Calculus II Lesson 9: Volumes

1. Presentations
   1. Scheduling
2. Questions?
3. Exam 1 Comments
4. Volumes
   1. Slicing a Sphere
   2. Slicing a Cone
5. Revolution
   1. Notice
   2. Example
   3. Solution
   4. Exercise
   5. Gabriel’s Horn
   6. Washer Method
   7. Cross Sections
   8. Example
   9. Solution
6. Shells
   1. Shell Method
7. Upcoming

Presentations

Scheduling

• Make the schedule for next Thursday (3/5) and the following Monday (3/9).

Questions?

• Improper integrals?
• Areas between curves?
• Homework?

Exam 1 Comments

• Interpreting graphs / FTC
• Other questions?

Volumes

General strategy:

• Take thin slices of 3D figure
• Find formula for $A(x)$, the area of one cross-section
• $V = \int_a^b A(x) dx$

Slicing a Sphere

Problem: Derive the formula for the volume of a sphere of radius $R$. ($R$ is a fixed constant.)

• What is the formula for the area of the “slice” that’s $x$-units from the origin?
• $A = \pi(R^2 - x^2)$.
• Multiply that area by a small “thickness” $\Delta x$: $\pi (R^2 - x^2) \Delta x$.
• Add all of these up and take the limit as $\Delta x$ goes to 0?

The formula we get is $V = \int_0^R \pi (R^2 - x^2) dx$. Evaluate this integral?

Slicing a Cone

Problem: Derive the formula for the volume of a (right-circular) cone whose base has a radius $R$ and whose height is $h$.

• Cross sections are circles
• Area of a circle: $\pi r^2$.
• Radius of circle at height $x$?

Notice the similar triangles:

Solution: First we see that $r = \frac{Rx}{h}$, so the area of that slice is $\pi (\frac{Rx}{h})^2$. Set up the volume integral:

\[\begin{align} V &= \int_0^h \pi (\frac{Rx}{h})^2 dx \\ &= \pi \frac{R^2}{h^2} \int_0^h x^2 dx \\ &= \left. \pi \frac{R^2}{h^2} \frac{x^3}{3} \right|_0^3 \\ &= \pi \frac{R^2}{h^2} \frac{h^3}{3} \\ &= \frac{1}{3} \pi R^2 h \end{align}\]

Revolution

Other three dimensional figures can be found by revolving 2D regions around axes.

Notice

When we revolve an entire region bounded by a curve $y = f(x)$ around the $x$-axis:

• Cross sections are circle: $A(x) = \pi r^2$.
• $r = f(x)$: the radius of each circle is $f(x)$!
• $V = \int_a^b \pi (f(x))^2 dx$

Example

Find the volume of the solid formed by revolving the region bounded above by $y = x^2$, below by the $x$-axis, between $x = 0$ and $x = 2$, around the $x$-axis.

Geogebra demo

Solution

\[\begin{align*} V &= \int_0^2 \pi (x^2)^2 dx \\ &= \pi \int_0^2 x^4 dx \\ &= \pi \left.\frac{x^5}{5} \right|_0^2 \\ &= \frac{32\pi}{5} \approx 20.1 \end{align*}\]

Exercise

1. Find the volume of the solid formed by revolving the region bounded by $f(x) = 1 - x^2$, $x = 0$, $x = 1$, and the $x$-axis around the $x$-axis.
2. Find the volume of solid formed by revolving the region bounded on the left by $x = 1$, and above by $y = \frac{1}{x}$, around the $x$-axis.

Try to sketch the 3D solids that are formed in each case.

Gabriel’s Horn

• “Gabriel’s Horn” (Who is Gabriel?)
• Finite Volume
• Infinite length?
• 2D Projection: infinite area
• Infinite surface area

Washer Method

What if we revolve a region bounded between two curves around the $x$-axis?

Cross Sections

• Cross sections are washers, outer radius $R$, inner radius $r$
• Area of a washer: $\pi (R^2 - r^2)$
• $R = f(x)$, $r = g(x)$.
• $V = \int_a^b \pi [(f(x))^2 - (g(x))^2] dx$

Example

Find the volume of the solid formed by revolving the region bounded by $y = \sin(x)$ and $y = \cos(x)$ from $x = 0$ to $x = \frac{\pi}{4}$ around the $x$-axis.

Solution

\[\begin{align} V &= \int_0^{\pi/4} \pi (\cos^2(x) - \sin^2(x)) dx \\ &= \pi \int_0^{\pi/4} \cos(2x) dx \\ &= \left. \pi \frac{\sin(2x)}{2} \right|_0^{\pi/4} \\ &= \pi \cdot \frac{\sin(\pi/2)}{2} = \frac{\pi}{2} \end{align}\]

Shells

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• Revolving regions around the y-axis?
• Which region?
• A: disk method. Convert to an integral using $y$ and $dy$
• B? Something else.

Shell Method

• Volume of one shell: $2\pi rh\Delta x$ (Why?)
• $r \approx x$
• $h \approx f(x)$.
• $V = \int_a^b 2 \pi x f(x) dx$

Upcoming

• MyOpenMath homework will be posted and due next Friday.
• Presentations continuing next week.