0
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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

$\endgroup$
4
  • $\begingroup$ Can you evaluate the integral what I have done using any tools? $\endgroup$
    – MAS
    Apr 28, 2018 at 17:52
  • $\begingroup$ That wouldn't be an answer to your original question. Are you asking a different question? $\endgroup$ Apr 28, 2018 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 ? $\endgroup$
    – MAS
    Apr 28, 2018 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, 2018 at 18:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.