Cylindrical tin has a 40° notch cut from the centre....

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Cylindrical tin has a 40° notch cut from the centre....

A 40 degree notch is cut to the centre of a solid cylindrical tin. The radius is 25cm. One plane face of the notch, is perpendicular to it's axis.

Find the volume of metal removed from the tin by using calculus.

Below is the diagram!
Remember radius = 25cm

cylindersm8.th.gif


I have looked at this a few ways, trying to look at a cross section drawing.. But still i don't understand because without the height we can't find the volume of the cylinder.. Perhaps, There is a formula i have overlooked?? Please help, any assistance would be greatly appreciated.

Calcwewlater!!
 
You do not need the height of the cylinder.

We can find the volume by slicing perpendicular to either axis.

This creates a bunch of triangles.

Perp. to y-axis:

\(\displaystyle tan({\theta})=\frac{h}{x}=xtan(\theta)=xtan(\frac{2\pi}{9})\)

\(\displaystyle A(y)=\frac{1}{2}hx=\frac{1}{2}x^{2}tan(\theta)=\frac{1}{2}(r^{2}-y^{2})tan(\theta)\)

because \(\displaystyle x^{2}=r^{2}-y^{2}\)

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

=\(\displaystyle tan(\theta)\int_{0}^{r}(r^{2}-y^{2})dy\)

Now, this is the general case. Enter in our given radius and angle.

\(\displaystyle \L\\\tan(\frac{2\pi}{9})\int_{0}^{25}(625-y^{2})dy\)

You can do the integration. Here's a diagram of the slices perp. to the y-axis:

wedgeof3.gif


~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Now, how about if we sliced perp. to the x-axis:

\(\displaystyle A(x)=xtan(\theta)(2\sqrt{r^{2}-x^{2})\)

=\(\displaystyle 2tan(\theta)x\sqrt{r^{2}-x^{2}}\)

\(\displaystyle V=2tan(\theta)\int_{0}^{r}x\sqrt{r^{2}-x^{2}}dx\)

\(\displaystyle \L\\V=2tan(\frac{2\pi}{9})\int_{0}^{25}x\sqrt{625-x^{2}}dx\)

You can do this integration, too. Okey-doke. You should get the same answer for both. You will.
 
awesome diagram galactus! :D

what did you use to create it?
 
How do i find the surface area of the notch?
I basically understand the formula for a surface of revoultion, but am unsure of my f(x) value to sub into that formula??
ANy assistance would be great.. Im currently looking at the area of a triangle and how my f(x) might = something like 1/2 radius times height, but im finding trouble expressing it..
Any help would be great!! Thanks
 
Just so you know what I have tried, and to show Im trying to work it out.. This is what I have done so far on surface area.. Im not sure if this is correct or not, maybe someone could verify how I am going??

S.A = \(\displaystyle \huge\\2{\pi}f(x)\int_a^{b}\sqrt{1+[f'(x)]^2}dx\\\)

Therefore


\(\displaystyle \huge\\2{\pi}x\tan\frac {2\pi}{9}\int_0^{25}\sqrt{1+(tan^2[\frac{2\pi}{9}])\)

Thus it equals
\(\displaystyle \huge\\2{\pi}x\tan\frac {2\pi}{9}.[x\sqrt{1+tan^2[\frac{2\pi}{9}])\)

Where x=25
\(\displaystyle \huge\\2{\pi}(25)\tan\frac{2\pi}{9}.[(25)\sqrt{1+tan^2[\frac{2\pi}{9}])\)

Thus i got the surface area of the notch to be Approximately 4301.495267\(\displaystyle cm^2\)
 
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