How to Find Volume by Spinning: Shell Method

Beyond handling typical revolution solids, the Shell Method shines in scenarios where internal structures vary significantly, like in hollow or layered objects. Its adaptability to both horizontal and vertical axes of rotation makes it a versatile tool in engineering and architectural design, enabling precise volume estimations for objects like pipes within walls or tunnels through hills. By focusing on the lateral surface area of thin shells, it also facilitates understanding material requirements for manufacturing and construction, bridging theoretical calculus with practical applications in material science and mechanical engineering.

For a solid generated by revolving a region around the y-axis, the Shell Method’s mathematical approach involves:

1. Identify the Function: Consider a function \( f(x) \) defined on the interval \([a, b]\).
2. Define Shell Radius and Height: The radius of each shell is \( x \), and the height is given by \( f(x) \).
3. Shell Volume Formula: The volume of a thin shell is approximated by the circumference of the shell \((2\pi x)\) times its height \((f(x))\) times its thickness \((dx)\).
4. Set Up the Integral: The total volume \(V\) is the integral of the shell volume formula over the interval \([a, b]\):
   \( V = \int_{a}^{b} 2\pi x f(x) \, dx \)

Let’s calculate the volume of the solid formed by revolving the region between the curve \( y = \ln(x) \) and the x-axis, around the y-axis from \( x = 1 \) to \( x = e \) (where \( e \) is the base of the natural logarithm).

Step-by-Step Solution Using the Shell Method:

1. Function and Interval:

• The function is \( y = \ln(x) \), defined from \( x = 1 \) to \( x = e \).

2. Shell Radius and Height:

• Radius \(( r )\) of each cylindrical shell is the x-coordinate, \( x \).
• Height \(( h )\) of each shell is given by the function \( y = \ln(x) \).

3. Volume of a Shell:

• The volume of each shell is \( 2\pi x \ln(x) \, dx \).

4. Set Up the Integral:

• The total volume \( V \) is the integral of the shell volume from \( x = 1 \) to \( x = e \):
  \( V = \int_{1}^{e} 2\pi x \ln(x) \, dx \)

5. Evaluate the Integral:

To solve the integral (\int 2\pi x \ln(x) dx) using integration by parts, we’ll follow the integration by parts formula, (\int u \, dv = uv – \int v \, du), step by step:

• Choose \(u\) and \(dv\):

• Let \(u = \ln(x)\), which means \(du = \frac{1}{x} dx\).
• Let \(dv = 2\pi x dx\), which means \(v = \pi x^2\).

• Apply the Integration by Parts Formula:

• According to the formula, \(\int 2\pi x \ln(x) dx = uv – \int v \, du\).
• Substituting \(u\), \(du\), and \(v\) gives: \(\pi x^2 \ln(x) – \int \pi x^2 (\frac{1}{x}) dx\).
• Simplify the integral: \(\pi x^2 \ln(x) – \pi \int x dx\).

• Solve the Simplified Integral:

• The integral \(\int x dx) simplifies to (\frac{1}{2}x^2\).
• Therefore, the expression becomes \(\pi x^2 \ln(x) – \pi \frac{1}{2}x^2\).

• Evaluate from \(x = 1\) to \(x = e\):

• The final step involves evaluating the expression from \(x = 1\) to \(x = e\).

Let’s perform these calculations for the correct evaluation.

After performing the integration by parts and evaluating from \(x = 1\) to \(x = e\), the volume of the solid formed by revolving the region between \(y = \ln(x)\) and the x-axis around the y-axis is approximately \(13.18\) cubic units.

Recommended EffortlessMath Books

For the calculus background behind volumes of revolution, the Pre-Calculus for Beginners pins down the function and integration prerequisites, and the Calculus for Beginners walks through every integration technique – including disk, washer, and shell methods – with full worked examples.

Pre-Calculus for Beginners The Ultimate Step by Step Guide to Acing Precalculus

Download

$27.99 Original price was: $27.99.$17.99Current price is: $17.99.

Frequently Asked Questions

What is the shell method?

The shell method computes the volume of a solid of revolution by adding up thin cylindrical shells. Each shell is like a hollow tube with circumference \(2\pi r\), height \(h\), and thickness \(dx\) (or \(dy\)). The total volume is \(V=2\pi\int_a^b r\,h\,dx\).

When should I use shells instead of disks?

Use shells when slicing parallel to the axis of rotation gives a cleaner setup. If your region is easy to describe in \(x\) but you’re rotating about a vertical axis, shells are usually faster. If the region is easier in \(y\) and rotated about \(y\)-axis, switch back to disks/washers in \(y\).

What’s the formula for rotating about the y-axis?

\(V=2\pi\int_a^b x\,f(x)\,dx\), assuming \(f(x)\geq 0\) on \([a,b]\) and the region sits between the curve and the \(x\)-axis. The radius is \(x\) (distance from the strip to the \(y\)-axis) and the height is \(f(x)\).

How do I rotate about a line other than an axis?

Replace the radius with the distance from the strip to the line. Rotating about \(x=3\)? Radius becomes \(|x-3|\). Rotating about \(x=-2\)? Radius becomes \(x-(-2)=x+2\). The integrand changes but the structure stays the same: \(2\pi\int r\,h\,dx\).

Walk me through a worked example?

Rotate the region bounded by \(y=x^2\) and \(y=0\) from \(x=0\) to \(x=2\) about the \(y\)-axis. Each shell at position \(x\) has radius \(x\), height \(x^2\), and thickness \(dx\). \(V=2\pi\int_0^2 x\cdot x^2\,dx = 2\pi\int_0^2 x^3\,dx = 2\pi[\frac{x^4}{4}]_0^2 = 2\pi\cdot 4 = 8\pi\).

What’s the difference between shell and disk methods?

Disk method slices perpendicular to the axis of rotation; each slice is a disk (or washer). Shell method slices parallel to the axis; each slice becomes a cylindrical shell. They give the same answer for the same region – they’re just different setups for the integral. Pick whichever has cleaner bounds.

Can I rotate about a horizontal axis?

Yes. For rotation about \(y=k\), use horizontal strips with thickness \(dy\). The radius is the distance from the strip to the line \(y=k\), and the height is the width of the strip in \(x\). Integrate in \(y\) with bounds equal to the \(y\)-values of the region.

What if the region has two curves bounding it?

The height of the shell is the difference of the two functions at \(x\). If the region is bounded above by \(y=f(x)\) and below by \(y=g(x)\), then \(h(x)=f(x)-g(x)\). Volume becomes \(V=2\pi\int_a^b x[f(x)-g(x)]\,dx\) for rotation about the \(y\)-axis.

Why does the formula have a \(2\pi\)?

Because each shell, when flattened, is a thin rectangle with width equal to the circumference (\(2\pi r\)) and length equal to the height \(h\). The thickness is \(dx\). Volume of one shell is \(2\pi r\cdot h\cdot dx\), and summing (integrating) gives the \(2\pi\) pulled outside.

Where does the shell method show up on tests?

AP Calculus AB and BC, college Calc II, and any course covering applications of integration. Free-response questions often give a region and let you choose your method – shells when the function is easier in one variable but the axis is parallel to that variable. Practice both methods so you can pick the easier setup.

Related EffortlessMath Lessons

If a topic on this page feels rusty, these short lessons go deeper:

• Volume by spinning: disk method
• Volume by spinning: washer method
• Volume by cross-sections
• How to calculate the volume of cubes and prisms
• How to use the Pythagorean theorem