Integrating & finding Volumes!

  • Thread starter bayan
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  • #1
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Main Question or Discussion Point

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 hight 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.
 

Answers and Replies

  • #2
VietDao29
Homework Helper
1,423
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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,
 
  • #3
VietDao29
Homework Helper
1,423
2
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,
 
  • #4
970
3

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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