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

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SUMMARY

The discussion focuses on calculating the area of the surface of revolution for the function \(y=\frac{1}{3}x^3\) from \(x=0\) to \(3\) about the y-axis. The integral used is \(S=\int_{0}^{9} 2 \pi x \sqrt{1+\left(\frac{dx}{dy}\right)^2}\,dy\), which is transformed through various substitutions to simplify the computation. Participants explore multiple approaches, including integrating with respect to \(x\) and utilizing substitutions like \(3y = u\) and \(u = z^3\), ultimately leading to a more manageable form of the integral.

PREREQUISITES
  • Understanding of surface area calculations for solids of revolution.
  • Familiarity with integral calculus, particularly substitution techniques.
  • Knowledge of derivatives and their applications in integration.
  • Proficiency in manipulating and simplifying integrals involving square roots and polynomial expressions.
NEXT STEPS
  • Study the application of the surface area formula \(S=2\pi\int_a^b y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx\) for different functions.
  • Learn about hyperbolic functions and their derivatives, particularly \(\sinh^{-1}(x)\) and its applications in integration.
  • Explore advanced integration techniques, including integration by parts and the product rule in calculus.
  • Practice solving surface area problems involving different axes of rotation and their respective integrals.
USEFUL FOR

Mathematicians, engineering students, and anyone interested in advanced calculus, particularly those focusing on surface area calculations and integration techniques.

  • #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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