Area of Solid: x=a(cos t)+(1/2)alntan^2[t/2],y=a(sin t)

[arjunverma]

What would be the area of the solid generated by revolving the curve x=a(cos t) + (1/2)alntan^2[t/2] ,y=a(sin t) about its asymptote?

I guess the answer is 4(pie)a^2?Could you please help?

I request you please show me some basic steps too.

Thank you

 

[Deleted member 4993]

Re: Area of Solid

What would be the area of the solid generated by revolving the curve x=a(cos t) + (1/2)alntan^2[t/2] ,y=a(sin t) about its asymptote?

I guess the answer is 4(pie)a^2?Could you please help?.....Really - Guess!!!

I request you please show me some basic steps too.

Thank you

Please show your work - so that we can help you better.

 

[galactus]

First of all, what is this 'pie' business?.

It would appear the x-axis is the asymptote.

What are the limits of integration?. Were any given?. I assume 0 to Pi

\(\displaystyle \L\\2{\pi}\int_{a}^{b}y(t)\sqrt{(x'(t))^{2}+(y'(t))^{2}}dt\)


I may venture to guess that this problem involves some sort of insight to solve, therefore, not requiring the brute force needed to evaluate the integral. As it is, it turns into a monstrosity.
Take note, the solution you have, \(\displaystyle 4{\pi}a^{2}\), is the surface area for a sphere. Can you relate this problem to a sphere?.
If you revolve \(\displaystyle x=a\cdot{cos(t)}, \;\ y=a\cdot{sin(t)}, \;\ 0\leq{t}\leq{\pi}\) about the x-axis, the surface area of a sphere is given. Which is \(\displaystyle 4{\pi}a^{2}\)