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.

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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.

Mathematics
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Ok, so how did you want to set this up? Disks?
? disks?
explain all my professor said was

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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).
this is all my professor gave me this is homework and i dont understand it
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).
explain all my professor said was
sorry, I thought you said volume, not surface area.
do you know how to do this
??
So how did you set up the integral
Or have you yet.
didnt do it yet but i know this much idk if its right
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))
\[\frac{d}{dx}x^{1/3} = \frac{1}{3\sqrt[3]{x^2}}\] That's right. So we plug in that for dy/dx in the integral (squaring it) and evaluate it as an improper integral.
~=1.885
can you explain how you got this because i need the work for it so i can try to understand it all
because ive got a final tomrrow and i sorta need to know this all
\[\lim_{b\rightarrow 0} \int_b^1 \sqrt{1+(\frac{1}{3\sqrt[3]{x^2}})^2} dx \]
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))
then its the limit as etc.. then its d/dx and then wa because now i still didnt get the answer
Hrm.. I'm not sure that's right.. Seems too awful
Ah. I forgot you multiply the radical by the original f(x). It should be \[\lim_{b\rightarrow 0} \int_b^1 \sqrt[3]{x}\sqrt{1+(\frac{1}{3\sqrt[3]{x^2}})^2} dx\]
That's much better.
so this is in order
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))
limb→0∫1bx−√31+(13x2−−−√3)2−−−−−−−−−−−−√dx
then
ddxx1/3=13x2−−√3
this doesnt come outto be the same wouldnt d/dx change since the limit changd?

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