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Volume and surface area of semi-circular groove. watch

1. I am looking at Stroud Engineering Mathematics Chapter 19 question 16.

"A groove, semi-circular in section and 1cm deep, is turned in a solid cylindrical shaft of diameter 6cm. Find the volume of material removed and the surface area of the groove."

So from frame 39 in the chapter, Pappus says:-

"If an arc of a plane curve rotates about an axis in its place, the area of the surface generated is equal to the length of the line multiplied by the distance travelled by its centroid"

and

"If a plane figure rotates about an axis in its plane, the volume generated is equal to the area of the figure multiplied by the distance travelled by its centroid."

So I am looking for the volume of material removed and think its something to do with the second part.

The area of the semi-circle is pi/2 (as r is 1).

The centroid of a semi-circle from the centre is 4/3pi. My semi circle is has the centre facing upwards and is on the surface of a 6cm diameter cylinder. So the distance of the centroid from the origin is 3-4/3pi.

The centroid therefore travels 2pi(3-4/3pi)

And so the volume removed is pi/2*2pi*(3-4/3pi) = 25.42 the answer in the back. So I am happy.

I then go to the surface area of the groove from the first pappus statement must be the length of the arc * distance travelled by its centroid.

So the length of my arc is pi (as r = 1). The distance travelled by the centroid is surely the same as in the first part and so I should have pi*2pi*(3-4/3pi) = 50.84, which isn't the answer in the book.

Any idea what I am missing?

Thanks
Anthony
2. (Original post by antlee)
Any idea what I am missing?
The centroid of a solid semicircle (an area) is different to the centroid of the arc of a semicircle (a line).
3. Thanks, I see.

I looked up the centroid of the semi-circular arc (line) and it's 2/pi. And I pop that into what I was doing before and get the answer in the book.

What I can't quickly find on web is the derivation of this.

I derived the centroid of the shape myself

Area*y(bar) = 1/2 int (y^2) dx between -1 and 1 (radius is 1 in this question) and this game me 4/3pi

Can you point me in the direction of how to derive 2/pi or suggest what I integrate and divide by the length of the line.

Thanks
Anthony
4. (Original post by antlee)

Can you point me in the direction of how to derive 2/pi or suggest what I integrate and divide by the length of the line.

Thanks
Anthony
I'd suggest using polar co-ordinates.

r is obviously 1, but I've put it in as r for clarity of the calculation.

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