Class 7 - Area and volume: disk method (Tuesday, 9/19/2017)

As always, we start by reviewing some due dates.

Drop deadline

9/25 Monday

Team HW 1

Submit your Team HW 1 write-ups now.

1st Gateway (Computer based Pass/Fail test)

Where? B069 EH

When? Sept 11 ~ Sept 25 (Hours are the same as Math Lab hours.) 

You have two trials a day, and if you don't pass during the above period, then you will loose 1/3 of your letter grade (i.e., B- to C). The contents will be about derivatives, which is prerequisite for this course.

Web Homework

Here are your web homework dues:

• 7.1 : 9/20 Wed
• 7.2 : 9/21 Thu

Let's now talk about mathematics now. Today, we will roughly cover Sections 8.1 and half of 8.2 of your book (but not all of them). 

Warm-up exercises

Since you have not computed integrals for a while, here are some exercises:

Exercise*. Compute

\[\int \frac{3x + 11}{x^{2} - x - 6} dx.\]

(Hint: First, factor the denominator $x^{2} - x - 6 = (x + 2)(x - 3)$. Second, try to find a partial fraction decomposition.)

Exercise. Compute $\int_{0}^{1} \sqrt{1 - x^{2}}dx.$

(Hint: Take $x = \sin(t).$ Then when $x$ goes from $0$ to $1$, we have $t$ going from $0$ to $\pi/2.$ We have $dx = \cos(t)dt,$ and use the Pythagorian theorem $\cos^{2}(t) + \sin^{2}(t) = 1$ to simplify the integral. You will need to use the identity $\cos(2t) = 2\cos^{2}(t) - 1$ to finish your computation.)

Review of some trigonometric identities

The following identities are often used are

• $\sin(2t) = 2\cos(t)\sin(t)$;
• $\cos(2t) = 2\cos^{2}(t) - 1 = 1- 2\sin^{2}(t)$.

Optional reading: Euler's magic formula. Euler's formula says 

$e^{it} = \cos(t) + i\sin(t)$,

where $i = \sqrt{-1}$ is the imaginary number. This identity works for all real number $t,$ so we may consider taking $2t$ instead of $t$ to get 

$\cos(2t) + i\sin(2t) = e^{2it} = (e^{it})^{2} = (\cos^{2}(t) - \sin^{2}(t)) + i(2\cos(t)\sin(t)).$

This implies that $\cos(2t) = \cos^{2}(t) - \sin^{2}(t)$ and $\sin(2t) = 2\cos(t)\sin(t).$ Now, to get what we had for $\cos(2t)$ above, just apply $\cos^{2}(t) + \sin^(t){2}(t) = 1$ to what we have derived.

Computing the area of a circle

Ever since you were born, you were told that the area of a circle whose radius is $r > 0$ is equal to $\pi r^{2}$. We have developed enough theory of integration to see why. In this section, we study how to get this answer rather closely.

Let us compute that the semi-circle (i.e., the half of the circle) has area $\pi r^{2} / 2$. Recall that the equation for the circle with radius $r$ whose center lies at the origin $(0, 0)$ is given by:

$x^{2} + y^{2} = r^{2}.$

We will be only thinking about the semi-circle, so we only need to think about everything above $x$-axis. This means that we are giving the condition $y \geq 0$. We can write the above equation as

$y^{2} = r^{2} - x^{2}.$

This is same as saying $y = \pm \sqrt{r^{2} - x^{2}},$ but we already have told ourselves that $y \geq 0,$ so we must have

$y = \sqrt{r^{2} - x^{2}}.$

That is $y = y(x)$ is now a function in $x.$ By our definition of (definite) integral, the area of the semi-circle is equal to

\[\int_{-r}^{r}\sqrt{r^{2} - x^{2}}dx.\]

We parametrize $x$ by the substitution $x = r\sin(t).$ As $t$ goes from $-\pi/2$ to $\pi/2$, our $x$ goes from $-r$ to $r$. Moreoever, we have $dx = r\cos(t)dt.$ By substitution method, the above integral is equal to

\begin{align*}\int_{-\pi/2}^{\pi/2}\left(\sqrt{r^{2}(1 - \sin^{2}(t))}\right)r\cos(t)dt &= \int_{-\pi/2}^{\pi/2}\left(\sqrt{r^{2}\cos^{2}(t)}\right)r\cos(t)dt \\ &= \int_{-\pi/2}^{\pi/2}r^{2}\cos^{2}(t)dt \\ &= r^{2}\int_{-\pi/2}^{\pi/2}\cos^{2}(t)dt \\ &= \int_{-\pi/2}^{\pi/2}r^{2}\cos^{2}(t)dt \\ &= \dfrac{r^{2}}{2}\int_{-\pi/2}^{\pi/2}(1 + \cos(2t))dt \\ &= r^{2}\int_{-\pi/2}^{\pi/2}\dfrac{1 + \cos(2t)}{2}dt \\ &= \dfrac{r^{2}}{2}\int_{-\pi/2}^{\pi/2}(1 + \cos(2t))dt \\ &= \dfrac{r^{2}}{2}\left.\left(t + \sin(2t)/2\right)\right|_{t = -\pi/2}^{t = \pi/2} \\ &= \dfrac{\pi r^{2}}{2}. \end{align*}

This verifies that the area of the semi-circle with radius $r > 0$ is equal to $\pi r^{2}/2$. Therefore, the area of the circle with radius $r > 0$ must be $\pi r^{2}$.

Computing the area of a volume: disk method

Consider the area given by the graph of $y = x^{2}$ from $x = 0$ to $x = 1$ under the graph and above the $x$-axis. Now, rotate the area with respect to the $x$-axis. You are going to get a top-looking figure.

Question. What is the volume of this figure?

Our strategy is to similar to who we computed the area when we did not know fundamental theorem of calculus. We slice the figure perpendicular to the $x$-axis in many pieces. Let $\Delta x$ to be thickness of each piece, which approximately looks like a cylinder. When we look at the $x$-coordinate, say $x$ much far away from $0,$ the radius of the cylinder at that position is given by $x^{2},$ because that's the function we use. Hence, the volume of the cylinder must be $\pi(x^{2})^{2} \Delta x = \pi x^{4} \Delta x,$ and we are adding this up to approximate the actual volume. The sum of the volumes of cylinders is given by $\sum_{x} \pi x^{4} \Delta x,$ where $x$ ranges over your partition of the interval $[0, 1]$. When the length of your partition goes to $0$, the same limiting process we discussed for the area says that the desired volume is equal to

\[\int_{0}^{1} \pi x^{4} dx.\]

Here, remember that 

1. $[0, 1]$ is the interval where your $x$ moves;
2. $\pi(x^{2})^{2} = \pi x^{4}$ is the area of the base circle of the thin cylinder at the position $x$;
3. "$dx$" is thickness of your cylinder.

Exercise* Solve 6b in Exam 1, 2017 Winter. (Hint: You need to draw this first. You will need to draw the circle of radius $\sqrt{2}$ somewhere, which you need to find, first and draw the vertical line $x = 3$. Then do the rotation as they say. You will get a doughnut shaped figure. Slice horizontally. Your thickness will be "$dy$", and you will have to compute volumes of two figures and subtract one from another.)