please help me with this math problem?

Question

The base of a certain solid is a circular disk with diameter AB of lenght 2a.Find the volume of the solid if each cross section perpendicular to AB is an equilateral triagle? i reall y dont get this


an observatory is shaped like a solid whose base is a circular disk with diameter AB of length 2a.Find thr volume of this solid if the cross section perpendicular to AB is a square? i dont know how to solve it

THANKS

Answer 1

Both problems you ask have something in common. You have a solid object, whose base is circular. If you look at any cross section perpendicular to the base, it is a regular polygon (I am generalizing, you have an equilateral triangle and a square, both are regular polygons).

If you imagine the base to be horizontal, any vertical cross section is a regular polygon . One side of this polygon is a chord of the circle that forms the base.

If you think of a thin vertical slice, of thickness t, its volume will be

Volume = area x t

But the area of the slice is proportional to the square of its side, which is a chord of the circle, so

Volume = k x (Chord)^2 x t

for the square case, k = 1.

The chord that makes the half angle h at the center will be

Chord = 2a sin h

Both your problems now depend on
1. Integrating (2a sin h)^2 da over a = 0 to pi.
2. Finding the appropriate k for the type of polygon.

I am not doing the rest of the work, but hope you will solve any similar problem, not just these two.

Note: If you do not know the concept of integration, you can still solve the first problem using a formula for the volume of the cone. And then use that to find the volume for other regular polygons.

Answer 2

in both they are describing the 3 dimensional object.

The first one is a inverted cone (if you visualise taking a slice of a cone from the circular base to the tip you get a equilateral triangle)

We know the diameter from A to B on the disk is 2a in length.
We are told the triangle is equilateral so we know all the sides are the same length.

We can work out using pythagorus that the height of the cone is Sqrt(3a^2)

Volume of a cone is
V = 1/3.πr^2h

So we get v = sqrt(3)/3.π(a^3)

The second one is a cylinder
Like in the previous problem we know the diameter of the disk is 2a from A to B.

Now as we know the cross section is square we know that the heing is 2a so we can work out the volume using the formular for the volume of a Cylinder which is πr^2h

we have r = a and h = 2a so we get
2πa^3

Edited: as I suddenley realised I had been being stupid.

Answer 3

This is two problems.In the first paragraph, you describe a solid CONE. In the second, the cone is truncated - i.e., shortened. And truncated to a specific height, which is AB or 2a, so that it looks like a cylinder instead of a cone.

A circle does not have length. It only has diameter. So 2a and AB are confusing.

It's a calculus problem. You need to write an equation for two different things. --the volume of a cone, by triangualar section, and that of a truncated cone, by square sections.

Answer 4

Question no 1 is an inverted cone, and question no. 2 is sphere.

Volume of cone = 1/2 of pi mulitply by square of radius and height of the triangle. r = 1/2 of ABcos60 = .5acos60
and height of triangle is equal to a also since the base of trianlge is equal to radius, therefore, Volume is equal to (1/2)pi a2 x .5acos60

for question no. 2

V = pi x square of radius x height of square

Answer 5

Take better notes in class