## 175 – Quiz 1

Here is quiz 1.

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 $1/2$ rather than 1.)

Using the disk method, the volume can be expressed as $\int_0^{\pi/2}\pi y^2\,dx.$ Here, $y^2=\sin^2 x\cdot \cos^2 x.$ 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:

• $\sin x\cdot \cos x=\displaystyle\frac{\sin 2x}2,$ so $y^2=\displaystyle\frac{\sin^2 2x}4.$ Again, the fact that the sine expression is squared makes things difficult. We use another trigonometric identity:
• $\sin^2\theta=\displaystyle\frac{1-\cos2\theta}2,$ so $y^2=\displaystyle\frac{1-\cos 4x}8.$ Now we can proceed to integrate:

$\displaystyle\int_0^{\pi/2}\pi y^2\,dx=\displaystyle\frac{\pi}8\int_0^{\pi/2} (1-\cos 4x)\,dx=\displaystyle\frac\pi8\left.\bigl(x+\frac{\sin 4x}4\bigr)\right|_0^{\pi/2},$ which simplifies to $\displaystyle\frac{\pi^2}{16}.$

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 $x$ is given as a function of $y,$ the most efficient route seems to be to use $y$ as the parameter, so the expression for the length of the curve takes the form $\int_2^3\sqrt{(x')^2+1}\,dy,$ where the derivative of $x$ is with respect to $y.$

We have $x'=\displaystyle\frac{y^2}2-\frac1{2y^2},$ so $(x')^2+1=\displaystyle \frac{y^4}4-\frac12+\frac1{4y^4}+1=\displaystyle\frac{y^4}4+\frac12+\frac1{4y^4}=\displaystyle\bigl(\frac{y^2}2+\frac1{2y^2}\bigr)^2.$

The expression for the length then reduces to $\displaystyle\int_2^3\bigl(\frac{y^2}2+\frac1{2y^2}\bigr)\,dy=\displaystyle\left.\bigl(\frac{y^3}6-\frac1{2y}\bigr)\right|_2^3$ $=\displaystyle\bigl(\frac{27}6-\frac16\bigr)-\bigl(\frac86-\frac14\bigr)=\displaystyle3+\frac14=3.25.$