Cross-sections

ProBasket

The base of a certain solid is an equilateral triangle with altitude 14. Cross-sections perpendicular to the altitude are semicircles. Find the volume of the solid, using the formula [tex]V= \int_a^{b} A(x)dx[/tex]

applied to the picture shown attached to this post (click for a better view), with the left vertex of the triangle at the origin and the given altitude along the -axis.

i figure a= 0 and b = 14 and i will be plugging it into the formula pi*r^2 right?

the problem i'm having is finding the cross-section. can someone lend me a hand?

Galileo

You'll be integrating from x=0 to x=14 right?

First, find the length of one side of the equilateral, let's call it 2R.

It's clear at x=0, the radius is zero and at x=14 the radius is R, and the radius varies linearly with x. So set up an equation for the radius in terms of x.
then you can find A(x) and integrate.

Jayboy

Think of a volume of revolution formed by a general line y = mx. What sort of shape does this yield and how is it related to the shape you are given?

Thinking about this will also provide a good check as to whether your answer is correct as there is a nice formula for its volume.

Similar Threads for: Cross-sections
Thread	Forum	Replies
cross sections	High Energy, Nuclear, Particle Physics	30
Cross sections on the web	High Energy, Nuclear, Particle Physics	4
extinction cross-sections	Advanced Physics Homework	1
Need some help with cross sections	General Physics	1

Galileo Recognitions: Homework Help Science Advisor	You'll be integrating from x=0 to x=14 right? First, find the length of one side of the equilateral, let's call it 2R. It's clear at x=0, the radius is zero and at x=14 the radius is R, and the radius varies linearly with x. So set up an equation for the radius in terms of x. then you can find A(x) and integrate.

Jayboy	Think of a volume of revolution formed by a general line y = mx. What sort of shape does this yield and how is it related to the shape you are given? Thinking about this will also provide a good check as to whether your answer is correct as there is a nice formula for its volume.