Question

Question about integrals

Answer 1

Any ideas Zarkon? I used green's theorem, but i don't get the right answer. The right answer is -3pi/2

Answer 2

i get the -3, but i can't convert the elipse into polar coordinates, so how are they getting the pi/2?

Answer 3

@wio

Answer 4

Uh... parametrize the elipse?

Answer 5

so like x = cost and y = 4 sint?

Answer 6

\[
x^2+4y^2=4
\]Divide both sides by 4: \[
\frac14x^2+y^2=1
\]Now consider: \[
\cos ^2 t+\sin^2t=1 \implies \frac 14x^2=\cos^2t, \, y^2=\sin^2t
\]

Answer 7

so that leaves me with -3[integral of (dxdy)] and then if i do a parametrization like you said, I'm still lost on the boundaries

Answer 8

i ouwld have x^2 = cos^2 /4, sow ould i just do [2pi, 0] as limits?

Answer 9

Since it is an ellipse, \(0\leq t<2\pi\)

Answer 10

Since you'd want to go the whole way around

Answer 11

but the integral is empty though

Answer 12

what would i get inside it?

Answer 13

Actually, hold that thought

Answer 14

\[\int\limits_{0}^{2\pi} dxdy\]

Answer 15

actually no!

Answer 16

cause if i did parametrize it, i would need a cross product to convert it like that

Answer 17

We are integrating through an ellipse over a vector field

Answer 18

This is a line integral

Answer 19

where is the vector field ocming from?

Answer 20

If you want to use GT: Find your partial derivatives-
\[\frac{\partial}{\partial y}(4y-3x)=4\\
\frac{\partial }{\partial x}(x-4y)=1\]
By Green's theorem (and converting to polar),
\[\Large\int_\gamma\cdots=\int_0^{2\pi}\int_0^{\frac{ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}}}(4-1)r~dr~d\theta\]
where
\[\frac{x^2}{a^2}+\frac{y^2}{b^2}=1~~\iff~~r(\theta)=\frac{ab}{\sqrt{(b\cos\theta)^2+(a\sin\theta)^2}}\]
is your ellipse.

Answer 21

The integrand is a differential 1 form, which often can be represented as a vector

Answer 22

um, @SithsAndGiggles, it'd actually be 1-4, wouldn't it?

Answer 23

I always mess that up... is it the partial wrt \(x\) minus the partial wrt to \(y\), or the other way around? In either case, use the right one :)

Answer 24

...

Answer 25

but using that polar transformation for elipse is complicated

Answer 26

they wouldn't ask us to use that transformation...

Answer 27

I'm not claiming it's the only way to go about this. It's just a suggestion.

Answer 28

yes, that's true, but there should be an easier way...

Answer 29

Well, if you did use greens theorem, you'll just pull out the integrand, and say "this is the area of our elipse anyway"

Answer 30

professor said these should be 5 min problems...

Answer 31

wait, what is the area of the elipse?

Answer 32

GT works best if it is on a closed interval and moving counterclockwise. In this case it is moving ANTIclockwise (counterclockwise), so it works.

Answer 33

you can't just pull out integrand using green's theorem.

Answer 34

Something like \(\pi ab\).

Answer 35

You can pull out a constant.

Answer 36

yes, that part is true

Answer 37

We had a constant before the conversion to polar

Answer 38

but if you were to calculate the area, and multiply to gether, it wouldn't come out wrong.

Answer 39

mhm, yeyeye, i get waht you're saying though

Answer 40

so bak to my main question, how do i get -3pi/2?

Answer 41

the polar transformation is one way, but how do you do it using the other methods?

Answer 42

Define the vector-valued function, \(\vec{r}(t)=\langle 2\cos t,\sin t\rangle\) (the ellipse).

Then
\[\int_\gamma \cdots=\int_0^{2\pi}[(4\sin t-6\cos t)(-2\sin t)+(2\cos t-4\sin t)(\cos t)]~dt\]
This might be the quickest way.

Answer 43

If you don't do the polar transformation. \[
\iint_D (1-4)~dA = -3\iint_DdA = -3 \pi (1)(2) = -6\pi
\]

Answer 44

I wonder why I get that instead...

Answer 45

The standard form would be: \[
\left(\frac{x}{2}\right)^2 + \left(\frac y1\right)^2=1
\]

Answer 46

I just did it and looked up at @wio 's -6pi and that's exactly what I did. I suppose we must all be doing it wrong lol.

Answer 47

Here's an approach using Green's theorem, integrating in both rectangular and polar coordinates.
\[\text{Rectangular:}\\\begin{align*}
\int_\gamma\cdots&=-3\int_{-2}^2\int_{-\sqrt{1-\frac{x^2}{4}}}^\sqrt{1-\frac{x^2}{4}}dy~dx\\\\
&=-6\int_{-2}^2\int_0^\sqrt{1-\frac{x^2}{4}}dy~dx\\\\
&=-6\int_{-2}^2\sqrt{1-\frac{x^2}{4}}~dx
\end{align*}\]
A trig sub follows from here.
\[\text{Polar:}\\\begin{align*}
\int_\gamma\cdots&=-3\int_{0}^{2\pi}\int_{0}^\frac{2}{\sqrt{\cos^2\theta+4\sin^2\theta}}r~dr~d\theta\\\\
&=-\frac{3}{2}\int_{0}^{2\pi}\frac{4}{\cos^2\theta+4\sin^2\theta}~d\theta\\\\
&=-6\int_{0}^{2\pi}\frac{d\theta}{1+3\sin^2\theta}
\end{align*}\]
which can be computed using the tangent-half-angle substitution, or possibly something much simpler.

Answer 48

I think the problem lies in there somehow being a singularity at the point (0,0) I guess?

Answer 49

All the integrals evaluate to \(-6\pi\). The given answer is probably wrong.

Answer 50

http://www.wolframalpha.com/input/?i=%5Cint%5B%284%5Csin+t-6%5Ccos+t%29%28-2%5Csin+t%29%2B%282%5Ccos+t-4%5Csin+t%29%28%5Ccos+t%29%5Ddt%2C+t+%3D+0++to+2pi

Answer 51

Since there is no division, I don't really see there being any singularity

Answer 52

um, first off, above, @SithsAndGiggles the parametrization is wrong i think?

Answer 53

theequation is x^2 + 4y^2 = 4.

Answer 54

so why would it be (2cost, sint)?

Answer 55

I'm working it out without using Green's Thm since it's really not that difficult to just do the line integral with <2cos t, sin t> as the parameterization.

Answer 56

Yeah that.

Answer 57

@Curry Because,
\[(2\cos t)^2+4\sin^2t=4\cos^2t+4\sin^2t=4\]

Answer 58

surely, the professor can't be wrong, i mean he's a universtiy of professor at UCB... and the T.A.s would have pointed it out ot be wrong if it was...

Answer 59

Even professors aren't infallible ;)

Answer 60

no i know, but other students would have pointed it out, but i guess if this is the answer we're all getting, i'll keep this one for now.

Answer 61

The professor is correct, it is the math which is lying to us.

Answer 62

Sometimes the numbers screw with the proletariat for their own well being.

Answer 63

Crazy how nature do that. Not only does directly solving the line integral but using Green's Theorem in rectangular and polar coordinates all give -6pi despite your professor being right.

Answer 64

lol, there are 250 students in the class, and 3 T.A.s, atleast one of them would have came up with a different answer than -3pi/2 is all I'm saying...

Answer 65

if that is answer is wrong...

Answer 66

3 TAs????

Answer 67

?

Answer 68

Although this is totally wrong, I'm tempted to say that they just did it for 1 quadrant and then forgot to multiply by 4. Obviously this is wrong because the line integral is different in each quadrant.

Answer 69

Look, if the ellipse was \[
x^2+4y^2=1
\]Then we would have: \[
\frac{x^2}{1^2}+\frac{y^2}{(1/2)^2}=1
\]And we would get: \[
-3\pi(1)(1/2) = -3\pi/2
\]

Answer 70

wait, back up there when Sith and the rectanglular approach of green's theorme, why is the bottom limit 0, instead of -sqrt(1-x^2/4)?

Answer 71

@wio

Answer 72

It's 0 because that's in polar coordinates and represents an integral from r=0, not y=0.

Answer 73

but it's in rectangular form, not polar though

Answer 74

Oh, when that occurs he's just using a fact that even functions obey this rule: \[\Large \int\limits_{-a}^af(x)dx=2 \int\limits_0^af(x)dx\]

Answer 75

For even functions?

Answer 76

Yes, because even functions are left-right symmetric instead of integrating the whole thing, you can just double the value of the integral on the positive half of the integral. Here the even function is f(x)=1. since f(x)=f(-x) it is even, right? Makes sense I hope.

Answer 77

\[\text{Rectangular:}\\\begin{align*}
\int_\gamma\cdots&=-3\int_{-2}^2\int_{-\sqrt{1-\frac{x^2}{4}}}^\sqrt{1-\frac{x^2}{4}}dy~dx\\\\
&=-6\int_{-2}^2\int_0^\sqrt{1-\frac{x^2}{4}}dy~dx\\\\

\end{align*}\]

So specifically this is what I'm talking about.