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

[bayan]

hi felles.

I am trying to find what is the volume of the $$y=\frac{a}{x^2}+b$$ 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 $$V=\frac{-Pi}{y^2+2Y+1}$$ where y is 3.

Is this right?

I did a U substitution to integrade $$x^2=\frac{1}{y+1}$$

Plz help.

 

[VietDao29]

It seems like your integral is wrong.
$$\int x ^ \alpha dx = \frac{x ^ {\alpha + 1}}{\alpha + 1} + C \mbox{, } \alpha \neq -1$$
$$\int \frac{1}{x} dx = \ln{x} + C$$
Viet Dao,

 

[VietDao29]

Looks good, or you can say:
$$V = \pi \int^3_0 \frac{1}{y + 1} dy = \pi(\ln{4} - \ln{1}) = \pi \ln{4}$$
Viet Dao,

 

[Orion1]

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?