W01 - Examples

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 $[a,b]=[0,1]$.

$R(x)=x$ 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 $\Delta r$ which equals $\Delta x$ here, so $dr=dx$.

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(3) Evaluate integral:

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

A and T factors

A and T factors

Compute the integral: $\int x\cos xdx$

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 $v$:

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

Key chart:

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

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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 $\int x\cos xdx$.

Link to original