Area of circle by Integration?? Watch

stargirlfran
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#1
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#1
ok... so if the general eqn is (x^2)+(y^2)=(r^2)
then y = sqroot(radius squared minus 'x' squared)
then integrate between limits zero and radius.

then i get stuck loll can someone plz explain what substitution i need to use and why...lol its the holidays...maths's gone out my head for the time being
dnt skip any steps plz im having a dumb day XD
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DFranklin
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#2
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x = r sin t.
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stevencarrwork
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You will also have to think about how much of the circle your integration represents.
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stargirlfran
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a quarter yesss i know that
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gummers
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#5
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why you doing this over summer holidays?
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stargirlfran
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(Original post by gummers)
why you doing this over summer holidays?
because all the cool kids do this :cool:

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gummers
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(Original post by stargirlfran)
because all the cool kids do this :cool:


**** son,
i better get my books out :woo: :woo:
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stargirlfran
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(Original post by gummers)
**** son,
i better get my books out :woo: :woo:

yeah..u do that.
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stargirlfran
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done it. proved the thing yeaahhhhh. dnt post solutions unless u reeli want to show off
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stargirlfran
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(Original post by DFranklin)
x = r sin t.
im sure u're right and it in fact works, but why??:o:
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DFranklin
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y = sqrt(r^2-x^2) = r sqrt(1-sin^2 t) = r sqrt(cos^2 t) = r cos t
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nuodai
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(Original post by stargirlfran)
im sure u're right and it in fact works, but why??:o:
How much do you know about coordinate geometry/polar coordinates?

If a circle has radius r and has centre (0, 0) then you can look at a point, call it P(x, y), on the circle as being a vertex of a right-angled triangle (you also have to draw a line from the point on the circle vertically down/up to a point, call it Q, on the x-axis. So, the angle OQP is a right-angle, and the length of OQ is x and the length of PQ is y. If you call the angle POQ θ, then using trigonometry you can derive the following rules:
\newline

(1):\ x = r\cos \theta \newline

(2):\ y = r\sin \theta \newline

(3):\ x^2 + y^2 = r^2 \newline

...and from there there are a load of other rules that you can derive, two of the most common ones are:
\newline

(4):\ \cos \theta = \dfrac{x}{r} = \dfrac{x}{\sqrt{x^2 + y^2}}\newline

(5):\ \sin \theta = \dfrac{y}{r} = \dfrac{y}{\sqrt{x^2 + y^2}}\newline

(6):\ \tan \theta = \dfrac{\sin \theta}{\cos \theta} = \dfrac{y/r}{x/r} = \dfrac{y}{x}
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stargirlfran
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(Original post by DFranklin)
y = sqrt(r^2-x^2) = r sqrt(1-sin^2 t) = r sqrt(cos^2 t) = r cos t
thank u
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My Alt
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Yes, this problem is so much easier using Polar integration

\displaystyle A = \frac{1}{2}\int_0^{2\pi}r^2 d\theta
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Totally Tom
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(Original post by My Alt)
Yes, this problem is so much easier using Polar integration

\displaystyle A = \frac{1}{2}\int_0^{2\pi}r^2 d\theta
errrm, where did you get that? don't you mean \displaystyle\int_0^{2\pi}\int_0  ^R\,rdrd\theta?

eugh. can't even do integration =(
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generalebriety
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(Original post by Totally Tom)
errrm, where did you get that? don't you mean \displaystyle\int_0^{2\pi}\int_0  ^R\,drd\theta? wait a second, why doesn't this give me the area of the circle? wtf.
Jacobian.

Edit: oh, you got it. Good boy. Now evaluate the r-integral to get what the poster above you had. It's also a standard A-level formula.
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Mr M
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(Original post by Totally Tom)
errrm, where did you get that?
I think he has reproduced the formula for the sector of a circle that is provided in the A Level formula book. Unfortunately, that doesn't really prove much at all!
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My Alt
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I got it from Bostock and Chandler. Basically you consider a triangle and find \frac{\delta A}{\delta\theta} and then integrate, iirc. You could potentially use it for any polar curve, if I understand correctly.
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around
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I heard it described by directly analogizing from the area of a sector: \dfrac{1}{2}r^2\theta becomes \dfrac{1}{2} \int r^2 d\theta
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My Alt
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Well the rate of change of area is equal to 1/2 r^2, integrating both sides with respect to theta gives the formula, so were approaching it from the same angle =)
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