## anonymous 5 years ago Let f(x) = the cube root of x. Find the area of the surface generated by revolving the curve y = f(x) around the x-axis, for x ranging from 0 to 1.

1. anonymous

Ok, so how did you want to set this up? Disks?

2. anonymous

? disks?

3. anonymous

explain all my professor said was

4. anonymous

Keep in mind that the derivative of the cube root of x tends to infinity as x tends to 0, so the area will be equal to an improper integral. Be sure that you express your answer so that it is an actual real number, not an expression which evaluates to infinity minus infinity. For example, if the indefinite integral were ln(x) + ln(2/x), you would not want to plug in x = 0, and write the answer as ln(0) + ln(2/0). Instead, you would want to simplify the indefinite integral to ln(x times 2/x) = ln(2).

5. anonymous

this is all my professor gave me this is homework and i dont understand it

6. anonymous

Keep in mind that the derivative of the cube root of x tends to infinity as x tends to 0, so the area will be equal to an improper integral. Be sure that you express your answer so that it is an actual real number, not an expression which evaluates to infinity minus infinity. For example, if the indefinite integral were ln(x) + ln(2/x), you would not want to plug in x = 0, and write the answer as ln(0) + ln(2/0). Instead, you would want to simplify the indefinite integral to ln(x times 2/x) = ln(2).

7. anonymous

explain all my professor said was

8. anonymous

sorry, I thought you said volume, not surface area.

9. anonymous

do you know how to do this

10. anonymous

??

11. anonymous

So how did you set up the integral

12. anonymous

Or have you yet.

13. anonymous

didnt do it yet but i know this much idk if its right

14. anonymous

S = 2π ∫(x = a,b) √(1 + (dy/dx)²) dx Differentiate f(x) = ∛(x) = x^(1/3) to get: f'(x) = (1/3)x^(-2/3) = 1/(3x^(2/3))

15. anonymous

$\frac{d}{dx}x^{1/3} = \frac{1}{3\sqrt[3]{x^2}}$ That's right. So we plug in that for dy/dx in the integral (squaring it) and evaluate it as an improper integral.

16. anonymous

~=1.885

17. anonymous

can you explain how you got this because i need the work for it so i can try to understand it all

18. anonymous

because ive got a final tomrrow and i sorta need to know this all

19. anonymous

$\lim_{b\rightarrow 0} \int_b^1 \sqrt{1+(\frac{1}{3\sqrt[3]{x^2}})^2} dx$

20. anonymous

so first its S = 2π ∫(x = a,b) √(1 + (dy/dx)²) dx Differentiate f(x) = ∛(x) = x^(1/3) to get: f'(x) = (1/3)x^(-2/3) = 1/(3x^(2/3))

21. anonymous

then its the limit as etc.. then its d/dx and then wa because now i still didnt get the answer

22. anonymous

Hrm.. I'm not sure that's right.. Seems too awful

23. anonymous

Ah. I forgot you multiply the radical by the original f(x). It should be $\lim_{b\rightarrow 0} \int_b^1 \sqrt[3]{x}\sqrt{1+(\frac{1}{3\sqrt[3]{x^2}})^2} dx$

24. anonymous

That's much better.

25. anonymous

so this is in order

26. anonymous

so first its S = 2π ∫(x = a,b) √(1 + (dy/dx)²) dx Differentiate f(x) = ∛(x) = x^(1/3) to get: f'(x) = (1/3)x^(-2/3) = 1/(3x^(2/3))

27. anonymous

limb→0∫1bx−√31+(13x2−−−√3)2−−−−−−−−−−−−√dx

28. anonymous

then

29. anonymous

ddxx1/3=13x2−−√3

30. anonymous

this doesnt come outto be the same wouldnt d/dx change since the limit changd?