Question

Find the volume of the solid generated by rotating the area bounded by the curves y=cotx and y=2cosx for 0<=x<=Pi/2, around the line y=-1

Answer 1

i come to this equation \[\int\limits_{0}^{\Pi/2} \Pi(2cosx+1)^2-\Pi(Cotx+1)^2\]

Answer 2

should this be right?

Answer 3

Is that capital pi or small pi?

Answer 4

you're gonna need two integrals I believe

Answer 5

its just pi

Answer 6

but...
x cannot be zero
another strange problem from your prof

Answer 7

haha really!? again?

Answer 8

thats why i post this to see if im doing it right

Answer 9

i try to do it on wolfram but its not letting me

Answer 10

cot is undefined at x=0
so the volume is infinity
unless this is some kind of improper integral problem I haven't figure out yet

Answer 11

look at the area between the graphs
http://www.wolframalpha.com/input/?i=plot+y%3Dcotx%2Cy%3D2cosx

Answer 12

if it was from pi/6 to pi/2 that would be ok...

Answer 13

maybe it's a right-hand improper integral, let me try that...

Answer 14

http://www.wolframalpha.com/input/?i=graph+y%3Dcotx+and+y%3D2cosx+for+0%3Cx%3CPi%2F2

Answer 15

seems like the two function doesnt intersect at 0

Answer 16

yes I was just coming to that conclusion from my own angle
http://www.wolframalpha.com/input/?i=integral+from+0+to+pi%2F6+cot%5E2x

Answer 17

(^this integral is in the answer i would figure)

Answer 18

and a right-hand limit is not helping me, because the integral of cot^2 is -x-cotx
so cotx is still in the problem :/

Answer 19

yea thats was my problems too

Answer 20

so in order to do this problem the limits have to be change?

Answer 21

yeah, or it needs a different boundary
unless I'm missing something major you have a very difficult problem, as even the CPV in this problem is infinity according to wolfram

Answer 22

ok another question to ask the prof about. thanks turing?

Answer 23

welcome, but feel free to get a second opinion

Answer 24

i did the whole problem but the answer.. im not confident with it

Answer 25

how did you integrate cot ?

Answer 26

btw, the way I see it the integral is this\[V=\pi\int_{0}^{\frac\pi6}\cot^2x-4\cos^2xdx+\pi\int_{\frac\pi6}^{\frac\pi2}4\cos^2x-\cot^2xdx\]

Answer 27

that's pi/6 in the middle there

Answer 28

yeah change your limits to pi/6 to pi/2 because that is where the functions intersect creating the region to be revolved around y=-1

i think the 0<x<pi/2 , is to restrict the domain of the intersecting points

Answer 29

if that is the case dumbcow the problem is very ill-written, you must agree
and it is not just 0<x<pi/2 but\[0\le x\le\frac\pi2\]if you want the region between the graphs in that interval you are in trouble.

Answer 30

hey turing just woke up and started working on this problem again.. so i will have to change the limits to Pi/6 and Pi/2

Answer 31

can you tell me how did you get the equation to intergrate?

Answer 32

i thought its outside radius^2 - inside radius^2

Answer 33

yeah I messed up and forgot the 1, so from pi/6 top pi/2 it should be\[\large V=\pi\int_{\frac\pi6}^{\frac\pi2}(2\cos x+1)^2-(\cot x+1)^2dx\]which I think is what you had

Answer 34

yea thats what i got hehe

Answer 35

and just integrate on right?

Answer 36

yep, nothing problematic-looking.

Answer 37

looks like you might save a little trouble by using the formula cot^2+1=csc^2 on the after you expand the second term

Answer 38

on the right*

Answer 39

ok thanks!

Answer 40

time to eat breakfast!

Answer 41

enjoy :D