# Thread: Volume of solid of revolution.

1. ## Volume of solid of revolution.

The curve $\displaystyle y = \frac{1}{1+x^2}$ is rotated about the X-axis. Find the volume enclosed between $\displaystyle x = \frac{1}{\sqrt{3}}$ and$\displaystyle x = \sqrt{3}$

How exactly would I do these types of questions? The only method I've learnt is the Simpson's rule. Is that applicable for this?

Could someone please show me how to do this???

2. It is possible to use the Simpson's rule to approximate the volume. Looking at your question though, I believe the volume is clearly meant to be calculated using calculus

Simpson's Rule Method: Lets assume each 'slice' of the area under the curve is rotated about the x-axis. This gives a circle with radius 'y' with thickness 'h', which varies based on each function value. So the idea is we add all these circles up to form the volume.

Since the area of a circle is $\displaystyle A_{circle} = \pi r^2$ we see that instead of adding up the areas of the function values (which is what the original simpson's rule does), we can square each value and multiply by $\displaystyle \pi$ to find the area. Then we multiply by h (as featured in the beginning of the formula) to find the volume for each slice. Using this piece of information we modify the Simpson's rule for area to one that calculates volume for n function values. This is:

$\displaystyle V = \pi \frac{h}{3} \big ( (f_0^2 + f_n^2) + 4 (f_1^2 + f_3^2...) + 2 (f_2^2 + f_4^2...))$

Calculus: Alternatively (and easier once you learnt it), you could use calculus which also uses the circle concept to calculate the volume.

For calculus, the simple, general formula is $\displaystyle V = \pi \int_{\frac{1}{\sqrt3}}^{\sqrt3} y^2 dx$

Answer is $\displaystyle \frac{\pi^2}{12}$

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### volumn of solid by simsons 1/3 rule

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