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# C4-Parametric Integration watch

1. An ellipse has the following parametric equations;

x = 5cosθ and y= 4sinθ

(b) by integration find the area enclosed by the ellipse. (4 marks)

I get to an equation, integrate it, and get to the stage where I stick in the definite integrals, but then get confuddled about how the whole thing works, needing to multiply by 2 et cetera (centre of ellipse at origin).

I will not say what the answer is, and I will just look at the method of the person who comes up with the correct answer (explaining the last bit thoroughly )

Thankyou
2. does this look right? Ok this is take 2

3. Theres a 99% chance of this being wrong but I get 20 pi.
4. (Original post by KAISER_MOLE)
An ellipse has the following parametric equations;

x = 5cosθ and y= 4sinθ

(b) by integration find the area enclosed by the ellipse. (4 marks)
Right I'll have a quick go.

x = 5cosθ, y = 4sinθ.

dx/dθ = - 5sinθ

∫y(dx/dθ)dθ = Area

=> ∫{(4sinθ)(-5sinθ)}dθ

=> ∫{-20sin²θ}dθ

Recall: 1 - 2sin²θ = cos2θ

==> sin²θ = ½ -½cos2θ

=> -20∫{½ -½cos2θ}dθ

=> -20{½θ + ¼sin2θ}dθ

Now limits (don't work very well in theis text box) but between 0 and pi/2:

=> -20{ [pi/4] - [ 0 ] } => 5pi

=> To give full area across 4 quadrants times by 4 ==> 20pi.
5. (Original post by samdavyson)
Working on it.
Ye i was totally wrong first, second, third times. Fourth time lucky Im really bad at the basics
6. there is also a 1 % chance of you be right ljfrugn can you show how you got this value please?
7. (Original post by KAISER_MOLE)
there is also a 1 % chance of you be right ljfrugn can you show how you got this value please?
look up i ammended my post
8. (Original post by Syncman)
Ye i was totally wrong first, second, third times. Fourth time lucky Im really bad at the basics
How do you integrate the sin²theta ?

Which identity is that?

EDIT: I've got it I think.
9. You use cos2 theta = 1 - 2 sin^2 theta.
10. o jus learnt the integral.
But i suppose sin²x = (1/2)-(1/2)cos(2x) ?
from cos2x = 1 - 2sin²x
11. I did it slightly differently, I will post it in a min.

EDIT: Well, I can't figure out MathType .
12. (Original post by Syncman)
o jus learnt the integral.
But i suppose sin²x = (1/2)-(1/2)cos(2x) ?
from cos2x = 1 - 2sin²x
Got you.

I was just confused as you wrote the integral as ... (sinxcosx)/2 which is the same as what I have now got ... (sin2x)/4.
13. (Original post by samdavyson)
Got you.

I was just confused as you wrote the integral as ... (sinxcosx)/2 which is the same as what I have now got ... (sin2x)/4.
ye i should have a minus sign in there doesnt matter in this case though.
14. (Original post by Syncman)
ye i should have a minus sign in there doesnt matter in this case though.
You doing C4 in a few Tuesday's time (28th) ?

I need some serious revision.
15. (Original post by ljfrugn)
I did it slightly differently, I will post it in a min.

EDIT: Well, I can't figure out MathType .
It is a bit of an irritation.
16. (Original post by samdavyson)
You doing C4 in a few Tuesday's time (28th) ?

I need some serious revision.
nope i doin P5
17. (Original post by Syncman)
nope i doin P5
Well Good Luck with it!

I get that pleasure ( ) next year.
18. Cheers all, it is that four quadrant bit at the end which gave me a little stumping

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