Day 41, Mat129: Calculus I

• Your 5.1 homework is due Monday.
• Your 4.5 homework is not yet graded, nor are your exam revisions. Monday....

Let's see how to compute volumes of some simple objects using integrals.

Before you may have memorized some of these formulas. Now you can derive them! This is the power of calculus.

In every case for what follows, we begin with what I consider the most important formula for integral calculus:

To calculate any quantity $V$, we simply divide it up into infinitesimal quantities $dV$, and add an infinity of them up.

Now for volumes we usually know the extent of the object, so we'll march along the x-axis from a to b, stop at x, and add in the little $dV(x)$ that is found there. So in each case what we need to do is find $dV(x)$

• The cube of side length $s$:
  $dV(x)=A(x)dx=s^2dx$
  $V=\int_{0}^{s}dV(x)$

• The right circular cylinder of radius $r$ and height $h$:
  $dV(x)=A(x)dx=\pi{r}^2dx$
  $V=\int_{0}^{h}dV(x)$

• The cone of radius $r$ and height $h$:
  $dV(x)=A(x)dx=\pi\left(r(1-\frac{x}{h})\right)^2dx=\pi{r^2}(1-\frac{2x}{h}+\frac{x^2}{h^2})dx$
  $V=\int_{0}^{h}dV(x)$

• The pyramid of height $h$, and base $s$:
  $dV(x)=A(x)dx=\left(s-sx/h\right)^2dx=s(1-\frac{2x}{h}+\frac{x^2}{h^2})dx$
  $V=\int_{0}^{h}dV(x)$

• The sphere of radius $r$:
  $dV(x)=A(x)dx=\pi\left(\sqrt{r^2-x^2}\right)^2dx=\pi(r^2-x^2)dx$
  $V=2\int_{0}^{r}dV(x)$

  (the "2" is out in front because we're using symmetry: we calculate the volume of half the sphere in running from 0 to r, then double it).

• Examples:
  • #48, p. 361
  • #52
  • #4 and 5, p. 360
  • #7, p. 360