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

[MarkFL]

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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:

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

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