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Midterm Review

TABLE OF CONTENTS

I. Sketch the solid

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II. Find the volume of a solid

1. Definition of volume:

Let $S$ be a solid that lies between $x=a$ and $x=b$. If the cross-sectional area of $S$ in the plane $P_x$, through $x$ and perpendcular to the $x$-axis, is $A(x)$, where $A$ is a continous function, then the volume of $S$ is:

\[V=\displaystyle\lim_{n \to \infty} \sum_{i=1}^n A(x_i^*) \Delta x = \int_a^b A(x) dx\]

2. Several formulas:

Given $y=f(x)$, $a \leq x \leq b$

  • Disk method: rotate about the $x$-axis
\[V=\int_a^b \pi \left[f(x)\right]^2 \ dx\]
  • Shell method: rotate about the $y$-axis
\[V=\int_a^b 2\pi xf(x) \ dx\]

III. Polar curves

1. Relevant information on Cartesian coordinate:

$x=r\cos\theta$

$y=r\sin\theta$

$r^2=x^2+y^2$

$tan \ \theta = \dfrac{y}{x}$

$\displaystyle \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=\cfrac{\cfrac{dr}{d\theta}\sin\theta+r\cos\theta}{\cfrac{dr}{d\theta}\cos\theta-r\sin\theta}$

2. Sketch the curve:

  • Lines
    • $\theta=\alpha \ \ \ \ $ The line from $(0,0)$ set at the angle $\alpha$
    • $r\cos\theta=a \ \ \ \ $ The vertical line through $x=a$
    • $r\sin\theta=b \ \ \ \ $ The horizontal line through $y=b$
  • Circles
    • $r=a \ \ \ \ $ The circle centered at $(0,0)$ with radius $a$
    • $r=2a\cos\theta \ \ \ \ $ The circle centered at $(a,0)$ with radius $\vert a \vert$
    • $r=2b\sin\theta \ \ \ \ $ The circle centered at $(0,b)$ with radius $\vert b \vert$
    • $r=2a\cos\theta+2b\sin\theta \ \ \ \ $ The circle centered at $(a,b)$ with radius $\sqrt{a^2+b^2}$

If we are given other form, we should investigate the function $r=f(\theta)$, which means calculating derivative, critical point, …, in order to have enough information for drawing the curve accurately.

3. Area:

The formula for the area $A$ of the polar region of the curve $r=f(\theta)$, $a \leq \theta \leq b$, is:

\[A=\int_a^b \frac{1}{2} r^2 d\theta\]

IV. Length of a curve

1. Cartesian coordinate:

If $f’$ is continuous on $[a,b]$, then the length of the curve $y=f(x)$, $a \leq x \leq b$, is:

\[L=\int_a^b \sqrt{1+[f'(x)]^2} \ dx\]

2. Polar coordinate:

If $f’$ is continuous on $[a,b]$, then the length of the curve $r=f(\theta)$, $a \leq \theta \leq b$, is:

\[L=\int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2} \ d\theta\]

V. The fundamental theorem of Calculus

1. Part 1:

If $f$ is continuous on $[a,b]$, then the function $g$ defined by

\[g(x)=\int_a^x f(t) \ dt \ \ \ \ \ \ a \leq x \leq b\]

is continuous on $[a,b]$ and differentiable on $(a,b)$, and $g’(x)=f(x)$.

2. Part 2:

If $f$ is continuous on $[a,b]$, then

\[\int_a^b f(x) \ dx = F(b)-F(a)\]

where $F$ is any antiderivative of $f$, that is, a function such that $F’=f$.