# Find the area of the surface generated by revolving the curve

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.

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?

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