replace-callouts

Prompt file imported from eboyden42/calc-2-quartz (.github/prompts/replace-callouts.prompt.md). Fill in {{selectedText}} before use. Copyright stays with the author.

You are a Markdown formatter.
Apply the following rules exactly to the selected text ({{selectedText}}):

1. Each main numbered step (like "1. ...") becomes a new [!abstract] callout block.

   • Format: [!abstract]- <step number>. <title or summary>
   • Title is the text before any sub-bullets.

2. Sub-bullets under each step:

   • Indent 4 spaces.
   • Use - as the bullet.

3. Special notes:

   • If a sub-bullet begins with ! treat it as a warning: move it into a new nested callout:

     > [!warning] ...
     

   • Use !! for more severe notes and convert similarly.

4. Display math (block equations):

   • Separate from the bullet with a blank line.
   • Wrap in triple dollar signs ($$ ... $$), indented to match the parent bullet.

5. Inline math ($...$) must remain inline.

6. Preserve all original wording, math content, and logical indentation.

7. Separate each with an entirely blank line (not even ">").

Example

Input:

1. & Define the cross section region.
   • Bounded above-right by $y=6-x$.
   • Bounded below-right by $y=3x+2$.
   • ! These intersect at $x=1$.
   • Bounded at left by $x=0$.
2. && Define range of integration variable.
   • Rotated around $y$-axis, therefore use $x$ for integration variable (shells!).
   • Integral over $x\in [0,1]$: $$V=\int_0^1 2\pi Rh,dr$$
3. & Interpret $R$.
   • Radius of shell-cylinder equals distance along $x$: $$R(x)=x$$
4. & Interpret $h$.
   • Height of shell-cylinder equals distance from lower to upper bounding lines: $$\begin{align*}h(x)&= (6-x)-(3x+2)\ &= 4-4x\end{align*}$$
5. & Interpret $dr$.
   • $dr$ is limit of $\Delta r$ which equals $\Delta x$ here so $dr=dx$.
6. & Plug data in volume formula.
   • Insert data and compute integral: $$\begin{align*}V&= \int_0^1 2\pi Rh,dr\\&= \int_0^1 2\pi\cdot x(4-4x),dx\\&=\left.2\pi\left(2x^2-\frac{4x^3}{3}\right)\right|_0^1 = \frac{4\pi}{3}\end{align*}$$

Output:

[!abstract]- 1. Define the cross section region.

• Bounded above-right by $y=6-x$.
• Bounded below-right by $y=3x+2$.

• [!warning] These intersect at $x=1$.

• Bounded at left by $x=0$.

[!abstract]- 2. Define range of integration variable.

• Rotated around $y$-axis, therefore use $x$ for integration variable (shells!).
• Integral over $x\in [0,1]$:

$$ V=\int_0^1 2\pi Rh,dr $$

[!abstract]- 3. Interpret $R$.

• Radius of shell-cylinder equals distance along $x$:

$$ R(x)=x $$

[!abstract]- 4. Interpret $h$.

• Height of shell-cylinder equals distance from lower to upper bounding lines:

$$ \begin{align*}h(x)&= (6-x)-(3x+2)\ &= 4-4x\end{align*} $$

[!abstract]- 5. Interpret $dr$.

• $dr$ is limit of $\Delta r$ which equals $\Delta x$ here so $dr=dx$.

[!abstract]- 6. Plug data in volume formula.

• Insert data and compute integral:

$$ \begin{align*}V&= \int_0^1 2\pi Rh,dr\\ &= \int_0^1 2\pi\cdot x(4-4x),dx\\ &=\left.2\pi\left(2x^2-\frac{4x^3}{3}\right)\right|_0^1 = \frac{4\pi}{3}\end{align*} $$

Apply the rules now to this text:
{{selectedText}}