# Thread: Papus theorem

1. ## Papus theorem

Question 1

Use the Theorem of Pappus to find the volume of the solid generated when the region bounded by the x-axis and the semicircle y= sqrt(a^2 -x^2) is revolved about

a) the line y=-a
b) the line y=x-a

Question 2
Use the Theorem of Pappus to find the centroid of the triangular region with vertices (0,0), (a,0), and (0,b) , where a> 0 and b>0 .
Hint Revolve the region about the x-axis to obtain y bar and about the y-axis to obtain x bar.

I don't quite understand the Pappus theorem. Please do help you know how to solve them. Thank you very much.

2. #1:

$\displaystyle \overline{x}=0$ from the symmetry of the region, $\displaystyle \frac{{\pi}a^{2}}{2}$ is the area of the semicircle, $\displaystyle 2{\pi}\overline{y}$ is the distnace traveled by the centroid to generate the sphere so $\displaystyle \frac{4}{3}{\pi}a^{3}=(\frac{{\pi}a^{2}}{2})(2{\pi }\overline{y}), \;\ \overline{y}=\frac{4a}{3{\pi}}$

Now for part a, $\displaystyle V=\left[\frac{1}{2}{\pi}a^{2}\right]\left[2{\pi}\left(a+\frac{4a}{3{\pi}}\right)\right]=\frac{1}{3}{\pi}(3\pi+4)a^{3}$

See if you can tackle part b.

#2:

Revolve the region about the x-axis to get $\displaystyle \overline{y}$ and about the y-axis to get $\displaystyle \overline{x}$.

The region generates a cone of volume $\displaystyle \frac{1}{3}{\pi}ab^{2}$

when revolved about the x-axis, the area of the region is

$\displaystyle \frac{1}{2}ab$, so $\displaystyle \frac{1}{3}{\pi}ab^{2}$$\displaystyle =\left(\frac{1}{2}ab\right)(2{\pi}\overline{y})$$\displaystyle , \;\ \overline{y}=\frac{b}{3}$.

A cone of volume $\displaystyle \frac{1}{3}a^{2}b$ is generated when the

region is revolved about the y-axis so $\displaystyle \frac{1}{3}{\pi}a^{2}b$$\displaystyle =\left(\frac{1}{2}ab\right)(2{\pi}\overline{x}), \;\ \overline{x}=\frac{a}{3}$.

The centroid is.........?

3. Hi galatus,

Thank you very much for your reply.

I don't fully understand this question.

I know the Pappus theorem is
V = area R * distance traveled by the centroid.

You have showed me

So the second part is the distance traveled by the centroid. Why equals 2pi ( a + (4a)/(3pi))? UM... just don't get it!

4. Hello again, Kittycat. As before, you are giving up too soon. There is nothing here that should be a surprise to you.
1. Your axis of revolution is y = -a.
2. Your Centroid is at $\displaystyle (0,\frac{4a}{3\pi})$
3. The distance from the Centroid to the Axis of Rotation is the Radius of the circle travelled. $\displaystyle \left(\frac{4a}{3\pi} - (-a)\right)$
4. $\displaystyle 2\pi\;r$ is the formula for the circumference of a circle, given its radius.
P.S. Pep talk - pep talk - pep talk. More confidence. There is more stuff in your brain than you think. Reach in a bring it out.

5. hi TKHunny,

I am not that great - don't have so much good stuff in my brain as you thought.

I think you are great - with brilliant intellects!

What about for part b - the line y=x-a?
Can you explaining to me how to get the distance from the centroid to the line y=x-a?

Thank you very much.

6. Originally Posted by kittycat
I am not that great - don't have so much good stuff in my brain as you thought.
Hogwash!

Can you explaining to me how to get the distance from the centroid to the line y=x-a?
Several ways. One of the most useful formulas ever, the Distance from a point (a,b) to a line cx + dy + e = 0 is:

$\displaystyle \frac{|c*a + d*b + e|}{\sqrt{c^{2}+d^{2}}}$

It is a beautiful thing.