really appreciate you helping me, by the way.

Answer 11

is volume from 0 to ...??

Answer 12

\[(-1+\sqrt{33})/2\]

Answer 13

guess you are right then.

Answer 14

the answer is pi \[((64x^3)/3) - 4x^4+ ((x^5)/5) - ((x^7)/7)\] upper limit of (-1+sqrt(33))/2 and i can't get it

Answer 15

agh. that's gross to look at. answer:

Answer 16

i just want to understand how to calculate the function part of it correctly. actual number answer is pretty self explanatory.

Answer 17

try with wolfram to verify
http://www.wolframalpha.com/input/?i=integrate+pi+*+%28x^6+-+%288x+-+x^2%29^2%29+from+0+to+%281-sqrt%2833%29%29%2F2

Answer 18

the answer did come out correctly in negative, but it doesn't show the work for the function

Answer 19

don't know ... i lost my glasses ... cannot concentrate at all.

Answer 20

ok! i caught my mistake. messed up when i squared (8x-x^2). you were right about having to (g^2(x))-(f^2(x)). how did you know to do that? traditional formula calls for (f^2(x))-(g^2(x)).

Answer 21

look for the top curve ... put it front ... the lower curve goes behind minus. :D

Answer 22

ah, ok.

Answer 23

what about the volume from rotating region A about the y-axis? i would use this? \[2 pi \int\limits_{a}^{b} (f(x)-g(x))dx\]

Answer 24

there are two ways ... shell method and disc method.

it would be integrate pi x^2 dy

Answer 25

never ask this type of question again LOL
http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/lecture-22-volumes/

Answer 26

i thought disc method was for rotation of region about x-axis?

Answer 27

i figured it out already. thanks so much for your time.