# Volumes, Strang

1. May 29, 2008

### rocomath

I have a simple question on the set-up of a triangular prism.

Strang, Ch. 8: Applications of the Integral

pdf page 4 (bottom) and correlates with example 5 on pdf page 5
http://ocw.mit.edu/ans7870/resources/Strang/Edited/Calculus/8.1-8.3.pdf [Broken]

How is he getting the length of $$1-\frac x h$$

Last edited by a moderator: May 3, 2017
2. May 29, 2008

### Vid

Look at the last figure on 8.3. 6(1-x/h)^2 is the area of a cross section at x. The line going through the height of each triangle is H = 3(1 - x/h). Using similar triangles the base of each triangle is B = 4(1-x/h). Thus, area = A(x) = 1/2*B*H = 6(1-x/h)^2. The volume is then obtained by adding together all the areas of each cross section which is the integral of A(x) from x = 0 to x =h.

3. May 29, 2008

### Dick

The bottom leg of the triangular section has a length of 4(1-x/h) and the side leg is 3(1-x/h).

4. May 29, 2008

### rocomath

That is a length, he says so in the next page. I don't get how he's setting it up by using similar triangles.

My main confusion comes from the 1 minus part, and the x/h part makes sense to me by just looking at the figures and what happens when x=0 & x=h.

5. May 29, 2008

### Vid

The slope of the line is -3/h and it intersects the z axis at z = 3. Thus z = -3/h*x + 3 = 3(1 - x/h). Let y be the base. 3/y*(1-x/h) = 3/4 by similar triangles. Solving for y, y= 4(1-x/h).

6. May 29, 2008

### rocomath

Oh that's easy! Lol, I didn't even think about that. Thanks Vid :)