Problem 1 is exercise 6.1.16 from the book. (The graph in the book has a typo. The maximum height of the graph is rather than 1.)

Using the disk method, the volume can be expressed as Here, If this expression were not squared, to find the integral would be easier, by a direct substitution. Being squared, we need to work harder. One possible approach is to use some trigonometric identities. For example:

so Again, the fact that the sine expression is squared makes things difficult. We use another trigonometric identity:

so Now we can proceed to integrate:

which simplifies to

There are other approaches. For example, other trigonometric identities could be used as well. Also, the book includes a formula for the integral of powers of sine and cosine; we will study this formula later. Trying to use the shell method leads to rather messy expressions, I didn’t work out the full details. I believe that the argument above is perhaps the most efficient.

Problem 2 is exercise 6.3.14 from the book. Since is given as a function of the most efficient route seems to be to use as the parameter, so the expression for the length of the curve takes the form where the derivative of is with respect to

We have so

The expression for the length then reduces to

43.614000-116.202000

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