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- anonymous

find the exact area of the regions bounded by y=x^3/(sqrt(4-x^2), y=0,x=0,x=sqrt(2)

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- anonymous

find the exact area of the regions bounded by y=x^3/(sqrt(4-x^2), y=0,x=0,x=sqrt(2)

- chestercat

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- anonymous

\[\int\limits_{0}^{\sqrt{2}} x ^{3}/\sqrt{4-x ^{2}}\]

- TuringTest

|dw:1328082060815:dw|all we're doing here is integrating up to the line x=sqrt2\[\int_{0}^{\sqrt2}\frac{x^3}{\sqrt{4-x^2}}dx\]which can be done with the substitution\[x=2\sin\theta\to dx=2\cos\theta d\theta\]

- anonymous

Or you can put 4-x^2=t and then do it..

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- TuringTest

...how would that work?
you have x^3 on top

- anonymous

Yeah so one x goes in dt and what is left is x^2 which is 4-t denominator is sqrt(t) so the net integral reduces to (4t^(-1/2)-t^(1/2))/2

- anonymous

it's a trigonometric integral u substituion wont work

- anonymous

And a minus sign outside.

- anonymous

The substitution works pretty well only....

- anonymous

if you use substitution your integral will be left in terms of both t and x, which will not work

- anonymous

if it were only x on top not x^3 you could but not in this form

- anonymous

Dude read what i have written if you put 4-x^2as t you are left with only x^2 in the numerator which is 4-t.

- anonymous

First do it and see you'll get the answer...

- anonymous

ok well what do you get by doing it your method?

- anonymous

Whatever you get by trignometric substitution you'll get the same thing only...

- anonymous

matt: Did you get what i'm trying to tell or should i elaborate further and solve the whole thing?

- anonymous

It's okay I think what you are saying does make sense, I don't actually have pen or paper right now but i'll cease to doubt you

- anonymous

\[\int\limits_{}^{}x^3/\sqrt{4-x ^{2}}\] put 4-x^2=t.
-2x dx=dt. dx=dt/2x\[\int\limits_{}^{}-x ^{3}/\sqrt{t} 2x\] cancelling x
you get x^2 on top which is 4-t\[\int\limits_{}^{}(4-t)/\sqrt{t} = (4/\sqrt{t})-\sqrt{t}\] This is easily integrable.

- anonymous

ah k, yeah i see. Well thank you

- anonymous

i forgot a minus sign. :D

- anonymous

ahha it's alright I see what you are doing now

- TuringTest

Never would have thought of that, cool.

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