# Lower and upper limits for a volume integral

## Homework Statement

I read an e-book about the classical mechanics and didn't know how to find the lower and upper limits for a volume integral.

## Homework Equations

Perhaps, it may be related to the use of similar triangle.

## The Attempt at a Solution

I calculated the x value only (y is the same as x because of symmetry)

z/h = x/(a/2) = 2x/a

x = az/2h (distance)

Lower Limit = -az/2h
Upper Limit = az/2h

Can anyone tell me how to obtain the same values as the one given by the author?

Thank you very much!

Related Introductory Physics Homework Help News on Phys.org
gneill
Mentor
f(z) = (1 - z/h) is a function that goes linearly from 1 down to zero as z goes from 0 to h:

z = 0 --> f(z) = 1
z = h --> z/h = 1 --> f(z) = 0

Multiply by the desired width at the base (a/2) and you've got a line that goes from x = a/2 when z = 0, to x = 0 when z = h.

You could also derive it from the equation of the line.

y = mx + b, with m = slope and b = y-intercept.

The right-hand line intercepts the y-axis at h, and the x-axis at a/2. So the slope is m = -2h/a, and b = h. Plug those into the equation and solve for x in terms of z.

Thank you very much!