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Integral Calculus

Blazing goIITian

Joined:	19 Mar 2007
Post:	585

23 Oct 2007 18:15:04 IST

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337

area strikes again !!!!!!!

None

the parabola y2=4ax and x2=4ay divide the square region bounded by the line x=4 and y=4 and the coordinates axes .if S1 and S2 and S3 r the
areas respectvely the areas of these parts from top to bottom then
S1 :S2 :S3 is ???????????

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Comments (4)

Rahul Sharma

Scorching goIITian

Joined: 29 Aug 2007

Posts: 222

23 Oct 2007 21:13:23 IST

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Hey I think there may be a mistake in the question, I think the lines should be x = 4a and y = 4a. If indeed there is a mistake then I am getting my answer as 1:1:1 ( each equal to 16a^2/3).

Plz correct me if I am wrong...

SPARTIAN

Blazing goIITian

Joined: 19 Mar 2007

Posts: 585

23 Oct 2007 22:34:01 IST

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no, u r wrong !!!!!!!!!
indeed the line is x=4 and y=4 as before !

Gaurav |spideyunlimited| Ragtah

Moderator

Joined: 16 Dec 2006

Posts: 3373

23 Oct 2007 23:50:50 IST

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Ans is 1:1:1 if a is 1.. im workin on the general solution.. watch this space soon! :D

Gaurav |spideyunlimited| Ragtah

Moderator

Joined: 16 Dec 2006

Posts: 3373

26 Oct 2007 01:19:57 IST

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ok.

S₁ = ₀⁴ x.dy = ₀⁴ y²/4a . dy

=| 1/4a. y³/3 | ₀⁴

= 16/3a

S₃ = ₀⁴ y.dx = ₀⁴ x²/4a .dx

= 16/3a

Now total area bounded by x axis, y axis, x = 4 and y = 4 is 4x4 = 16 sq. units

S ₂= 16 - ( S₃+ S₁)

= 16 - 32/3a = 16/3a

Hence S₁: S₂ : S₃ = 1 : 1 : 1

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