Surface Area of Revolved Curve: An Intriguing Challenge

[G01]

Find the surface area when this curve is revolved around the x ax

x = 1/3(y^2 + 2)^3/2 [1,2]

I set it up both ways and i get two really complicated integrals.

$$2 \pi/3 \int_1^2 \sqrt{\frac{y^2+y^4}{(y^2+2)^{2/3}}} dy$$

Yeah I can't figure out how to do this integral, and I am thinking there must be an easier way. The other way, integrating with x, seemed even more complicated If anyone can give me some hints, it'll be appreciated. Thanks.

 

[G01]

sry i seem to be having trouble getting the latex to work. I'll try fix it, but I am not to good at it so i can't promise anything.

 

[pizzasky]

Reply

Did you use the correct formula?

I tried part of the problem and it seems the final expression you needed to integrate is quite a simple one!

To find the surface area generated when a curve is rotated completely about the x-axis, you may use this formula...

$$2 \pi \int y \sqrt{1 + (\frac{dx}{dy})^2} dy$$

Please figure out the limits by yourself..

All the best!

 

[G01]

thanks a lot