Hardcore (I think) Volumes Q (1 Viewer)

[monaro-man]

Hi all, I came across this Question in my coaching homework and have no idea how to go about it. It was under the"Volumes of solids with Similar Cross-Sections" heading.

Q: Show that the volume common to two right circular cylinders intersecting at right angles and both having the same radius, 'a' units, is (16a^3)/3 units cubed.

Have fun!


PS: I will post a diagram when I get home, so try and visualise for now.

 

[wogboy]

Take one cylinder to be z^2 + y^2 = a^2 (going horizontally in xyz space), and the other cylinder to be x^2 + z^2 = a^2 (going into the page):

Now slice this region of intersection, making each slice in the xz plane. Symmetry allows us to ignore the half where z<0, and the half where y<0, and the half where x<0 (so x,y,z are all positive) for now, to get the eighth volume (we'll multiply the result by 8 at the end).

See the attached diagram to see what the cross section looks like (I can't describe it in words!) Dividing up the cross sectional region as shown in my diagram with the blue line to get the total area:

Triangle area:
y/2 * sqrt(a^2 - y^2)

Sector area:
pi*a^2 * (pi/2 - arcsin(y/a))
= pi*a^2 * arccos(y/a) / 2*pi

so A(y) = y/2 * sqrt(a^2 - y^2) + a^2 * arccos(y/a) / 2 (this is the cross sectional area as a function of y)

The eighth-volume is Int{0->a} A(y) dy. Integrating the first term is easy, it works out to a^3 / 6 (try it for yourself). Integrating the second term is somewhat trickier, because it's an inverse cosine.

Graph y = arccos(x) for 0<=x<=1. The area under this graph is simply 1 (The integral from x=0 to x=pi/2 of cos(x). So Int{0->1} arccos(x) dx = 1

so to integrate arccos(y/a) dy from y=0 to y=a, do a substitution, let y=ab -> dy = a*db, so Int{0->a} arccos(y/a) dy = a * Int{0->1} arccos(b) db = a

So to integrate the second term in A(y), it works out to a^3 / 2. Add this to the integral of the first term (a^3 / 6) and we get:

V/8 = a^3 / 2 + a^3 / 6 = 2a^3 / 3
so V = 16*a^3 / 3
(hope it made sense)

 

[wogboy]

The diagram of the cross section is attached to this post ...