find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the -axis.
y= 3/(x+1) y=0 x=0 x=8

Question

find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the -axis.
y= 3/(x+1) y=0 x=0 x=8

amistre64 · Answer

your missing part of the question

anonymous · Answer

Am I? let me check.

jhannybean · Answer

Does your question tell you which axis it wants the function revolved around? Or can it be either one?

anonymous · Answer

Thats all they gave me. I think i know how to set it up though. I believe its the x axis

amistre64 · Answer

and what are your thoughts on how to approach it?

anonymous · Answer

I believe I know how to set this problem up, you use the disk method and the boundaries are [0,8]

amistre64 · Answer

good, sound ideas

anonymous · Answer

I just forgot some basic steps

amistre64 · Answer

such as?

amistre64 · Answer

|dw:1431286157228:dw|

anonymous · Answer

$$pie \int\limits_{0}^{8} [3/(x+1]^2$$

amistre64 · Answer

teh area of a circle is pi r^2,  and r is defined as y for a given value of x

yep

anonymous · Answer

Why question is do i square the denominator or just the 3

anonymous · Answer

*one

amistre64 · Answer

pi is a greek letter

pie is an edible disc :)

amistre64 · Answer

y^2,  what does y=?

anonymous · Answer

I know, sorry my bad

anonymous · Answer

I think i am finding the area under y=3/x+1 bounded by y=0 x=0 x=8

anonymous · Answer

So i use the disk method since there inst a . Right?

amistre64 · Answer

$$\int \pi r^2$$
$$\int \pi (y)^2~dx$$
$$\pi\int  (\frac{3}{x+1})^2~dx$$
$$\pi\int  \frac{3}{x+1}*\frac{3}{x+1}~dx$$
$$\pi\int  \frac{9}{(x+1)^2}~dx$$

jhannybean · Answer

Not quite. You're finding the area of the revolved function... which is your y$^2$

jhannybean · Answer

Basically your volume.

anonymous · Answer

Thats what i first had, but i forgot how to integrate that. Do i have to use ln?

jhannybean · Answer

$$V=\int A(x)dx$$ Remember? :)

amistre64 · Answer

try it,  whats the derivative of ln((x+1)^2) ??

anonymous · Answer

2(x+1)/(x+10^2?

amistre64 · Answer

2x/(x+1)^2

so ln aint gonna cut it,  but its simler than that

amistre64 · Answer

$$\pi\int  \frac{9}{(x+1)^2}~dx$$
$$\pi\int  9(x+1)^{-2}~dx$$
its just a power rule

anonymous · Answer

okay , sorry i had a brain fart. Let me try to work it out by myself first

amistre64 · Answer

whats the derivative of (x+1)^(-1) ??

anonymous · Answer

Derivative or anti derivative

amistre64 · Answer

well im jsut trying to get you to recall the power rule, and since we drop down to -2,  it must come from -1

anonymous · Answer

I remember how to do it. I will try to solve it from here, but can u just check my answer?

amistre64 · Answer

im feeling pretty lucid, so sure :)

anonymous · Answer

8 pi?

amistre64 · Answer

good job

anonymous · Answer

Thanks for your help, I over complicate things some times

amistre64 · Answer

lol, i used to make up my own problems during a test, and solve them ... got the right answer, but not for the actual problem on the test :)

amistre64 · Answer

good luck :)