Question

Find area of elipse using calculus.
(4x^2)+y^2=9

Answer 1

What i've done so far
\[f(x)=y=\sqrt{9-4x^2}\]\[x=(3/2)\sin(\theta)\]\[dx=(3/2)\cos(\theta)d \theta\]

Answer 2

solve for y, then integrate in the upper right, then multiply by 4

Answer 3

\[4\int\limits_{0}^{1/2}\sqrt{9-4(\frac{ 3 }{ 2 }\sin(\theta))^2}*3\cos(\theta)d \theta\]

Answer 4

Yeah so I got that and then did a bunch of calculations and jazz and then at the end I got to
\[36\int\limits_{0}^{1/2}\cos^2d \theta\]\[36(\frac{ \cos \theta \sin \theta+ \theta }{ 2 })|^{1/2}_0\] and that's whee i'm stuck. Because I don't want a decimal answer.

Answer 5

@satellite73

Answer 6

wow that looks way complicated

Answer 7

maybe we can do it an easier way, maybe not, but first off how did you get your limits of integration?

Answer 8

graphing xD

Answer 9

Well its not all that complicated because things cancel out pretty quickly..

Answer 10

http://www.wolframalpha.com/input/?i=ellipse+4x^2%2By^2%3D9

Answer 11

oops!! I knew it was 1.5 I was just being dumb heh. 3/2 it is then xD

Answer 12

maybe it is easier if we write it like an ellipse in the form of \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\]

Answer 13

in fact i know it is

Answer 14

easiest of all is just to do it that way, get a formula with \(a,b\) in it

Answer 15

haha ok.
but remember we have to use integrals and calc. I already did it using the area of an elipse and the prof said "must use integrals!"

Answer 16

yeaa using the pi*a*b would be epic. but not allowed :/

Answer 17

yeah i mean derive the formula using integrals

Answer 18

cause these constants are killing me

Answer 19

hmm. what do you think then?
I mean, this way isn't too bad. Like everything cancels out. It's just the evaluating at 3/2 with sin and cos that I don't quite get.

Answer 20

why not try \[f(x)=y=\sqrt{9-4x^2}=\sqrt{1-\frac{4}{9}x^2}\]

Answer 21

then at least the integral will work nicer using \[x=\frac{2}{3}\sin(\theta)\]

Answer 22

oh wait that is wrong, sorry

Answer 23

i see the problem, hold on we can fix it

Answer 24

oh my gosh I see an error that I made :o

Answer 25

it's (3/2) cos
not 3cos
for the dx

Answer 26

ok lets do it this way
i will assume that all your work above is correct, since you already did it
but you should change the limits of integration for sure, to 0 to \(\frac{\pi}{2}\)

Answer 27

ooo why can we change it to that?

Answer 28

these constants are stressing me out, are you sure we can't use \(a\) and \(b\)?

Answer 29

got it. I got the answer, you genius you.
Let me write out the process for you xD

Answer 30

ok
change the limits when you make the substitution

Answer 31

btw i made a typo, meant to write \[f(x)=y=\sqrt{9-4x^2}=3\sqrt{1-\frac{4}{9}x^2}\]

Answer 32

then make \(x=\frac{3}{2}\sin(\theta)\)

Answer 33

then the change in limits is clear

Answer 34

\[y=\sqrt{9-4x^2}\]\[x=(3/2)\sin \theta\]\[dx=(3/2)\cos \theta d \theta\]\[4\int\limits_{0}^{\pi/2}\sqrt{9-4(\frac{ 3}{ 2 }\sin \theta)^2}*(3/2)\cos d \theta\]\[4\int\limits_{0}^{\pi/2} \sqrt{9-9\sin^2 \theta}*(3/2)\cos \theta d \theta\]\[4\int\limits_{0}^{\pi/2}3 \sqrt{1-\sin^2 \theta}*(3/2)\cos \theta d \theta\]\[18\int\limits_{0}^{\pi/2}\cos^2 \theta d \theta\]\[18(\frac{ \cos \theta \sin \theta + \theta }{ 2 })|^{\pi/2}_0\]\[18(\pi/2 - 0)\]\[\frac{ 9\pi }{ 2 }\]

Answer 35

it is right?

Answer 36

Sorry, a bunch of little steps and calculations that I left out, but I trust you can follow that process. It takes a frikin amount of time to type those out heh
Yeah that's the answer!

Answer 37

do I have to show how I got the limits of integration to be 0 and pi/2?

Answer 38

i usually use \[\int\cos^2(\theta)d\theta =\frac{1}{2}\int(\cos(2\theta)+1)d\theta = \frac{1}{2}(\sin(2\theta)+\theta)\]

Answer 39

no you don't i know how they got there

Answer 40

the limits i mean
you have \[x=\frac{3}{2}\sin(\theta)\] that gives the change in limits

Answer 41

ooh..should I use that then?
I used this version of the answer for cos because I had done it in a previous question so I just wrote "see question 3" (we were asked to solve int of cos^2 by integration by parts)

Answer 42

unless you are unclear about the limits?

Answer 43

makes no difference i am sure

Answer 44

okay :)
mm i'm a little unsure about the limits. But...do you think I would be expected to write out the process for finding the limit? It's a project homework type thing.

Answer 45

lets do it, can't take but a second, and it will make it clear

Answer 46

take \[x=\frac{3}{2}\sin(\theta)\] and solve for \(\theta \)

Answer 47

i guess i should ask if you know how to do that

Answer 48

arcsin(2x/3) = theta ?

Answer 49

yes

Answer 50

so if \(x=0\) you get arcsin(0)=0

Answer 51

that is \[\theta=\sin^{-1}(0)=0\]

Answer 52

and if \(x=\frac{3}{2}\) you get \[\theta =\sin^{-1}(\frac{2}{3}\times \frac{3}{2})=\sin^{-1}(1)=\frac{\pi}{2}\]

Answer 53

now, if it was not clear before, you see where the \(\frac{\pi}{2}\) comes from right?

Answer 54

Yes!
one little question just for clarification. We set x = 3/2 because that's the x intercept, right?

Answer 55

yes

Answer 56

for \[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\] it is \(\pm a\) in your case if you divide by 9 to put in in standard form you get the rather unwieldy \[\frac{x^2}{\left(\frac{3}{2}\right)^2}+\frac{y^2}{3^2}=1\]

Answer 57

oh I see :D

Answer 58

ok good
we done? all the i's dotted and t's crossed?

Answer 59

Absopositiveloutely. Thank you so much :)

Answer 60

yw
you did almost all the work!