Wedge

[Dr. Flim-Flam]

What happen to galactus's problem in regards to the wedge?

However, not to be dismayed, here is another one.

A wedge is cut from a right circular cylinder of radius r inches by a plane through the diameter of the base,

which makes a 45 degree angle with the plane of the base. Find the volume of the wedge cut out. I don't

know if this is the same as galactus's problem or not, however it should be similiar.

 

[galactus]

I deleted it because it didn't seem anyone was too interested. I appear to have been hasty.

I will repost it.

Here is a diagram. What you have is a specific angle. Mine was a general case.

With slices perp. to the y-axis we can use a triangle as per the diagram. 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:

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

Since your wedge has a 45 degree angle it would be \(\displaystyle \frac{2}{3}r^{3}tan(\frac{\pi}{4})=\frac{2}{3}r^{3}\)

Now, do it perp. to the x-axis.

 

[Dr. Flim-Flam]

Trick Question,as as long as we are dealing with a wedge cutting through a right circular cylinder of radius r by a plane through the diameter of the base making an angle of theta, then the volume only depends on the angle and the radius.