Setting Up the Region
A power function can make a useful model for learning how shell radius and height work together. When a region under its graph is revolved around the \(y\)-axis, each vertical slice sweeps out a cylindrical shell. The shell's radius depends on its horizontal position, while its height is determined by the power function.
There is one boundary detail to settle first. The graph \(y=x^p\) and the coordinate axes alone do not enclose a finite region extending to the right. We will use the region under the graph from \(x=0\) to a specified endpoint \(x=a\), where \(a>0\). The familiar unit-interval region is the case \(a=1\).
The condition \(p>0\) ensures that \(x^p\) is defined for \(x\ge 0\) and that the graph meets the origin. The shell method uses vertical slices here: each slice is parallel to the axis of rotation, so revolving it produces a shell. The shell radius is the distance from the slice at \(x\) to the \(y\)-axis, namely \(x\). Its height is the vertical distance from the \(x\)-axis to the curve, namely \(x^p\).
The Power-Function Shell Formula
A cylindrical shell of radius \(r\), height \(h\), and thickness \(\Delta x\) has approximate volume \(2\pi r h\,\Delta x\). Adding the shells across the interval and taking the limit gives the shell integral. For the region \(R\), the radius is \(x\), the height is \(x^p\), and the bounds are \(0\) and \(a\).
The first expression is the shell setup. Since \(x\cdot x^p=x^{p+1}\), evaluating its integral gives the stated exact formula. In particular, for the unit-interval region, the formula simplifies to \(V=2\pi/(p+2)\).
Why the Integral Has This Form
Divide \([0,a]\) into narrow intervals. A typical interval near \(x_i\) has width \(\Delta x_i\). Its vertical slice lies about \(x_i\) units from the axis and has height about \(x_i^p\). On rotation, that slice contributes approximately the volume of a thin cylindrical shell.
Here \(2\pi x_i\) is the shell's circumference, \(x_i^p\) is its height, and \(\Delta x_i\) is its thickness. Adding these approximate volumes gives a Riemann sum. As the interval widths shrink, the sum approaches the volume integral.
At position \(x\), its distance from the \(y\)-axis is \(x\).
The top boundary is \(y=x^p\) and the bottom boundary is \(y=0\), so height is \(x^p-0=x^p\).
The shell contribution is approximately \(2\pi x\cdot x^p\cdot\Delta x\).
The slices cover \(0\le x\le a\), so the limiting sum is \(2\pi\int_0^a x^{p+1}\,dx\).
Since \(p>0\), the exponent \(p+1\ne -1\), and the power rule gives \(2\pi[a^{p+2}/(p+2)-0]\).
This also justifies the exact formula: the shell contributions add in the limit, and the power rule evaluates their integral. The lower endpoint contributes zero because \(0^{p+2}=0\) for \(p>0\).
Worked Examples
Worked Example: The Unit Region Under \(y=x^2\)
Rotate the region bounded by \(y=x^2\), the coordinate axes, and \(x=1\) about the \(y\)-axis. Here \(p=2\) and \(a=1\). A slice at \(x\) has radius \(x\), height \(x^2\), and thickness \(dx\), with bounds \(0\le x\le1\). Thus $$ V=2\pi\int_0^1 x(x^2)\,dx =2\pi\int_0^1 x^3\,dx. $$ An antiderivative of \(x^3\) is \(x^4/4\), so $$ V=2\pi\left[\frac{x^4}{4}\right]_0^1 =2\pi\left(\frac{1^4}{4}-\frac{0^4}{4}\right) =\frac{\pi}{2}. $$ Check the evaluation by differentiating: \(\frac{d}{dx}(x^4/4)=x^3\), the integrand. Also, the power-function formula gives \(2\pi(1^{2+2})/(2+2)=\pi/2\), in agreement.
Worked Example: A Fractional Power on a Longer Interval
Rotate the region under \(y=\sqrt{x}=x^{1/2}\), for \(0\le x\le4\), about the \(y\)-axis. The shell radius is \(x\), and the height is \(\sqrt{x}\). Therefore $$ V=2\pi\int_0^4 x\sqrt{x}\,dx =2\pi\int_0^4 x^{3/2}\,dx. $$ An antiderivative of \(x^{3/2}\) is \(\frac{2}{5}x^{5/2}\), so $$ V=2\pi\left[\frac{2}{5}x^{5/2}\right]_0^4 =\frac{4\pi}{5}\left(4^{5/2}-0^{5/2}\right). $$ Because \(4^{5/2}=(\sqrt{4})^5=2^5=32\), this becomes $$ V=\frac{4\pi}{5}(32)=\frac{128\pi}{5}. $$ Check the antiderivative: \(\frac{d}{dx}(\frac{2}{5}x^{5/2})=\frac{2}{5}\cdot\frac{5}{2}x^{3/2}=x^{3/2}\). The formula with \(p=1/2\) and \(a=4\) also gives \(2\pi\,4^{5/2}/(5/2)=128\pi/5\).
Worked Example: A Cubic Power with Endpoint \(x=2\)
Rotate the region under \(y=x^3\) on \(0\le x\le2\) about the \(y\)-axis. The radius is \(x\), and the height is \(x^3\). The bounds come from the left endpoint \(x=0\) and the right endpoint \(x=2\): $$ V=2\pi\int_0^2 x(x^3)\,dx =2\pi\int_0^2 x^4\,dx. $$ An antiderivative of \(x^4\) is \(x^5/5\). Evaluating, $$ V=2\pi\left[\frac{x^5}{5}\right]_0^2 =2\pi\left(\frac{2^5}{5}-\frac{0^5}{5}\right) =2\pi\cdot\frac{32}{5} =\frac{64\pi}{5}. $$ Check by differentiation: \(\frac{d}{dx}(x^5/5)=x^4\). Independently, the general formula gives \(2\pi\cdot2^{3+2}/(3+2)=2\pi\cdot32/5=64\pi/5\).
A Common Height Error
A frequent mistake is to use \(1-x^p\) as the shell height because it is the difference between \(y=1\) and \(y=x^p\). That height would apply to a region whose top boundary is \(y=1\), not to the region under \(y=x^p\) and above the \(x\)-axis. For the region defined here, the top is \(x^p\) and the bottom is \(0\).
The example highlights why identifying the region before integrating matters. A shell integral must encode the actual radius and the actual vertical distance between the region's upper and lower boundaries, not merely produce a plausible number.
Using the Formula Carefully
For the unit interval, increasing \(p\) makes the graph \(y=x^p\) lower for \(0<x<1\), so the region becomes thinner over most of the interval. The formula reflects this: \(2\pi/(p+2)\) decreases as \(p\) increases. For an endpoint \(a\) other than \(1\), retain the factor \(a^{p+2}\); omitting it would change the volume and usually the scale of the region.
Before evaluating, check three features of the setup:
- Axis: rotation about the \(y\)-axis makes the vertical shell radius \(x\).
- Height: the region is between \(y=x^p\) and \(y=0\), so the height is \(x^p\).
- Bounds: the shells run from \(x=0\) to the specified endpoint \(x=a\).
These checks keep the geometric model separate from the algebra. Once the setup is correct, the exact evaluation follows directly from the power rule.
Check Your Understanding
For each question, identify the shell quantities before evaluating the integral.
- For the region under \(y=x^4\) on \(0\le x\le1\), what are the radius, height, and shell-method integral when rotating about the \(y\)-axis?
- Find the exact volume when the region under \(y=x\) on \(0\le x\le3\) is rotated about the \(y\)-axis.
- For \(y=x^{2/3}\) on \(0\le x\le8\), state the shell integral and evaluate it exactly.
- Why is \(1-x^p\) not the shell height for the region under \(y=x^p\) and above the \(x\)-axis?
- In the formula \(V=2\pi a^{p+2}/(p+2)\), what geometric information does \(a\) represent?
