Math 142 - Week 7 Notes

Monday, April 5

Today we looked at two applications of integrals. First we looked at finding areas between curves. We did these two examples:

  1. Find the area between the parabola \(y = 2-x^2\) and the line \(y = x\).

  2. Find the area between \(y = \sin x\) and \(y = \cos x\).

Then we discussed 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 \(V = \displaystyle \int \pi r^2 \, dx\).

Here are the examples we did in class:

  1. Find the volume of a square pyramid if the base is 100 meters by 100 meters, and the height is 50 meters.

  2. 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.

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

  4. Find the volume of a sphere with radius \(R\) by revolving \(y = \sqrt{R^2 - x^2}\) around the x-axis. Here is a pretty good video explanation.

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:

In addition to the two examples in the videos, we did the following other examples in class:

  1. Write down an integral that represents the length of the parabola \(y = x^2\) from \(x = -2\) to \(x=2\).

  2. Calculate the arc length of the parametric curve \(x(t) = \cos t\) and \(y(t) = t + \sin t\) from \(t = 0\) to \(t = \pi\). Hint: This one needed the half-angle formula \(\displaystyle \cos^2 \left(\frac{t}{2} \right) = \frac{1+\cos t}{2}\).

  3. Write down an integral that represents the perimeter of the ellipse with parametric equations \(x(t) = 2 \cos t\), \(y(t) = \sin t\).

  4. Use a Riemann sum with \(N=1000\) rectangles to estimate the perimeter of the ellipse in the previous problem.

  5. The parametric curve \(x(t) = \ln(\sec t + \tan t)\), \(y(t) = \sec t\) is a catenary which is the shape of a hanging cable connecting two points. Find the length of this catenary from \(t = -\frac{\pi}{4}\) to \(t = \frac{\pi}{4}\).