anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
Ok, so how did you want to set this up? Disks?
anonymous
  • anonymous
? disks?
anonymous
  • anonymous
explain all my professor said was

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

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