Question

Find the surface area of revolution generated by revolving the cardioid about the x-axis:

Answer 1

\[x=2\cos2\theta-\cos \theta,y=2\sin2\theta+\sin2\theta,0\le \theta \le \Pi\]so now i got\[\frac{dx}{d \theta}=\sin \theta-4\sin2\theta\]\[\frac{dy}{d \theta}=2\cos \theta+2\cos2\theta\]and the surface area of revolution about the x-axis is given by\[A=2\Pi \int\limits_{0}^{\Pi}y \sqrt{(\frac{dx}{d \theta})^{2}+(\frac{dy}{d \theta})^{2}}d \theta\]substituting everything given, \[A=2\Pi \int\limits_{0}^{\Pi}(2\sin \theta+\sin2\theta)\]\[\sqrt{\sin^{2}\theta-8\sin \theta \sin2\theta+16\sin^{2}2\theta+4\cos^{2}\theta+8\cos \theta \cos2\theta+4\cos^{2}2\theta}d \theta\]what is the possible simplified form before doing the integration?

Answer 2

i reached\[A=2\Pi \int\limits_{0}^{\Pi}(2\sin \theta \sin2\theta)\sqrt{\sin^{2}\theta+4\cos^{2}\theta+8\cos3\theta+4(4\sin^{2}2\theta+\cos^{2}2\theta)}d \theta\]

Answer 3

I don't think I've ever done a surface area revolution in polar coordinates, but this looks wrong for a few reasons.
are you sure ds is given by\[\sqrt{(\frac{dx}{d\theta})^2+(\frac{dy}{d\theta})^2}\]?

Answer 4

in polar coordinates, the arc length \(ds\) is given by\[\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta\]so it looks like you gotta get the x an y into a form such that you can determine what the formula for r is

Answer 5

its not polar coordinates. its parametric equations, whereby x and y are in terms of t

Answer 6

o.o or u mean this question i should use polar coordinate to sovle?

Answer 7

I see you're right.... ok then the goal is to make this all less ugly somehow

Answer 8

but im not sure whether im correct or not. coz i haven learn polar form yet. if this question really requires polar form formula then i can skip this question

Answer 9

well it's not defined in terms of r=blah, nor is it a cartesian we can convert into polar form, so... I'm not so sure. just seems parametric.
I'm afraid I'm sort of multi-tasking right now, so I can only say I'll let you know if an idea comes to me.

Answer 10

@amistre64 wanna lend a hand here?

Answer 11

ive seen this thing floating about for the past 13 hours or so lol

Answer 12

yeah as you can see I gave it a brief look this morning, but I'm not really sure what can be dont to simplify it

Answer 13

surface are integrals i aint comfy with yet; i know it has something to do with adding up all the curves as we swing about the origin

Answer 14

x and y are parametric, not polar to me

Answer 15

http://www.wolframalpha.com/input/?i=parametric+x%3D2cos2t%E2%88%92cost%2Cy%3D2sin2t%2Bsin2t

is what i get

Answer 16

@edr1c btw there must be a typo somewhere 'cuz you got two sin(2theta)'s in your original post, and then you change it
what is the original problem again?

Answer 17

ooo ou din realized tat. its 2 sin theta + sin 2 theta

Answer 18

the polar rendition doesnt gives even less of a cardoid and more of a petal formation

http://www.wolframalpha.com/input/?i=polar+x%3D2cos2t%E2%88%92cost%2Cy%3D2sin2t%2Bsin2t

but it plots it individually

Answer 19

\[x=2\cos2\theta-\cos \theta,y=2\sin\theta+\sin2\theta,~~~~~~~~~0\le \theta \le \pi\]about x

Answer 20

it might be best to start over with a complete correct posting of the question at hand

Answer 21

oh kay.
Find the surface of revolution generated by revolving
\[x=2\cos2\theta-\cos \theta\]\[y=2\sin \theta+\sin 2\theta\]from theta=0 to theta=Pi
sorry for the confusion

Answer 22

surface area revolution \[A=2\Pi \int\limits_{a}^{b}y \sqrt{(\frac{dx}{d \theta})^{2}+(\frac{dy}{d \theta})^{2}}d \theta\]

Answer 23

if we split this up and go cartesian ... not that its a better idea; but its just an idea

Answer 24

im just not confident in the surfaces of revolution