# Thread: volume of wedge cut from cylinder

1. ## volume of wedge cut from cylinder

Here's a cool calc problem you may wish to tackle. Perhaps you've seen it before.

"A wedge is cut from a cylinder of radius r by two planes, one perpendicular to the axis of the cylinder and the other making an angle @ with the first. Find the volume of the wegde by slicing perpendicular to the y-axis".

Then find the volume by slicing perpendicular to the x-axis.

Here's a diagram to help. I used @ instead of theta because I used paint to construct the diagram.

2. From the diagram below, it looks like

$\displaystyle V\ =\ 2\int_0^r{x\tan{\alpha}\sqrt{r^2-x^2}}\,\mathrm{d}x$

Which will come out to be $\displaystyle \frac{2}{3}\,r^3\tan{\alpha}$.

3. Good Jane. That's certainly it. It appears you done it perp. to the x-axis.

Here's the perp. to the y-axis version:

Let the height of the triangle be h

Then we have $\displaystyle \frac{h}{x}=tan({\theta})$

$\displaystyle h=xtan({\theta})$

The area of the triangle is $\displaystyle \frac{1}{2}hx=\frac{1}{2}x^{2}tan({\theta})=\frac{ 1}{2}(r^{2}-y^{2})tan({\theta})$

because $\displaystyle x^{2}=r^{2}-y^{2}$.

And our integral is:

$\displaystyle V=\frac{1}{2}tan({\theta})\int_{-r}^{r}(r^{2}-y^{2})dy=\frac{2}{3}r^{3}tan({\theta})$

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# volume of wedge of cylinder

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