Volume of a solid...

[Jaskaran]

The base of a certain solid is an equilateral triangle with altitude 11. Cross-sections perpendicular to the altitude are semicircles. Find the volume of the solid, using the formula
\(\displaystyle {{V=\int_a^b A(x)\,dx}}\)
applied to the picture shown above (click for a better view), with the left vertex of the triangle at the origin and the given altitude along the x-axis.

How do I figure out what the lower and upper limits of integration are?

The diameter 2r of the semicircular cross-section is the following function of x: ?

And finally, knowing the radius, how do I integrate this problem?


Many thanks,

Cuanzo.

 

[daon]

"With altitude 11" -- I'll take that to mean the height of the triangle is 11, so we'll be integrating from 0 to 11.

The volume will be the surface area of each semicircular disk times a small change in depth (dx, dh, whatever).

The bases of the smaller equilateral triangles can be found like so: Given h is the height of the smaller triangle

\(\displaystyle s^2 = h^2 + (\frac{1}{2}s)^2 \implies s^2 = h^2 + \frac{s^2}{4} \implies s^2 = \frac{4(h)^2}{3} \implies s = \frac{2\sqrt{3}h}{3}\)

The base of the smaller triangle is 2r, hence 2r=s and that gives:

\(\displaystyle 2r = \frac{2\sqrt{3}h}{3} \implies r = \frac{h\sqrt{3}}{3} \implies \pi r^2 = \frac{\pi}{3}h^2\)

So we integrate from h=0 to h=11, noting that theyre only HALF-cricles.

\(\displaystyle \frac{1}{2}\int_0^{11}\pi r^2 dh = \frac{\pi}{6} \int_0^{11}h^2dh\)