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

1. anonymous

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

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$x=2\sin\theta\to dx=2\cos\theta d\theta$

3. anonymous

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

4. TuringTest

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

5. 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

6. anonymous

it's a trigonometric integral u substituion wont work

7. anonymous

And a minus sign outside.

8. anonymous

The substitution works pretty well only....

9. anonymous

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

10. anonymous

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

11. 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.

12. anonymous

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

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. 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

17. 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.

18. anonymous

ah k, yeah i see. Well thank you

19. anonymous

i forgot a minus sign. :D

20. anonymous

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

21. TuringTest

Never would have thought of that, cool.