Question

Integral problem regarding finding the area of a solid via the shell method, the rotational axis being the y-axis. (Posting below, one moment.)

Answer 1

The bounds for this object are \[x = 1\]\[x = -1\]\[x-axis\]\[y = 3x ^{4}\]So you have an object that has the entirety of its volume being non-negative, so the integral \[V = \int\limits_{a}^{b}2 \pi (r)(h)dx \]Should produce the actual volume of the solid in this case, and given the bounds of the object, you can express that as \[V = \int\limits_{0}^{1}2 \pi (x)f(x)dx\]
The answer should be pi^2, but i'm not getting that:

Answer 2

\[V = \int\limits_{0}^{1}2 \pi (x)(3x ^{4})dx = 2 \pi \int\limits_{0}^{1}(3x ^{5})dx = 2 \pi \left[ \frac{ 3x ^{6} }{ 6 } \right]\]
(Evaluated at 0-1, lower-upper bound, respectively)\[2 \pi \left[ \frac{ 3(1)^{6} }{ 6 } \right] = 2 \pi \frac{ 1 }{ 2 } = \pi\]
What am I doing wrong?

Answer 3

For starters, the formula is \[\pi \times \int\limits_{b}^{a}(\Big R squared) - (Little R squared) dx\]

Answer 4

Your formula didn't turn out right.

Answer 5

It is pi times the integral of the big radius squared minus the little radius squared.

Answer 6

But there is no little radius in this situation; at exactly the origin, there is nothing, but at literally every other section has a volume.

Answer 7

And I assume you're trying to find the VOLUME of the solid and not the area?

Answer 8

Yes, the "VOLUME". Thanks for that.

Answer 9

It says you're doing it about the y-axis, so your integral will need to be with respect to y.

Answer 10

Nope, unless Salman Khan is wrong. And my calculus textbook.

Answer 11

Oh, sorry, it's been awhile since Calc I. And FYI, Salman Khan frequently gets semantics incorrect.

Answer 12

Sure, but I don't think he did it wrong in this one. https://www.khanacademy.org/math/calculus/solid_revolution_topic/shell-method/v/shell-method-for-rotating-around-horizontal-line No problem, thanks regardless.

Answer 13

Did you attempt to do it with another method?

Answer 14

No, but this should still work with this method as far as i'm aware;I can do this otherwise, but I want to get very competent with the shell method to make sure i've got it under my hands.

Answer 15

the method and the formulae are correct

Answer 16

Did you produce that using a program, or was that grabbed off the internet? If it was done by yourself through a program, i'd really like to find out which.

Answer 17

Do you still need some explanation?

Answer 18

Yes.

Answer 19

I'd like to know how to do this using the shell method, specifically. Not another one, but just the shell method.

Answer 20

the tool is called "GeoGebra" http://lmgtfy/?q=geogebra

Answer 21

http://www.lmgtfy/?q=geogebra

Answer 22

wow, I never really understood this program, I didnt even know it could do that, I prefer to use mathematica

Answer 23

No, no, sorry, not at all, lol, it was in fact Volume. I posted that wrong and corrected it later.

Answer 24

http://www.lmgtfy.com/?q=geogebra

Answer 25

all hail open source

Answer 26

Mathematica aint free and too much un-neccasry clutter, sluggish, requires a good computer, etc
I am an oldschool guy loves the terminal, and typing my stuff with keyboard and no mouse.

This creates nice vector graphics and PStricks code for use in LaTeX. The main use for me

Answer 27

I'm super terrible at using Mathematica, lol. But yes, all hail OS. I still don't understand either what I did wrong, or why i'm doing it wrong; my book lists pi^2 as the answer, and i'm not getting that at all.

Answer 28

(To prove my sanity) Part b is what we're looking at.

Answer 29

Oh. Ohgod. I'm sorry. Uh, yeah. My bad. Nope. It wasn't pi^2.

Answer 30

*Fail of the week*

Answer 31

Wait, is it the volume then?

Answer 32

Lol, yes, again, I corrected it, if anybody read the whole thread, I posted a little after that it was volume, not area

Answer 33

\[\Large \color{red}{V=\pi{\rm cu.unit}}\]

Answer 34

the above pdf file I attached gives the same answer using disc method

Answer 35

Yeah, you're dead right, and I agree. I just somehow repeatedly misread what the answer was when I got the right one in the first place, lol. Jesus. Pardon, guys. My bad.

Answer 36

I never liked the washer method and will not attempt it

Answer 37

^yes.

Answer 38

hehe, never saw this coming. By the way, I also prefer opensource, but since my college made mathematica available for us, and I need to run some simulations I use it, and it does not require mouse use.

Answer 39

whats the washer method?

Answer 40

http://www.lmgtfy.com/?q='washer method'

Answer 41

hahaha, ok

Answer 42

Broken link

Answer 43

Got it.