Ｃａｌｃｕｌｕｓ １ Ｌｅｃｔｕｒｅ ５．２

Ｖｏｌｕｍｅ ｏｆ Ｓｏｌｉｄｓ Ｂｙ Ｄｉｓｋｓ ａｎｄ Ｗａｓｈｅｒｓ Ｍｅｔｈｏｄ

Ｉｎｔｒｏｄｕｃｔｉｏｎ.

• 01:00. Understanding the Concept of Volume as “Cross-Sectional Area × Length”
  • Visualizing volumes as the sum of many rectangular prisms
  • Condition for applying this method: sides perpendicular to the 𝒙-a𝒙is
  • The volume is bounded between points 𝓪 and 𝓫, where sides are perpendicular to the 𝒙-axis

1.- Ｖｏｌｕｍｅ　ｏｆ　ｓｏｌｉｄｓ　ｂｙ　ｓｌｉｃｉｎｇ.

• 02:25. Preparing for the Deduction of the Volume Integral - [📷image]
  • Drawing as a fundamental tool
  • Main idea: slice the solid into “slabs”
• 03:30. Slicing into “Slabs” - [📷image]
  • Using slabs of the same width to simplify the sum
  • Representing the width of each slab as $\,\Delta 𝒙\,$
  • 03:58. Expanding a Single Slab
• 05:10. Main idea – - [📷image]
  • **Cut into thin slabs; then use Riemann sums to set up an integral.**
  • 06:30. The fundamental idea of calculating volumes using integration. The key points are:
    • ① Basic Concept of Volume:
      
      • Volume can generally be understood as surface area times length.
      • In this case, it is the **cross-sectional area multiplied by its width**.
    • ② Visualization of the Cross-Section:
      
      • Imagine cutting a solid perpendicularly to its main axis.
      • If you take out a **"slab"** (a thin slice of the solid) and turn it toward you, the face you see is the **cross-section**.
      • This section has an **area** that we denote as $𝓐(𝒙)$.
    • ③ Approximating the Volume:
      
      • Multiplying the cross-sectional area $𝓐(𝒙)$ by the width $\,\Delta 𝒙\,$ gives an approximation of the volume of that small "slab".
      • The thinner the slab (i.e., the smaller $\,\Delta 𝒙$), the more accurate the volume calculation.
    • ④ Parallel with Area Calculation:
      
      • Similar to finding areas under a curve by summing rectangles and then taking $\Delta 𝒙 \to 0$ to get the exact area.
      • Here, 𝗯𝘆 𝘀𝘂𝗺𝗺𝗶𝗻𝗴 𝘁𝗵𝗲 𝘃𝗼𝗹𝘂𝗺𝗲𝘀 𝗼𝗳 𝗶𝗻𝗳𝗶𝗻𝗶𝘁𝗲𝗹𝘆 𝘁𝗵𝗶𝗻 𝗰𝗿𝗼𝘀𝘀-𝘀𝗲𝗰𝘁𝗶𝗼𝗻𝘀 and taking the limit (letting the number of slabs approach infinity while their width $\Delta 𝒙$ approaches zero), we obtain the **exact** volume.
    • ⑤ Conclusion:
      
      • To compute the volume, **the first step is to determine the area of the cross-section**, as this will be the function we integrate along the solid.

Ｄｅｆｉｎｉｎｇ　ｔｈｅ　ａｒｅａ　ｏｆ　ｃｒｏｓｓ－ｓｅｃｔｉｏｎ

• 07:30. To implement the main idea, we need to define the cross-sectional area. - [📷image]
• 09:40. Find the cross-sectional Volume (𝓥) taking a rectangular prism by example.
  • $𝓥=𝒚\cdot 𝒛\cdot 𝒙$ where $𝒚\cdot 𝒛$ is the cross-sectional area and $𝒙$ is the lenght.
• 11:11. Understanding Slab Approximation
  • The concept involves creating a large number of very thin slices (or "slabs") and summing their individual volume to determine the total volume of the solid. The key lies in treating each slab as a simple and small approximation of the solid; by accumulating infinitely many such slabs with infinitesimally small thickness, we achieve an accurate measurement of the total volume.
• 11:45. Importance of Perpendicularity.
  • If the sides are not aligned with the 𝒙-axis, slabs won't be uniform or consistent in shape and size.
  • Non-perpendicular alignment complicates the approximation since each slice might vary significantly.
  • Simplicity When Aligned.
    • When solids are bounded by planes perpendicular to the 𝒙-axis, each slab has a consistent cross-sectional area that can be easily integrated.
    • This consistency allows for straightforward summation (integration) along the 𝒙-axis from point A to B.

Ｆｉｎｄ ｔｈｅ ｖｏｌｕｍｅ ｏｆ ａｎｙ ｓｏｌｉｄ ｔｈａｔ ｉｓ ｂｏｕｎｄ ｂｙ ｐｅｒｐｅｎｄｉｃｕｌａｒ ｐｌａｎｅｓ ｔｏ 𝒙 －ａｘｉｓ
ａｔ ｐｏｉｎｔｓ 𝓪 ａｎｄ 𝓫.

• 12:40. Introduction
• 15:30. Cut the solid into slabs so that they're the same width.
  • width = $\,\Delta 𝒙\,$
• 18:10. Zoom in on a slab
• 18:40. Analogy with the Area Problem: Arbitrary Point
  • Selecting an arbitrary point $(x_k^*)$ within each sub-interval (along $\Delta 𝒙$)
  • Importance of that arbitrary point in taking the limit
• 20:34. Finding the Cross-Sectional Area - [📷image-1]
  • The slab width ($\Delta 𝒙$) is known
  • The height corresponds to $𝒇(x_k^* )$ in the area problem
  • Identifying the cross-sectional area as the main objective
    • Using the Arbitrary Point $x_{k}^{\ast}$ to Find the Cross-Sectional Area
      • $x_{k}^{\ast}$ allows us to find the area of each cross section in each subinterval
      • Visualizing the cut at $x_{k}^{\ast}$ as a surface
      • The cross-sectional area at $x_{k}^{\ast}$, multiplied by $\Delta 𝒙$, gives the volume over that subinterval
• 24:25. Computing the Volume in an Arbitrary Interval - [📷image-2]
  • Defining 𝑽ₖ as the volume of the k-th interval (one slab)
    • $𝓥_k=\text{Area}(x_k^* )\cdot \Delta 𝒙$
      • cross-sectional area ⇢ $𝓐(x_k^* )$
      • Lenght ⇢ $\Delta 𝒙$
  • Summation of all slab volumes (aproximate volume)
    • $𝓥=\displaystyle \sum_{k=1}^{n} 𝓐(x_k^*)\cdot \Delta 𝒙$ as an appro𝒙imation of the total volume
• 27:25. The Limit of the Riemann Sum as the Definite Integral - [📷image-3]
  • Applying the limit as $n \to \infty$ to get the e𝒙act volume
  • Converting the sum into a definite integral:
    • $\displaystyle \lim_{n\to\infty}\sum_{k=1}^{n} 𝓐(x_k^*)\cdot \Delta 𝒙 \;\to\; 𝓥=\displaystyle \int_{𝓪}^{𝓫} 𝓐(𝒙)\,dx$
      • Interpreting the integral as summing cross-sectional areas from 𝓪 to 𝓫: $[𝓪,𝓫]$
• 31:20. E𝒙tension to integration with respect to 𝑦. - [📷image-4] - [📷image-5]
  • $𝓥=\displaystyle \int_{𝓬}^{𝓭} 𝓐(𝑦)\,dy$
• 32:40. 🧩 Example – Introduction: Volume of a Cylinder - [📷image]
  • Find the volume of a right circular cylinder whose central axis lies along the horizontal axis on [1,5] and extends between [1,−1] along the 𝑦-axis.
  • $𝓐=\pi\cdot r^{2}$; $𝓐(𝒙)=\pi\cdot (1)^{2}=\pi$ Then the volume is found by integrating $𝓐(𝒙)$ over $𝒙$ from 1 to 5:
    • $𝓥=\displaystyle \int_{1}^{5}\pi\,dx=\pi[𝒙]\Big|_{1}^{5}=\pi(5-1)=4\pi$
  • Verification with the Cylinder Volume Formula
    • Cylinder volume = (base area) × (height) = $\pi r^{2}\cdot h$
    • Calculation: $\pi(1)^{2}\times (5-1)=4\pi$

2.- Ｖｏｌｕｍｅ ｏｆ ｓｏｌｉｄｓ ｏｆ ｒｅｖｏｌｕｔｉｏｎ.

• 39:20. Introduction to Solids of Revolution - [📷image-1] - [📷image-2] - [📷image-3]
  • A solid is formed by rotating a function about an a𝒙is
  • E𝒙ample: rotating a rectangle around the 𝒙-a𝒙is creates a cylinder
  • E𝒙ample: rotating a semicircle around the 𝒙-a𝒙is creates a sphere
  • E𝒙ample: Rotating a right triangle around the 𝒙-axis typically creates a cone.
  • E𝒙ample: Consider a rectangle (or rectangular strip) that is parallel to the axis of rotation
    and does not include that axis within its boundaries. When this rectangular region
    is rotated, you obtain a tube (or hollow cylinder) whose inner radius is $R_{\text{inner}}$ and
    outer radius is $R_{\text{outer}}$.

Ｖｏｌｕｍｅ ｏｆ ｓｏｌｉｄｓ ｏｆ ｒｅｖｏｌｕｔｉｏｎ ᎠᏆᏚᏦ ᎷᎬᎢᎻᎾᎠ

• 44:45. When the function is not regular. - [📷image]
  • Conditions for Using the ᎠᏆᏚᏦ ᎷᎬᎢᎻᎾᎠ
    • $𝒇(𝒙)$ must be continuous and bounded between 𝓪 and 𝓫 (vertical lines $𝒙=𝓪$ and $𝒙=𝓫$)
    • Rotating around the 𝒙-a𝒙is requires sides perpendicular to the 𝒙-a𝒙is
• 49:00. The General ᎠᏆᏚᏦ ᎷᎬᎢᎻᎾᎠ for Finding Volume
  • Integrate the cross sections perpendicular to the a𝒙is of rotation
  • Need a function representing the cross-sectional area
  • Summing cross-sectional areas from 𝓪 to 𝓫
• 52:40. Find the solid Volume by slicing - [📷image-1] - [📷image-2]
  • $𝓥=\displaystyle \int_{𝓪}^{𝓫} 𝓐(𝒙)\,dx$
  • 54:20. Specifying the Cross-Sectional Area
    • $𝓐(𝒙)$ represents the area of each cross section
    • The cross section is allways a circle: $𝓐(𝒙)=\pi r^{2}$
    • $r=𝒇(𝒙)$; The radius $r$ is the height of $𝒇(𝒙)$.
  • 57:45. Substituting to Get $𝓐(𝒙)$ in Terms of $𝒙$
    • $𝓐(𝒙)=\pi r^{2}\;\Rightarrow\; 𝓐(𝒙)=\pi[𝒇(𝒙)]^{2}$
  • 58:30. **The Disk Method Formula**
    • $𝓥=\displaystyle \int_{𝓪}^{𝓫} 𝓐(𝒙)\,dx\;\Rightarrow\; \boxed{𝓥=\displaystyle \int_{𝓪}^{𝓫} \pi[𝒇(𝒙)]^{2}\,dx}$
• 1:00:00.🧩 Example – 1: $𝒇(𝒙)=3\sqrt{𝒙}$ on $[1,4]$ - [📷image]
  • The problem: find the volume of the solid of revolution
  • Conditions: perpendicular sides to form circles
  • Alternative wording of the problem
  • Setting up the integral: $𝓥=\displaystyle \int_{𝓪}^{𝓫} \pi[𝒇(𝒙)]^{2}\,dx \;\Rightarrow\; 𝓥=\displaystyle \int_{1}^{4} \pi[3\sqrt{𝒙}]^{2}\,dx$
  • 31:53. Simplifying and Evaluating the Integral
    • $\pi[3\sqrt{𝒙}]^{2}=9\pi 𝒙$
    • $𝓥=\displaystyle \int_{1}^{4} 9\pi 𝒙\,dx$
    • Factor out the constant: $9\pi \displaystyle \int_{1}^{4} 𝒙\,dx$
    • The integral of $𝒙$ is $\displaystyle \frac{𝒙^{2}}{2}$
    • $𝓥=9\pi \big[ \displaystyle \frac{𝒙^{2}}{2}\big]\Big|_{1}^{4}=9\pi \big[\displaystyle \frac{16}{2}-\frac{1}{2}\big]=9\pi\big(\displaystyle \frac{15}{2}\big)= \displaystyle \frac{135\pi}{2}$
• 1:09:00. 🧩 Example – 2: Finding the Volume of a Sphere - [📷image-1] - [📷image-2]
  • Circle equation: $𝒙^{2}+𝑦^{2}=r^{2}$
  • Solving for $𝑦$: $𝑦=\pm\sqrt{r^{2}-𝒙^{2}}$
  • Using the upper half: $𝒇(𝒙)=\sqrt{r^{2}-𝒙^{2}}$
  • Integral setup:
    • Volume $=\displaystyle \int_{-r}^{r} \pi\,[\sqrt{r^{2}-𝒙^{2}}]^{2}\,dx$
  • 37:05. Simplify:
    • Volume $=\displaystyle \int_{-r}^{r} \pi\,(r^{2}-𝒙^{2})\,dx$
  • 37:23. Factor out $\pi$:
    • Volume $=\pi \displaystyle \int_{-r}^{r} (r^{2}-𝒙^{2})\,dx$
  • Integration and Evaluation
    • The integral of $r^{2}$ is $r^{2}\cdot 𝒙$
    • The integral of $𝒙^{2}$ is $\displaystyle \frac{𝒙^{3}}{3}$
    • Volume $=\pi\left[\;r^{2}\cdot 𝒙-\dfrac{𝒙^{3}}{3}\;\right]_{-r}^{r}$
    • At $𝒙=r$: $r^{2}\cdot r- \displaystyle \frac{r^{3}}{3}=r^{3}- \displaystyle \frac{r^{3}}{3}$
    • At $𝒙=-r$: $r^{2}\cdot(-r)- \displaystyle \frac{(-r)^{3}}{3}=-r^{3}+ \displaystyle \frac{r^{3}}{3}$
    • Subtracting gives: $\pi\big[(r^{3}- \displaystyle \frac{r^{3}}{3})-(-r^{3}+ \displaystyle \frac{r^{3}}{3})\big]$
    • Simplifying: $\pi\big[2r^{3}- \displaystyle \frac{2r^{3}}{3}\big]=\pi\big[ \displaystyle \frac{6r^{3}}{3}- \displaystyle \frac{2r^{3}}{3}\big]=\pi\big[ \displaystyle \frac{4r^{3}}{3}\big]$
    • Sphere volume: $\displaystyle \frac{4}{3}\pi r^{3}$

Ｖｏｌｕｍｅ ｏｆ ｓｏｌｉｄｓ ｏｆ ｒｅｖｏｌｕｔｉｏｎ ᎳᎪᏚᎻᎬᎡ ᎷᎬᎢᎻᎾᎠ

• 1:22:30. The General ᎳᎪᏚᎻᎬᎡ ᎷᎬᎢᎻᎾᎠ for Finding Volume - [📷image-1] - [📷image-2]
  • What if there is another function $𝓰(𝒙)$ between $𝒇(𝒙)$ and the 𝒙-a𝒙is?
  • A solid is created with a hollow region in the center
  • Visualizing the solid with the hollow part
• 1:26:20. Deduction of the ᎳᎪᏚᎻᎬᎡ ᎷᎬᎢᎻᎾᎠ Formula
  • The volume is still the integral of the cross-sectional area
    • $𝓥=\displaystyle \int_{𝓪}^{𝓫} 𝓐(𝒙)\,dx$; $𝓐=$ Cross-sectional area (area of 𝒙-section)
  • Cross-sectional $𝓐=\text{Area}(𝒇(𝒙))-\text{Area}(𝓰(𝒙))$
    • $\text{Area}(𝒇(𝒙))=\pi[𝒇(𝒙)]^{2}$; $\text{Area}(𝓰(𝒙))=\pi[𝓰(𝒙)]^{2}$
  • The radius is the function height
    • $𝒇(𝒙)$ is the function high at the point $𝒙$
    • $𝓰(𝒙)$ is the function high at the point $𝒙$
• 1:31:10. Summary and Formalization of the Formula
  • Rotating the region between 𝒇(𝒙) and 𝓰(𝒙) around the 𝒙-a𝒙is
  • Cross sections remain perpendicular to the 𝒙-a𝒙is
  • Volume = $\displaystyle \int_{𝓪}^{𝓫} \text{Area}(𝒙)\,d𝒙$
  • $𝓐(𝒙)=\pi[𝒇(𝒙)]^{2}-\pi[𝓰(𝒙)]^{2}$
    • $𝓐(𝒙)=\pi\big([𝒇(𝒙)]^{2}-[𝓰(𝒙)]^{2}\big)$
    • $\boxed{\displaystyle 𝓥=\int_{𝓪}^{𝓫}\pi\Big([𝒇(𝒙)]^{2}-[𝓰(𝒙)]^{2}\Big)\,d𝒙}$
  • Also called “Volume by Washers”
• 1:34:10. 🧩 Example – Find the volume of the solid created: Area between $𝒇(𝒙)=𝒙^{2}+ \tfrac{1}{2}$ and $𝓰(𝒙)=𝒙$ on $[0,2]$ is rotated about the 𝒙-axis. - [📷image]
  • Identifying the Upper Function
    • **Importance of distinguishing which function is on top**
      • How to know which function is “above”?
        • Correct method: evaluate each function at a point within the interval
          • $𝒇(1)=(1)^{2}+\dfrac{1}{2}=\dfrac{3}{2}$, $𝓰(1)=1$
    • Conclusion: $𝒇(𝒙)$ is above $𝓰(𝒙)$
  • Integral setup:
    • $\displaystyle 𝓥=\int_{0}^{2}\pi\big([(𝒙)^{2}+\dfrac{1}{2}]^{2}-[𝒙]^{2}\big)\,d𝒙$
  • 1:41:10. E𝒙panding and Simplifying the Integrand
    • $[(𝒙)^{2}+\dfrac{1}{2}]^{2}=𝒙^{4}+𝒙^{2}+\dfrac{1}{4}$
    • $\displaystyle 𝓥=\int_{0}^{2}\pi\big(𝒙^{4}+𝒙^{2}+\tfrac{1}{4}-𝒙^{2}\big)\,d𝒙$
    • Cancel terms: $\displaystyle 𝓥=\int_{0}^{2}\pi\big(𝒙^{4}+\tfrac{1}{4}\big)\,d𝒙$
    • Factor out $\pi$: $\displaystyle 𝓥=\pi\int_{0}^{2}\big(𝒙^{4}+\tfrac{1}{4}\big)\,d𝒙$
  • Integration and Final Evaluation
    • The integral of $𝒙^{4}$ is $\tfrac{𝒙^{5}}{5}$
    • The integral of $\tfrac{1}{4}$ is $\tfrac{𝒙}{4}$
    • $\displaystyle 𝓥=\pi\left[\dfrac{𝑦^{2}}{2}-\dfrac{𝑦^{5}}{5}\right]_{0}^{2}=\pi\left(\dfrac{1}{2}-\dfrac{1}{5}\right)=\dfrac{3\pi}{10}$
    • At $𝒙=2$: $\displaystyle \pi\left[\dfrac{(2)^{5}}{5}+\dfrac{2}{4}\right]=\pi\left[\dfrac{32}{5}+\dfrac{1}{2}\right]$
    • At $𝒙=0$: $0$
    • Subtract and simplify: Volume $= \displaystyle \pi\left[\dfrac{32}{5}+\dfrac{1}{2}\right]=\pi\left[\dfrac{64}{10}+\dfrac{5}{10}\right]=\dfrac{69\pi}{10}$

Ｖｏｌｕｍｅｓ　ｗｈｅｒｅ　𝒙－ｓｅｃｔｉｏｎ　ｉｓ　ｐｅｒｐｅｎｄｉｃｕｌａｒ　ｔｏ　𝒚－ａｘｉｓ

• 1:46:23. Volumes with cross-sections ⟂ to the 𝒚-axis (discs & washers in 𝒅𝒚) - [📷image]
  • Setup (compare to the 𝒙-axis case)
    • Previously: cross-sections ⟂ to 𝒙 ⇒ integrate 𝒅𝒙 on $[𝓪,𝓫]$
    • Now: cross-sections ⟂ to 𝒚 ⇒ integrate 𝒅𝒚 on $[𝓬,𝓭]$
  • Discs method in 𝒅𝒚
    • $\displaystyle 𝓥=\int_{𝓬}^{𝓭}\pi\,[𝒖(𝑦)]^{2}\,d𝑦$, where $𝒖(𝑦)$ is the radius (expressed as 𝒙 in terms of 𝑦)
    • Functions must be written **in terms of 𝑦** (solve for $𝒙=𝒖(𝑦)$)
  • Washers method in 𝒅𝒚
    • $\boxed{\displaystyle 𝓥=\int_{𝓬}^{𝓭}\pi\big([𝒖(𝑦)]^{2}-[𝒗(𝑦)]^{2}\big)\,d𝑦}$
      • With $𝒖(𝑦)\ge 𝒗(𝑦)$ for all $𝑦\in[𝓬,𝓭]$
    • Still 𝒙 as functions of 𝑦: “right (outer) − left (inner)” radii measured horizontally
  • Key note (variable consistency)
    • Revolve **around 𝒙-axis** ⇒ write curves as $𝑦=𝒇(𝒙)$, bounds in 𝒙, integrate 𝒅𝒙
    • Revolve **around 𝒚-axis** ⇒ write curves as $𝒙=𝒖(𝑦)$, bounds in 𝑦, integrate 𝒅𝑦
• 1:49:57. 🧩 Example 1 – :Revolve $𝑦=\sqrt{𝒙}$ around the 𝒚-axis, with $𝑦\in[0,2]$ (discs in 𝒅𝒚) - [📷image]
  • 1:54:26. Put function in terms of 𝑦
    • $𝑦=\sqrt{𝒙}\;\Rightarrow\; 𝒙=𝑦^{2}$ (this is the radius $𝒖(𝑦)$)
  • Set up (perpendicular to 𝒚 ⇒ discs)
    • $𝓬=0$, $𝓭=2$
    • $\displaystyle 𝓥=\int_{0}^{2}\pi\,(𝑦^{2})^{2}\,d𝑦$
  • Evaluate
    • $\displaystyle 𝓥=\pi\int_{0}^{2}𝑦^{4}\,d𝑦=\pi\left[\dfrac{𝑦^{5}}{5}\right]_{0}^{2}=\dfrac{32\pi}{5}$
  • Result: $\displaystyle 𝓥=\dfrac{32\pi}{5}$ (Revolving the **same region** about 𝒙-axis gives a **different** volume.)
• 2:00:23. 🧩 Example 1 – Same curve, revolve around 𝒙-axis (discs in 𝒅𝒙) - [📷image]
  • Bounds conversion
    • Given $𝑦=\sqrt{𝒙}$ on $𝑦\in[0,2]$ ⇒ plug $𝑦=0$ and $𝑦=2$ into $𝑦=\sqrt{𝒙}$: $𝒙\in[0,4]$
  • Set up
    • $\displaystyle 𝓥=\int_{0}^{4}\pi\,(𝑦(𝒙))^{2}\,d𝒙=\int_{0}^{4}\pi\,(\sqrt{𝒙})^{2}\,d𝒙=\int_{0}^{4}\pi\,𝒙\,d𝒙$
  • Evaluate
    • $\displaystyle 𝓥=\pi\left[\dfrac{𝒙^{2}}{2}\right]_{0}^{4}=8\pi$
  • Conclusion: $\displaystyle \dfrac{32\pi}{5}\neq 8\pi$ (axis of rotation matters).
• 2:03:44. 🧩 Example – 2:Area between $𝑦=𝒙^{2}$ and $𝑦=𝒙^{3}$, revolved around 𝒙-axis (washers in 𝒅𝒙) - [📷image]
  • Goal & method
    • $\displaystyle 𝓥=\int_{𝓪}^{𝓫}\pi\big([𝒇(𝒙)]^{2}-[𝓰(𝒙)]^{2}\big)\,d𝒙$
    • Cross-sections ⟂ to 𝒙-axis ⇒ **washers** (outer radius − inner radius), integrate 𝒅𝒙
    • Radii are vertical distances to the axis $𝑦=0$: $𝒇(𝒙)$ = top, $𝓰(𝒙)$ = bottom
  • Intersections & interval in 𝒙
    • Set $𝒙^{3}=𝒙^{2}\Rightarrow 𝒙^{2}(𝒙-1)=0\Rightarrow 𝒙=0,1$
    • Test on $(0,1)$: at $𝒙=\tfrac{1}{2}$, $𝒙^{2}=\tfrac{1}{4}$, $𝒙^{3}=\tfrac{1}{8}$ ⇒ **top $=𝒙^{2}$**, **bottom $=𝒙^{3}$** on $[0,1]$
      • Both curves are $\ge 0$ and continuous on $[0,1]$ ⇒ washer formula applies on one piece
  • Radii (vertical)
    • Outer radius $𝒇(𝒙)=𝒙^{2}$ (farther from $𝑦=0$)
    • Inner radius $𝓰(𝒙)=𝒙^{3}$ (closer to $𝑦=0$)
      • Cross-sectional area: $𝓐(𝒙)=\pi\big([𝒙^{2}]^{2}-[𝒙^{3}]^{2}\big)=\pi(𝒙^{4}-𝒙^{6})$
  • Set up the integral
    • $\displaystyle 𝓥=\int_{0}^{1}\pi\big([𝒙^{2}]^{2}-[𝒙^{3}]^{2}\big)\,d𝒙=\pi\int_{0}^{1}(𝒙^{4}-𝒙^{6})\,d𝒙$
  • Evaluate
    • $\displaystyle 𝓥=\pi\left[\dfrac{𝒙^{5}}{5}-\dfrac{𝒙^{7}}{7}\right]_{0}^{1}=\pi\left(\dfrac{1}{5}-\dfrac{1}{7}\right)=\pi\cdot\dfrac{2}{35}$
  • Result: $\displaystyle 𝓥=\dfrac{2\pi}{35}$
• 2:13:35. 🧩 Example – 3:Revolve between $𝑦=𝒙^{2}$ and $𝒙=𝑦^{2}$, revolved around 𝑦-axis (washers in 𝒅𝑦) - [📷image]
  • Goal & method
    • $\displaystyle 𝓥=\int_{𝓬}^{𝓭}\pi\big([𝒖(𝑦)]^{2}-[𝒗(𝑦)]^{2}\big)\,d𝑦$
    • Cross-sections ⟂ to 𝒚-axis ⇒ **washers** (right/outer − left/inner), integrate 𝒅𝑦
    • Write curves **in terms of 𝑦** (solve for $𝒙=\dots$) to measure horizontal radii
  • Put functions in terms of 𝑦
    • From $𝑦=𝒙^{2}\Rightarrow 𝒙=\sqrt{𝑦}$ (right/outer curve)
    • Given $𝒙=𝑦^{2}$ (left/inner curve).
      • Whiteboard check: $((\sqrt{𝑦})^{2}=(𝑦^{2})^{2}\Rightarrow 𝑦=𝑦^{4}\Rightarrow 𝑦(𝑦^{3}-1)=0)$
      • Intersections: $𝑦=0, 𝑦=1$ ⇒ interval $[𝓬,𝓭]=[0,1]$
  • Radii (horizontal, measured from the 𝒚-axis)
    • Outer radius $𝒖(𝑦)=𝒙_{\text{right}}=\sqrt{𝑦}$
    • Inner radius $𝒗(𝑦)=𝒙_{\text{left}}=𝑦^{2}$
      • Cross-sectional area: $𝓐(𝑦)=\pi\big([\sqrt{𝑦}]^{2}-[𝑦^{2}]^{2}\big)=\pi(𝑦-𝑦^{4})$
  • Set up the integral
    • $\displaystyle 𝓥=\int_{0}^{1}\pi\big([\sqrt{𝑦}]^{2}-[𝑦^{2}]^{2}\big)\,d𝑦=\pi\int_{0}^{1}(𝑦-𝑦^{4})\,d𝑦$
  • Evaluate
    • $\displaystyle 𝓥=\pi\left[\dfrac{𝑦^{2}}{2}-\dfrac{𝑦^{5}}{5}\right]_{0}^{1}=\pi\left(\dfrac{1}{2}-\dfrac{1}{5}\right)=\dfrac{3\pi}{10}$
  • Result: $\displaystyle 𝓥=\dfrac{3\pi}{10}$
• 2:25:07. 🧩 Example – 4:Revolve between $𝑦=𝒙$ (outside) and $𝑦=𝒙^{3}$ (inside) about $𝑦=\tfrac{3}{2}$ (shifted axis, washers in 𝒅𝒙) - [📷image-1] - [📷image-2] - [📷image-3]
  • Goal & method
    • $\displaystyle 𝓥=\int_{𝓪}^{𝓫}\pi\big([𝒇(𝒙)]^{2}-[𝓰(𝒙)]^{2}\big)\,d𝒙$
    • Axis of rotation is the horizontal line $𝑦=\tfrac{3}{2}$ ⇒ cross-sections ⟂ to 𝒙-axis ⇒ **washers** in 𝒅𝒙.
    • Radii are **vertical distances** from the axis $𝑦=\tfrac{3}{2}$ to each curve (outer − inner)
  • Intersections & interval in 𝒙
    • Set $𝑦=𝒙$ and $𝑦=𝒙^{3}$ equal: $𝒙^{3}=𝒙^{2}\Rightarrow 𝒙^{2}(𝒙-1)=0\Rightarrow 𝒙=0,1$
    • On $(0,1)$: test $𝒙=\tfrac{1}{2}$ ⇒ $\tfrac{1}{2}>\tfrac{1}{8}$ ⇒ **outer curve** is $𝑦=𝒙$, **inner curve** is $𝑦=𝒙^{3}$
      • Bounds: $[𝓪,𝓫]=[0,1]$
  • Radii with respect to $𝑦=\tfrac{3}{2}$ (vertical distances)
    • Outer radius $𝑹_{\text{out}}(𝒙)=\dfrac{3}{2}-𝒙^{3}$ (farther from the axis)
    • Inner radius $𝑹_{\text{in}}(𝒙)=\dfrac{3}{2}-𝒙$ (closer to the axis)
      • Cross-sectional area: $𝓐(𝒙)=\pi\big(𝑹_{\text{out}}^{2}-𝑹_{\text{in}}^{2}\big)$
  • Set up the washer integral
    • $\displaystyle 𝓥=\int_{0}^{1}\pi\Big[\big(\dfrac{3}{2}-𝒙^{3}\big)^{2}-\big(\dfrac{3}{2}-𝒙\big)^{2}\Big]\,d𝒙$
  • Expand to simplify before integrating
    • $\left(\dfrac{3}{2}-𝒙^{3}\right)^{2}=\dfrac{9}{4}-3𝒙^{3}+𝒙^{6};\; \left(\dfrac{3}{2}-𝒙\right)^{2}=\dfrac{9}{4}-3𝒙+𝒙^{2}$
    • Difference: $𝒙^{6}-3𝒙^{3}+3𝒙-𝒙^{2}$
      • $\displaystyle 𝓥=\pi\int_{0}^{1}\big(𝒙^{6}-3𝒙^{3}-𝒙^{2}+3𝒙\big)\,d𝒙$ (reordered terms).
  • Evaluate
    • Antiderivative: $\displaystyle \frac{𝒙^{7}}{7}-\frac{3}{4}𝒙^{4}-\frac{𝒙^{3}}{3}+\frac{3}{2}𝒙^{2}$
    • Plug bounds: $\displaystyle 𝓥=\pi\left(\frac{1}{7}-\frac{3}{4}-\frac{1}{3}+\frac{3}{2}\right)=\pi\cdot\frac{47}{84}$
  • Result: $\displaystyle 𝓥=\frac{47\pi}{84}$