Question

Find the volume of the solid obtained by rotating the region enclosed by the graphs of y=12-x & y=3x-4, and x-axis about the y-axis.

Answer 1

What are your efforts? Is this your first "rotate the region" problem?

Answer 2

no, we are doing them in calc 2. I have worked the problem out twice already and it says that i am wrong, so i don't know what i am doing wrong or what other ways to do it

Answer 3

the answer i got was (11392/27)pi

Answer 4

This looks like applications of integration.

Answer 5

yeah, i know the application. it's a matter of the process
I drew the graphs/region, changed the eqs to x=... because it asks for rotation about the y axis, therefore making it 'dy', and then integrated using the equation: pi S(from 0 to 8) [R^2outer-R^2inner]dy

Answer 6

Please show your work. Anything. Rotating around the x-axis is \(\dfrac{3139}{9}\pi\)

Answer 7

yeah no problem. can i take a phone pic and put on here?

Answer 8

Rotating around the x-axis can be either dy or dx.
Rotating around the y-axis can be either dx or dy.

Don't get lost in one methodology or one way of thinking.

Answer 9

yes it can, sorry for not clarifying. we are supposed to use disk/shell method. (when the slice is perpendicular to the axis of rotation)
But i was going to do the shell method if i could not manage to get the answer using disk/washer

Answer 10

first is supposed to say disk/washer

Answer 11

ok here is a picture of my work

Answer 12

You can give clues...

1) Which axis are we using as the axis of rotation?
2) Are you using disks/washers or shells?
3) What's in front? \(\pi\;or\;2\pi\)
4) What are the limits?
5) What are your arguments?

Answer 13

I ALWAYS do it both ways. This builds great confidence in the result.

Answer 14

i am rotating about the y-axis using the disk/washer method, pi is in front because it is the area of a circle (the slice we are cutting out of the graph), it is bounded by 0 to 8 on the y-axis, and we use those bc we are using dy

Answer 15

https://mail-attachment.googleusercontent.com/attachment/?ui=2&ik=eb3fdf3750&view=att&th=13c986cc58ea03ba&attid=0.1&disp=inline&realattid=1425818936295817216-1&safe=1&zw&saduie=AG9B_P9W7iwryvZl_LePcVvsYnIR&sadet=1359767137548&sads=x8fCl6AIrrTCD5_wTZ3nwDc-Ef8

Answer 16

can you view that?

Answer 17

No, sorry.

Answer 18

ok hold on and i will try to upload it a different way

Answer 19

You can just answer my five questions.

Answer 20

i did. It was above my http link

Answer 21

try that.

Answer 22

Ah! First thing, I read the problem wrong. You have the right region. Forget the answer I gave above.

About he y-axis should be \(\dfrac{13312\pi}{9}\)

Answer 23

it says that the answer is incorrect. I appreciate the effort though. I am just completely lost as to what to do

Answer 24

Why on earth would you expand those simple arguments?

\(\int (12-y)^{2}\;dy = -\dfrac{(12-y)^{3}}{3} + C\)

Use a little Chain Rule, rather than a LOT of algebra. This will prevent errors.

Answer 25

Whoops. I mean "27" in the denominator. Not "9". Trouble paying attention, i guess.

Answer 26

umm because i don't think that you can do integration that way without a u-substitution. I put in 27 and it still says i am incorrect. haha i wasn't joking when i said i had tried just about everything and it denied me

Answer 27

\(\int\left(\dfrac{y+4}{3}\right)^{2}\;dy = \dfrac{3}{3}\cdot\left(\dfrac{y+4}{3}\right)^{3} + C\).

Call it what you will, Chain Rule or u-substitution, it will make your life easier.

Around the y-axis, defintely 13312\(\pi\)/27, assuming we have the irght region. I do believe you picked the right region.

You have it right all the way down to the quadratic expression all over 9.

Answer 28

chain rule is derivatives
u sub is for integration

Answer 29

Two steps down is no good, though. Your 112 magically turned into 122.

p'-tay-to p'-tah-to.

Remember when I said not to get stuck in one way of thinking?

Answer 30

btw, you appear to have a very good, clear style. Excellent work!

Answer 31

yeah thanks for catching that. i will look it over. thanks for that compliment and the help!

Answer 32

x-axis shoud be \(\dfrac{2048}{9}\cdot\pi\)

Answer 33

Shells, y-axis:

\(2\pi\int\limits_{4/3}^{4}x\cdot (3x-4)\;dx + 2\pi\int\limits_{4}^{12}x\cdot (12-x)\;dx = \dfrac{13312}{27}\cdot\pi\)

Answer 34

\[\Delta V = \pi(R^2 - r^2)\Delta y\]
where

\[R=12-y,\quad r=\frac{y+4}{3}\]

The volume of the solid of revolution is
\[\Large V=\pi\int_{-4}^8\left[ (12-y)^2 - \left(\frac{y+4}{3}\right)^2\right]\,\mathrm dy\]