MHB Surface of revolution - y = (1/3)x^3

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The discussion revolves around finding the area of the surface of revolution for the function y = (1/3)x^3 from x = 0 to 3, about the y-axis. Participants explore various integral forms and substitutions, expressing frustration over the complexity of the integrals involved, particularly with the tangent substitution leading to difficult integration by parts. There is a debate on whether to integrate with respect to x or y, with suggestions that differentiating with respect to x might simplify the process. The conversation highlights the importance of recognizing known derivatives and the potential for rewriting integrands to facilitate integration. Ultimately, the discussion underscores the challenges and strategies in solving surface area integrals in calculus.
  • #31
Rido12 said:
...
At the end of your solving, did you end up plugging $\tan^{-1}(9)$ in? Or did you convert the bounds to $z$? I've always avoided changing bounds to that of radians because of situations like this.

Yes, I used:

$$\theta=\tan^{-1}(9)\implies\tan(\theta)=9,\,\sec(\theta)=\sqrt{9^2+1}=\sqrt{82}$$

$$\theta=0\implies\tan(\theta)=0,\,\sec(\theta)=\sqrt{0^2+1}=1$$
 

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