W01 Notes

Volume using cylindrical shells

Videos

Review Videos

Review

• Volume using cross-sectional area
• Disk/washer method - 01
• Disk/washer method - 02
• Disk/washer method - 03

Shells

• Shell method - 01
• Shell method - 02
• Shell method - 03

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01 Theory

Theory 1

Take a graph $y=f(x)$ in the first quadrant of the $xy$-plane. Rotate this about the $y$-axis. The resulting 3D body is symmetric around the axis. We can find the volume of this body by using an integral to add up the volumes of infinitesimal shells, where each shell is a thin cylinder.

The volume of each cylindrical shell is $2\pi Rh\mathrm{\Delta }r$:

In the limit as $\mathrm{\Delta }r\to dr$ and the number of shells becomes infinite, their total volume is given by an integral.

Volume by shells - general formula

$$V={\int }_a^{b}2\pi Rhdr$$

In any concrete volume calculation, we simply interpret each factor, ‘$R$’ and ‘$h$’ and ‘$dr$’, and determine $a$ and $b$ in terms of the variable of integration that is set for $r$.

Shells vs. washers

Can you see why shells are sometimes easier to use than washers?

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02 Illustration

Example - Revolution of a triangle

Revolution of a triangle

A rotation-symmetric 3D body has cross section given by the region between $y=3x+2$, $y=6-x$, $x=0$, and is rotated around the $y$-axis. Find the volume of this 3D body.

Solution

(1) Cross-section region:

Bounded above-right by $y=6-x$. Bounded below-right by $y=3x+2$. These intersect at $x=1$.

Bounded left by $x=0$.

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(2) Set up integral:

Rotated around $y$-axis, therefore use $x$ for integration variable (shells!). Formula:

$$V={\int }_a^{b}2\pi Rhdr$$

Domain is $[a,b]=[\mathrm{0,1}]$.

$R(x)=x$ because shell radius is the $x$-distance from $x=0$ to the shell position.

Height:

$$\begin{array}{c}h(x)\gg \gg (6-x)-(3x+2)\\ \\ \gg \gg 4-4x\end{array}$$

$dr$ is limit of $\mathrm{\Delta }r$ which equals $\mathrm{\Delta }x$ here, so $dr=dx$.

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(3) Evaluate integral:

$$\begin{array}{c}V={\int }_a^{b}2\pi Rhdr\gg \gg {\int }_0^{1}2\pi \cdot x(4-4x)dx\\ \\ \gg \gg {2\pi (2x^2-\frac{4x^3}{3})|}_0^1\gg \gg \frac{4\pi }{3}\end{array}$$ Link to original

Practice exercise

Revolution of a sinusoid

Consider the region given by revolving the first hump of $y=\sin (x)$ about the $y$-axis. Set up an integral that gives the volume of this region using the method of shells.

Solution

(1) Set up the integral for shells:

Integration variable: $r=x$, the distance of a shell to the $y$-axis.

Then $dr=dx$ and $h=\sin x$, the height of a shell.

Bounds: one hump is given by $x\in [0,\pi ]$. Thus:

$$V={\int }_0^{\pi }2\pi x\sin xdx$$

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(2) Perform the integral using IBP:

Choose $u=2\pi x$ and $v^{\prime }=\sin x$ since $x$ is A and $\sin x$ is T.

Then $u^{\prime }=2\pi$ and $v=-\cos x$.

Use IBP formula:

$$\begin{array}{c}\int u{v}^{\prime }dx=uv-\int u^{\prime }vdx\\ \\ \gg \gg {\int }_0^{\pi }2\pi x\sin xdx\gg \gg (2\pi x)(-\cos x){|}_0^{\pi }-{\int }_0^{\pi }2\pi (-\cos x)dx\end{array}$$

Compute first term:

$$\begin{array}{c}\gg \gg -2\pi x\cos x{|}_0^{\pi }\\ \\ \gg \gg -2\pi (\pi )(-1)--2\pi (0)(+1)\gg \gg 2{\pi }^2\end{array}$$

Compute integral term:

$$-{\int }_0^{\pi }2\pi (-\cos x)dx\gg \gg 2\pi \sin {|}_0^{\pi }\gg \gg 0$$

So the answer is $2{\pi }^2$.

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Integration by substitution (review only)

Videos

Review Videos

[Note: this section is non-examinable. It is included for comparison to IBP.]

• Integration by Substitution 1: $\int \frac{-x}{(x+1)-\sqrt{x+1}}dx$
• Integration by Substitution 2: $\int \frac{x^5}{{(1-x^3)}^3}dx$
• Integration by Substitution 3: ${\int }_0^1{x}^2{(1+x)}^{4}dx$
• Integration by Substitution 4: $\int \frac{2x+3}{\sqrt{2x+1}}dx$
• Integration by Substitution 5: $\int \frac{\sin x}{{\cos }^{3}x}dx$
• Integration by Substitution: Definite integrals, various examples

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03 Theory

Theory 1

The method of $u$-substitution is applicable when the integrand is a product, with one factor a composite whose inner function’s derivative is the other factor.

Substitution

Suppose the integral has this format, for some functions $f$ and $u$:

$$\int f(u(x))\cdot u^{\prime }(x)dx$$

Then the rule says we may convert the integral into terms of $u$ considered as a variable, like this:

$$\int f(u(x))\cdot u^{\prime }(x)dx\gg \gg \int f(u)du$$

The technique of $u$-substitution comes from the chain rule for derivatives:

$$\frac{d}{dx}F(u(x))=f(u(x))\cdot u^{\prime }(x)$$

Here we let $F^{\prime }=f$. Thus ${\displaystyle \int f(x)dx=F(x)+C}$ for some $C$.

Now, if we integrate both sides of this equation, we find:

$$F(u(x))=\int f(u(x))\cdot u^{\prime }(x)dx$$

And of course $F\left(u\right)={\displaystyle \int f(u)du-C}$.

Extra - Full explanation of $u$-substitution

(1) Chain rule for derivatives:

Let $F(x)$ be a function and $F^{\prime }=f$ its derivative. Let $u(x)$ be another function.

Using primes:

$$(F\circ u{)}^{\prime }=(F^{\prime }\circ u)\cdot u^{\prime }$$

Using differentials:

$$\frac{d}{dx}F(u(x))=f(u(x))\cdot u^{\prime }(x)$$

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(2) Integrate both sides:

$$\begin{array}{cc}\frac{d}{dx}F(u(x))=f(u(x))\cdot u^{\prime }(x)& {\gg \gg }^{\int }\int \frac{d}{dx}F(u(x))=\int f(u(x))\cdot u^{\prime }(x)\\ & \\ & {\gg \gg }^{\text{FTC}}F(u(x))=\int f(u(x))\cdot u^{\prime }(x)\end{array}$$

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(3) Introduce ‘variable’ $u$ from the $u$-format of the integral:

Treating $u$ as a variable, the definition of $F$ gives:

$$F(u)=\int f(u)du+C$$

Set the ‘variable’ $u$ to the ‘function’ $u$ output:

$$F(u){|}_{u=u(x)}=F(u(x))$$

Combine these:

$$\begin{array}{cc}F(u(x))& =F(u){|}_{u=u(x)}\\ & \\ & ={\int f(u)du|}_{u=u(x)}+C\end{array}$$

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(4) Substitute for $F(u(x))$:

$$\int f(u(x))\cdot u^{\prime }(x)dx=F(u(x))={\int f(u)du|}_{u=u(x)}+C$$

This is “$u$-substitution” in final form.

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Integration by parts

Videos

Review Videos

Videos:

• Integration by Parts 1: $\int e^{x}dx$ and $\int \ln xdx$
• Integration by Parts 2: $\int {\tan }^{-1}xdx$ and $\int x\sec xdx$
• Integration by Parts 3: Definite integrals
• Example: $\int e^{3x}\cos 4xdx$, two methods:
  • Double IBP
  • Fast Solution
• Integration by Parts 6: $\int {\sec }^{5}xdx$

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04 Theory

Theory 1

The method of integration by parts (abbreviated IBP) is applicable when the integrand is a product for which one factor is easily integrated while the other becomes simpler when differentiated.

Integration by parts

Suppose the integral has this format, for some functions $u$ and $v$:

$$\int u\cdot v^{\prime }dx$$

Then the rule says we may convert the integral like this:

$$\int u\cdot v^{\prime }dx\gg \gg u\cdot v-\int u^{\prime }\cdot vdx$$

This technique comes from the product rule for derivatives:

$$(u\cdot v{)}^{\prime }=u^{\prime }\cdot v+u\cdot v^{\prime }$$

Now, if we integrate both sides of this equation, we find:

$$u\cdot v=\int u^{\prime }\cdot vdx+\int u\cdot v^{\prime }dx$$

and the IBP rule follows by algebra.

Extra - Full explanation of integration by parts

(1) Product rule for derivatives:

$$(u\cdot v{)}^{\prime }=u^{\prime }\cdot v+u\cdot v^{\prime }$$

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(2) Integrate both sides:

$$\begin{array}{cccc}& (u\cdot v{)}^{\prime }=u^{\prime }\cdot v+u\cdot v^{\prime }& \gg \gg \int (u\cdot v{)}^{\prime }dx=\int u^{\prime }\cdot vdx+\int u\cdot v^{\prime }dx& \\ & & & \\ & & \gg \gg u\cdot v=\int u^{\prime }\cdot vdx+\int u\cdot v^{\prime }dx& \text{(FTC)}\\ & & & \\ & & \gg \gg \int u\cdot v^{\prime }dx=u\cdot v-\int u^{\prime }\cdot v& \text{(Rearrange)}\end{array}$$

Definite IBP

Definite version of FTC:

$$\begin{array}{c}{\int }_a^b(u\cdot v{)}^{\prime }dx=u\cdot v{|}_a^b\\ \\ \gg \gg {\int }_a^{b}u\cdot v^{\prime }dx=u\cdot v{|}_a^b-{\int }_a^b{u}^{\prime }\cdot v\end{array}$$

Choosing factors well

IBP is symmetrical. How do we know which factor to choose for $u$ and which for $v$?

Here is a trick: the acronym “LIATE” spells out the order of choices – to the left for $u$ and to the right for $v$:

$$\begin{array}{c}\text{LIATE}:\\ \\ u\leftarrow \text{Logarithmic – Inverse\_trig – Algebraic – Trig – Exponential}\to v\end{array}$$

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05 Illustration

Example - A and T factors

A and T factors

Compute the integral: ${\displaystyle \int x\cos xdx}$

Solution

(1) Choose $u$ and $v^{\prime }$:

Set $u(x)=x$ because $x$ simplifies when differentiated.

(By the trick: $x$ is Algebraic, i.e. more “$u$”, and $\cos x$ is Trig, more “$v$”.)

Remaining factor must be $v^{\prime }=\cos x$.

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(2) Compute $u^{\prime }$ and $v$:

$$\begin{array}{cc}u=x& \gg \gg u^{\prime }=1\\ & \\ v^{\prime }=\cos x& \gg \gg v=\sin x\end{array}$$

Key chart:

$$\begin{array}{cc}u=x& v^{\prime }=\cos x\\ u^{\prime }=1& v=\sin x\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }\text{original}\\ ⟶\int u^{\prime }\cdot v\text{final}\end{array}$$

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(3) Evaluate IBP formula:

$$\begin{array}{c}\int u^{\prime }vdx=uv-\int u{v}^{\prime }dx\\ \\ \gg \gg \int x\cos xdx=x\sin x-\int 1\cdot \sin xdx\\ \\ \gg \gg x\sin x+\cos x+C\end{array}$$

Why IBP?

The rationale of IBP is that $\int 1\cdot \sin xdx$ is easier to compute than $\int x\cos xdx$.

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Exercise - Hidden A

Hidden A

Compute the integral:

$$\int \ln xdx$$

Solution

(1) Choose $u=\ln x$:

Because Log is farthest left in LIATE.

Therefore we must choose $v^{\prime }(x)=1$.

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(2) Compute $u^{\prime }$ and $v$:

We have $u^{\prime }=\frac{1}{x}$ and $v=x$. Obtain chart:

$$\begin{array}{cc}u=\ln x& v^{\prime }=1\\ u^{\prime }=1/x& v=x\end{array}\begin{array}{c}⟶\int u\cdot v^{\prime }\text{original}\\ ⟶\int u^{\prime }\cdot v\text{final}\end{array}$$

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(3) Evaluate IBP formula:

$$\begin{array}{c}\int u^{\prime }vdx=uv-\int u{v}^{\prime }dx\\ \\ \gg \gg \int \ln x\cdot 1dx=x\ln x-\int \frac{1}{x}\cdot xdx\end{array}$$

Perform new integration:

$$-\int \frac{1}{x}\cdot xdx\gg \gg -\int 1dx\gg \gg -x+C$$

Final answer is: $x\ln x-x+C$

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