Title: Math 2260 Chapter 6 (Volume, Length, Area, Work)

Q: [latex]The volume of the portion of solid of cross-sectional area $A(x)$ with $a \leq x \leq b$[/latex]
[latex]is[/latex]
A: [latex]$$\operatorname{Vol} = \int_a^b A(x) \, dx.$$[/latex]

Q: [latex]The volume of the portion of the solid created by rotating $y = f(x)$ around the $x$-axis (with $a \leq x \leq b$) is computed by the disk method using the integral[/latex]
A: [latex]$$\operatorname{Vol} = \int_a^b \pi [R(x)]^2 \, dx = \int_a^b \pi f(x)^2 \, dx.$$[/latex]

Q: [latex]The volume of a solid of revolution with inner radius $r(x)$ and outer radius $R(x)$ (rotated around the $x$ axis) is computed using the \emph{washer method} by the integral [/latex]
A: [latex]$$\operatorname{Vol} = \int_a^b \pi[R(x)]^2 - \pi [r(x)]^2 \, dx.$$[/latex]

Q: [latex]The volume of a region can be computed by the shell method [/latex]
[latex]using the integral[/latex]
A: [latex]$$\operatorname{Vol} = \int 2 \pi (\text{radius of shell})(\text{height of shell}) \, dr$$[/latex]

Q: [latex]The length of the portion of the graph of $y = f(x)$ from $a \leq x \leq b$ is[/latex]
A: [latex]$$\operatorname{Length} = \int_a^b \sqrt{1 + f'(x)^2} \, dx.$$[/latex]

Q: [latex]The length of the portion of the \emph{parametrized} curve $(x(t),y(t))$ with $a \leq t \leq b$ is[/latex]
A: [latex]$$\operatorname{Length} = \int_a^b \sqrt{ x'(t)^2 + y'(t)^2} $$[/latex]

Q: [latex]The area of portion of the surface of revolution swept out by the graph $y = f(x)$ for $a \leq x \leq b$ as it is revolved around the x axis is given by[/latex]
A: [latex]$$\operatorname{Area} = \int_a^b 2 \pi f(x) \sqrt{ 1 + f'(x)^2 } \, dx.$$[/latex]

Q: [latex]The \emph{work} done by a variable force $F(x)$ in the direction of motion along the $x$ axis is given by [/latex]
A: [latex]$$\operatorname{Work} = \int F(x) \, dx$$[/latex]

Q: [latex] (Hooke's law for springs.) \\
The force required to hold a spring $x$ units from its natural (unstretched or uncompressed) length is [/latex]
A: [latex]$k x$, \\
where $k$ is the \emph{spring constant} [/latex]

Q: [latex]Let $x$ be the depth below a pump. \\
A tank under the pump has cross-sectional area $A(x)$, and contains fluid in the region $a \leq x \leq b$. \\
How much work is required to pump out the tank?[/latex]
A: [latex]$$\int_a^b x A(x) (\text{density}) g \, dx$$
if density is given in units of mass (eg. grams, kilograms), and
$$\int_a^b x A(x) (\text{density}) \, dx$$
if density is given in units of weight  (eg. lbs).[/latex]

Q: [latex]The \emph{total mass} of the plate between $y = f(x)$ and $y = g(x)$ for $x \in [a,b]$:[/latex]
[latex]if the plate has density $\delta(x)$:[/latex]
A: [latex]$$\text{mass} = \int_a^b (f(x) - g(x)) \delta(x) \,\, \mathrm{d} x $$[/latex]

Q: [latex]The x-centroid or x-coordinate of the center of mass $\bar{x}$ of the thin plate between $y = f(x)$ and $y = g(x)$ for $x \in [a,b]$:[/latex]
[latex]if the plate has density $\delta(x)$:[/latex]
A: [latex]$$\bar{x} = \frac{1}{\text{mass}} \int_a^b x (f(x) - g(x)) \delta(x) \, \mathrm{d}x.$$
Why? Because $$(f(x) - g(x)) \delta(x) \, \mathrm{d}x$$ is the total mass of a thin vertical strip at $x$, which has $x$ coordinate of its center of mass at $x$. We divide by the total mass to get the (weighted) average $x$-coordinate of the whole plate.[/latex]

Q: [latex]The y-\emph{centroid} or y-coordinate of the center of mass $\bar{y}$ of the plate between $y = f(x)$ and $y = g(x)$ for $x \in [a,b]$:[/latex]
[latex]if the plate has density $\delta(x)$:[/latex]
A: [latex]$$ \bar{y} = \frac{1}{\text{mass}} \int_a^b \frac{f(x) + g(x)}{2} (f(x) - g(x)) \delta(x) \, \mathrm{d}x.$$
Why? The quantity $\frac{f(x) + g(x)}{2}$ is the $y$-coordinate of the center of mass of each vertical strip. \\
The strip is weighted by $$(f(x) - g(x)) \delta(x) \, \mathrm{d}x$$ which is the total mass of the vertical strip. We then integrate to get the (weighted) average $y$-coordinate of the centers of mass of all the strips, and divide by total mass as usual.[/latex]

Q: [latex]The \textbf{work-energy} theorem is:[/latex]
A: [latex]$$W = \Delta E_k = \frac{1}{2} mv_2^2 - \frac{1}{2} mv_1^2$$ where $E_k$ (kinetic energy) is given by $\frac{1}{2} m v^2$, $v_2$ is the velocity after the work is performed and $v_1$ is the velocity before the work is performed.[/latex]

Q: [latex] The unit for work is ... [/latex]
A: [latex] The \textbf{Joule}.
$$1 \operatorname{Joule} = 1 \operatorname{Newton} \operatorname{meter} = 1 \operatorname{kg} \frac{\operatorname{meter}^2}{\operatorname{sec}^2}.$$
[/latex]

Q: [latex] The unit for \textbf{power} is ... [/latex]
A: [latex] the \textbf{watt}.
$$ 1 \operatorname{watt} = 1 \frac{\operatorname{Joule}}{\operatorname{sec}} $$
Power is the rate at which work is done.
$$ 1 \operatorname{horsepower} = 746 \operatorname{watts}
[/latex]