Question

Find the volume of the solid created by the bounded region of (x=y-y^2) and (x=y^2 -3) when it is spun around the line (x=-4).

Answer 1

Ah this question still o.O

Answer 2

Yeah :) but I found out what I was doing wrong.... other than trying to drink and derive :)

Answer 3

So we at least know you can add 4 to all terms to change the bounds and have it revolved around x = 0

Answer 4

And it's a torus, right? So you need to integrate the area of the torus around the line in a full circle.

Answer 5

A cross section of the torus.

Answer 6

yep...

Answer 7

Can't you just integrate it?

Answer 8

its pi (S) [4+F(y)]^2 - [4+G(y)]^2 | -1, 3/2

I kept trying:

pi (S) [(4+F(y)) - (4+G(y))]^2 | -1, 3/2

Answer 9

math: yes, but couldnt figure out how to last night :)

Answer 10

\[x = y - y^2 + 4\]
\[x = y^2 + 1\]

Answer 11

Okay so the bounds are y = -3, y = 1/2

Answer 12

No...

Answer 13

What?

Answer 14

Krb: goog good. the bounds of integration are -1 and 3/2

Answer 15

Yeah those are the bounds ^

Answer 16

i need to figure out how to use the equations to find the solids of revolution..if this is x=-4, then the line we get is a vertical one right? so its been revolved upwards?

Answer 17

you got the equation right; now square them individually before you subtract them..

Answer 18

So \[\int\limits_{(-1)}^{(3/2)}(y - y^2 + 4) - (y^2 + 1)\]

Answer 19

That's what i would think

Answer 20

Krb you're always on this site. I like how when I look at paul's notes, you were the first one on open study.

Answer 21

think; the rotation is around the "y axis" so to speak

Answer 22

yeah..thats what i mean, so its like integrating the difference of squares of the equations right? and these equations signify -----?

Answer 23

Krb: getting closer. we need to integrate [F(y)]^2 - [G(x)]^2 in the bounds

Answer 24

Yes, normally when integrating with respect to x you are finding the are underneath the curve that drops to the x axis. So in this case you are finding the area "to the side of the curve" that drops to the y axis

Answer 25

Think; exactly :)

Answer 26

I'm confused as to why it would be squared

Answer 27

typo...[G(y)]^2

Answer 28

i guess it should be squared...shouldn't it?

Answer 29

Krb: if we do it seperately for each on, there is no real "area" so to speak until we combine them. Does that make sense?

Answer 30

pi* (S) F(y)^2 dy - (S) G(y)^2 dy

The equations ARE our Radiusessess

Answer 31

OH

Answer 32

if y1 & y2 are the curves then ,
\[\int\limits_{}^{}[y1^{2}-y2^{2}]\] gives the volume of the solid? where y1=upper curve & y2 the lower

Answer 33

multiplied by a pi

Answer 34

SO we are looking at it from a top down view, and then you need pi * r^2 for both "circles"

Answer 35

exactly

Answer 36

on efor inner and the other for outer 'circle' i guess

Answer 37

think: thats the gyst of it yeah

Answer 38

thnx!!

Answer 39

\[[\pi \int\limits_{-1}^{(3/2)}(y - y^2 + 4)^2] - [\pi \int\limits_{-1}^{(3/2)}(y^2 + 1)^2]\]

Answer 40

Krb: your a genius :)

Answer 41

No, you are :)

Answer 42

u both are :)

Answer 43

we should start a mensa club right here lol

Answer 44

the open study math mensa

Answer 45

:) :) so krb ur'e pursuing your masters too?

Answer 46

C1 - C2 is what we were missing last night.

2pi(R) - 2pi(r) does not equal 2pi(R-r)

Answer 47

it does, but it doesnt when you integrate lol

Answer 48

wer those constants that u'll missed out yesterday?

Answer 49

Ends up being 85.9029

Answer 50

think: I'm honestly not sure

Answer 51

krb: that's alright

Answer 52

i dunno why it gets posted twice always, when i click only once!

Answer 53

I think I probably would like to get a Masters

Answer 54

good