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Find the area of the surface generated by revolving the curve $ \ x(t)=t^2+\frac{1}{2t} , \ y(t)=4 \sqrt t \ $ , $ \ \frac{1}{\sqrt 2} \leq t \leq 1 \ $ about the y-axis.

Answer:

The formula is

$ 2 \pi \int_{\frac{1}{\sqrt 2}}^{1} x(t) \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} dt \\ \\ = 2 \pi \int_{\frac{1}{\sqrt 2}}^{1} (t^2+\frac{1}{2t} ) \sqrt{4t^2-\frac{7}{4t}+\frac{1}{4t^4}} \ dt$

But this becomes complicated to evaluate the integral and even I can not evaluate it using computer also..

Can someone evaluate the integral even using computer ?

Is there shortest way?

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Hint...the expression inside the square root should be $$(2t+\frac 12t^{-2})^2$$

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  • $\begingroup$ Can you evaluate the integral what I have done using any tools? $\endgroup$
    – Why
    Apr 28 '18 at 17:52
  • $\begingroup$ That wouldn't be an answer to your original question. Are you asking a different question? $\endgroup$ Apr 28 '18 at 18:18
  • $\begingroup$ Why not ? The integral which I wrote is the formula for parametric representation . But the problem is I can not evaluate it. Do you help me ? $\endgroup$
    – Why
    Apr 28 '18 at 18:22
  • $\begingroup$ As I have said in my answer, the expression inside the square root should be something different to what you have written. Your integral is not correct. $\endgroup$ Apr 28 '18 at 18:25

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