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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 xaxis, for x ranging from 0 to 1.
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 xaxis, for x ranging from 0 to 1.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ok, so how did you want to set this up? Disks?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0explain all my professor said was

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Keep 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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this is all my professor gave me this is homework and i dont understand it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Keep 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).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0explain all my professor said was

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, I thought you said volume, not surface area.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you know how to do this

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So how did you set up the integral

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0didnt do it yet but i know this much idk if its right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0S = 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))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{d}{dx}x^{1/3} = \frac{1}{3\sqrt[3]{x^2}}\] That's right. So we plug in that for dy/dx in the integral (squaring it) and evaluate it as an improper integral.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0can you explain how you got this because i need the work for it so i can try to understand it all

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because ive got a final tomrrow and i sorta need to know this all

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{b\rightarrow 0} \int_b^1 \sqrt{1+(\frac{1}{3\sqrt[3]{x^2}})^2} dx \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then its the limit as etc.. then its d/dx and then wa because now i still didnt get the answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Hrm.. I'm not sure that's right.. Seems too awful

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Ah. I forgot you multiply the radical by the original f(x). It should be \[\lim_{b\rightarrow 0} \int_b^1 \sqrt[3]{x}\sqrt{1+(\frac{1}{3\sqrt[3]{x^2}})^2} dx\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 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))

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0limb→0∫1bx−√31+(13x2−−−√3)2−−−−−−−−−−−−√dx

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0this doesnt come outto be the same wouldnt d/dx change since the limit changd?
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