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# Area of circle by Integration?? watch

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
2. x = r sin t.
3. You will also have to think about how much of the circle your integration represents.
4. a quarter yesss i know that
5. why you doing this over summer holidays?
6. (Original post by gummers)
why you doing this over summer holidays?
because all the cool kids do this

7. (Original post by stargirlfran)
because all the cool kids do this

**** son,
i better get my books out
8. (Original post by gummers)
**** son,
i better get my books out

yeah..u do that.
9. done it. proved the thing yeaahhhhh. dnt post solutions unless u reeli want to show off
10. (Original post by DFranklin)
x = r sin t.
im sure u're right and it in fact works, but why??
11. y = sqrt(r^2-x^2) = r sqrt(1-sin^2 t) = r sqrt(cos^2 t) = r cos t
12. (Original post by stargirlfran)
im sure u're right and it in fact works, but why??
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:

...and from there there are a load of other rules that you can derive, two of the most common ones are:
13. (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
14. Yes, this problem is so much easier using Polar integration

15. (Original post by My Alt)
Yes, this problem is so much easier using Polar integration

errrm, where did you get that? don't you mean ?

eugh. can't even do integration =(
16. (Original post by Totally Tom)
errrm, where did you get that? don't you mean ? 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.
17. (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!
18. I got it from Bostock and Chandler. Basically you consider a triangle and find and then integrate, iirc. You could potentially use it for any polar curve, if I understand correctly.
19. I heard it described by directly analogizing from the area of a sector: becomes
20. 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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