How to Use Integration to Find the Volume of a Rotated Function in Mathematica

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    Integrating Volumes
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Discussion Overview

The discussion revolves around calculating the volume of the solid formed by rotating the function \( y = \frac{a}{x^2} + b \) around the y-axis using integration techniques in Mathematica. Participants explore integration methods, evaluate specific cases, and inquire about graphical representations of the volume.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant presents a volume calculation resulting in \( V = \frac{-\pi}{y^2 + 2y + 1} \) for \( y = 3 \) and seeks validation.
  • Another participant challenges the correctness of the integral used in the volume calculation, providing the general form of integration for power functions and logarithmic functions.
  • A different participant suggests an alternative volume expression using integration: \( V = \pi \int^3_0 \frac{1}{y + 1} dy \) leading to \( V = \pi \ln{4} \).
  • A participant inquires about the capability of Mathematica to generate a 3D image of the rotated function and requests example code and a graphic.

Areas of Agreement / Disagreement

There is no consensus on the correctness of the initial volume calculation. Multiple viewpoints on the integration methods and results are presented, indicating ongoing debate and exploration of the topic.

Contextual Notes

Participants express uncertainty regarding the integration steps and the application of U-substitution. The discussion includes various interpretations of the volume calculation without resolving the discrepancies.

Who May Find This Useful

Students and practitioners interested in calculus, specifically in applications of integration for volume calculations, as well as those looking to utilize Mathematica for graphical representations of mathematical functions.

bayan
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hi felles.

I am trying to find what is the volume of the [tex]y=\frac{a}{x^2}+b[/tex] is when it is rotated in y-axis.

The values of a is 1 and b is -1.

max height is 3 and min is 0.

I was trying to integrade and ended up with [tex]V=\frac{-Pi}{y^2+2Y+1}[/tex] where y is 3.

Is this right?

I did a U substitution to integrade [tex]x^2=\frac{1}{y+1}[/tex]

Plz help.
 
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It seems like your integral is wrong.
[tex]\int x ^ \alpha dx = \frac{x ^ {\alpha + 1}}{\alpha + 1} + C \mbox{, } \alpha \neq -1[/tex]
[tex]\int \frac{1}{x} dx = \ln{x} + C[/tex]
Viet Dao,
 
Looks good, or you can say:
[tex]V = \pi \int^3_0 \frac{1}{y + 1} dy = \pi(\ln{4} - \ln{1}) = \pi \ln{4}[/tex]
Viet Dao,
 

Is it possible for Mathematica to generate a 3D image from a function rotated around an axis using the integration method?

Please demonstrate some source code and a graphic?

 

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