## anonymous 4 years ago **Calc2 Help** Surface Area generated by the curve y=x^3/2 from x = 0 to 3 about the x-axis.

1. amistre64

so whats our integral look like?

2. amistre64

$2pi\int \sqrt{y^2(1+dy^2)}dx$

3. anonymous

der is 3y^2/3

4. anonymous

i mean x..

5. amistre64

3-2 = 1 3/2 x^1/2 is our derivative

6. anonymous
7. amistre64

$2pi\int \sqrt{(x^{3/2})^2(1+(\frac{3}{2}x^{1/2})^2)}dx$ $2pi\int_{0}^{3} \sqrt{x^3(1+\frac{9}{4}x)}dx$

8. anonymous

you can square the 4, so now it's just pi outside the integral.

9. amistre64

is that:$x^{3/2}$or $\frac{1}{2}x^3$???

10. anonymous

where??

11. amistre64

12. amistre64

is it:$y=x^{3/2}$ or $y=\frac12 x^3$???

13. anonymous

oh it is y = (x^3)/2

14. amistre64

ahh, ok

15. anonymous

16. amistre64

let me fix up my int then :) $2pi\int \sqrt{(\frac{x^{3}}{2})^2(1+(\frac{3}{2}x^{2})^2)}dx$ $2pi\int_{0}^2 \sqrt{(\frac{x^{6}}{4})(1+(\frac{9}{4}x^{4}))}dx$ $2pi\int_{0}^2 \sqrt{(\frac{x^{6}}{4})(\frac{4+9x^4}{4})}dx$ $2pi\int_{0}^2 \sqrt{(\frac{4x^6+9x^{10}}{16})}dx$ $\frac{pi}{2}\int_{0}^2 \sqrt{4x^6+9x^{10}}dx$

17. amistre64

doesnt really pretty up does it :)

18. anonymous

ok so how did you bring the x^3/2 inside the sqrt??

19. anonymous

haha all these things are freaking ugly!!

20. amistre64

i undid it, for example $\sqrt{a^2b}=a\sqrt{b}$so to put it back in; square it

21. TuringTest

why did you not finish prettying it amistre?

22. anonymous

I tried it and got an ugly big number..

23. amistre64

lol, cause this math latex is slowing my computer like mole asses

24. TuringTest

lol$\frac{\pi}2\int_{0}^{2}x^3\sqrt{4+9x^4}dx$now it's beautiful!

25. anonymous

$2/3 * (4x^6 + 9x^(10))^(3/2)$

26. amistre64

still a bit yucky :) but im sure it trigs up nice http://www.wolframalpha.com/input/?i=integrate+%28pi%2F2%29%28sqrt%284x%5E6%2B9x%5E10%29%29+from+0+to+2

27. amistre64

sqrt(tan^2) if im not mistaken

28. anonymous

@amistre.. 52.1416 didn't work. :/

29. amistre64

im not the expert at plugging into programs .... :)

30. TuringTest

what trig?$u=4+9x^4\to du=36x^3dx$$xdx=\frac{du}{72}$$\frac{\pi}{72}\int\sqrt udu$

31. anonymous

@Turing, that is what I had..

32. TuringTest

let me see what I get

33. anonymous

I had pi/2 integral 2/3 * (-9x^4 + 1 )^ (3/2)

34. anonymous

pi/72****

35. anonymous

with limits 144 and..

36. anonymous

1.. lol

37. TuringTest

yeah I got 52.14... as well so I would have to read back and see what you guys were doing

38. anonymous

I HATE THIS STUFF!!!!!!!

39. amistre64

does yout program say to round it to some deci spot?

40. TuringTest

hey wait, I integrated from 0 to 2, not 3 let me try again...

41. anonymous

It likes fractions..

42. TuringTest

577.041397813276 let me try to get a fraction...

43. anonymous

it says 4 or 5 sig figs..

44. amistre64

http://www.wolframalpha.com/input/?i=integrate+%282pi%29*%28x%5E3%2F2%29*%28sqrt%281%2B%283x%5E2%2F2%29%5E2%29%29+from+0+to+2 even without prettying it up and just plugging in the formula I get the same 52.

45. amistre64

52.14164475991909058699913948569216045125036230794221370364473673799360642806811915388818213592188470 see if itll do that lol

46. TuringTest

but the integral is from 0 to 3, no?

47. anonymous

52.14164475991909058699913948569216045125036230794221370364473673799360642806811915388818213592188470 HAHAHAHAHAAH.. THAT WORKED!!!

48. amistre64

0 <= x <= 2

49. amistre64

:)

50. anonymous

Sorry for the caps, my excitement overtook my fingers.

51. TuringTest

well it is obviously right, but the post says 0 to 3... I put in 0 to 2 at first as well, I must have picked up on it subconsciously

52. anonymous

You guys are badass! Now it's time for me to study and to this crap on a test tomorrow morning.. Fun....

53. amistre64

good luck :)

54. anonymous

Thank ya! Take care guys.. See you next week with a new set of problems.. This time they will not be solids of revolution, but spring related force.