Math 142 - Week 7 Notes

Monday, October 5

Today we looked at how to use integrals to find volumes. We started with the problem of finding the volume of a pyramid, and came up with the central idea: that the volume of a solid is the integral of the areas of its cross sections.

\[\text{Volume} = \int \text{Area} \, dx.\]

If the cross sections are disks, this formula becomes \(V = \displaystyle \int \pi r^2 \, dx\).

We used this idea to find the volume of a sphere with radius \(R\). Here is a pretty good video explanation.

Here are the examples we did in class:

  1. Find the volume of the solid formed by revolving the region between \(y = \sqrt{x}\) and the x-axis from \(x=0\) to \(x=4\) around the \(x\)-axis.

  2. Use the disk method to find the volume of a cone that is \(h\) units tall and has a base radius of \(R\) units.

The last example we did was a hollow shape. To get the volume, you use the washers method since the cross sections look like flat washers. The formula for the washers method is:

\[ V= \int \pi R^2 - \pi r^2 \, dx\]

  1. Let \(\mathcal{R}\) be the region between the curve \(y = 2-x^2\) and the line \(y = 1\). When you revolve this region around the x-axis, you get a ring shape. Use the washer method to find the volume of this ring.

Friday, October 9

Today we talked about how to find the length of a curve using integrals. There are two approaches. For a function with an explicit formula \(y = f(x)\), use:

\[\text{Length} = \int_a^b \sqrt{1+(y')^2} \, dx.\]

For functions with a parametric formula where both the x and y coordinates are functions of a third parameter \(t\), use:

\[\text{Length} = \int_a^b \sqrt{(dx/dt)^2+(dy/dt)^2} \, dt.\]

This formula has a nice interpretation if \(t\) represents time. Then \((dx/dt)\) is the horizontal velocity, \((dy/dt)\) is the vertical velocity, and \(\displaystyle \sqrt{(dx/dt)^2+(dy/dt)^2}\) is the speed of the object. Then distance traveled is the integral of speed with respect to time.

Here are some video examples: