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Disk Volumes About a Vertical Axis

Use horizontal distances from the y-axis to build and evaluate disk-method volume integrals.

Intermediate 9 min read

What You'll Learn

  • Explain why rotation of a horizontal segment that reaches the y-axis forms a disk.
  • Express a disk radius as a horizontal distance, not as a vertical coordinate.
  • Choose y-bounds and write a volume integral for rotation about the y-axis.
  • Evaluate disk-method integrals for curved and linear boundaries.
  • Split an integral when the radius has different formulas on different y-intervals.
  • Check units and identify a common radius error.

From Horizontal Slices to Disks

When a plane region is rotated around a vertical axis, a horizontal slice sweeps out a circular cross section. If that slice begins at the axis of rotation and extends outward to the region’s boundary, the cross section is a disk. Its radius is the slice’s horizontal length.

This tutorial considers rotation about the y-axis, whose equation is \(x=0\). The region must touch the y-axis along the horizontal slices being used, with each slice extending from the axis in one direction. This makes every cross section a solid disk rather than a disk with a hole.

Definition: If a horizontal slice at height \(y\) extends from the y-axis to \(x=r(y)\), where \(r(y)\ge 0\), then rotation about the y-axis produces a disk of radius \(r(y)\). Its area is \(A(y)=\pi[r(y)]^2\).

The radius is a horizontal distance. For a region to the right of the y-axis, it is the right-hand \(x\)-coordinate of the slice. For example, if the boundary is described by \(x=\sqrt{y}\), the radius at height \(y\) is \(\sqrt{y}\), not \(y\).

The Disk-Method Formula

To add the areas of these disks, integrate with respect to \(y\). If the region extends from \(y=a\) to \(y=b\), and the radius \(r(y)\) is continuous or piecewise continuous on that interval, then the volume is

$$ V=\int_a^b A(y)\,dy =\pi\int_a^b [r(y)]^2\,dy. $$

Here \(dy\) represents a small thickness in the vertical direction. The area of a disk has units of square units, so multiplying by a thickness gives cubic units, as a volume should.

The bounds \(a\) and \(b\) are the lowest and highest \(y\)-values in the region. The formula for the radius comes from the horizontal distance between the y-axis and the boundary at each height. If that boundary changes, the integral must be split where the radius formula changes.

Why the Integral Gives the Volume

The disk formula follows from approximating the solid with thin layers. Consider a short vertical interval of thickness \(\Delta y\) near height \(y_i\). The cross section there has area approximately \(\pi[r(y_i)]^2\). The layer’s volume is therefore approximately its cross-sectional area times its thickness.

1
Choose a height.
At height \(y_i\), the horizontal slice has radius \(r(y_i)\).
2
Find the disk area.
Rotating the slice gives a disk with area \(\pi[r(y_i)]^2\).
3
Approximate a thin layer.
For thickness \(\Delta y\), its volume is approximately \(\pi[r(y_i)]^2\Delta y\).
4
Add the layers and take a limit.
The sum of these approximations approaches the integral as the largest thickness approaches zero.

More precisely, divide \([a,b]\) into subintervals and choose a sample height \(y_i\) in each one. The resulting sum is

$$ \sum_i \pi[r(y_i)]^2\Delta y_i. $$

As the subintervals become arbitrarily thin, this Riemann sum converges to \(\int_a^b \pi[r(y)]^2\,dy\), provided the cross-sectional area is integrable. Thus the integral adds the volumes of the circular layers.

Worked Examples

Worked Example: Rotating a Region Bounded by a Parabola

The region is bounded by \(x=0\), \(y=x^2\), and \(y=4\). Find its volume when it is rotated about the y-axis.

First find the vertical limits. The region begins at \(y=0\), where \(y=x^2\) meets \(x=0\), and ends at \(y=4\). Solve the parabola equation for \(x\), taking the nonnegative branch because the region is to the right of the y-axis:

$$ y=x^2 \quad\Longrightarrow\quad x=\sqrt{y}. $$

At height \(y\), the horizontal slice runs from \(x=0\) to \(x=\sqrt{y}\). Therefore its radius is \(r(y)=\sqrt{y}\). Set up the disk integral:

$$ V=\pi\int_0^4(\sqrt{y})^2\,dy =\pi\int_0^4 y\,dy. $$

Evaluate:

$$ V=\pi\left[\frac{y^2}{2}\right]_0^4 =\pi\left(\frac{4^2}{2}-\frac{0^2}{2}\right) =\pi(8-0) =8\pi. $$

Check the integration by differentiating the antiderivative: \(\frac{d}{dy}(y^2/2)=y\), the integrand. The volume is \(8\pi\) cubic units.

Worked Example: A Triangular Region

The region is bounded by the y-axis, the line \(x+y=3\), and the x-axis. Find the volume when it is rotated about the y-axis.

The line gives \(x=3-y\). Its intersection with the x-axis occurs at \(y=0\), and its intersection with the y-axis occurs at \(y=3\). Thus the bounds are \(0\le y\le 3\). At each height, the horizontal segment runs from \(x=0\) to \(x=3-y\), so its radius is \(r(y)=3-y\).

$$ V=\pi\int_0^3(3-y)^2\,dy. $$

Expand and integrate:

$$ \begin{aligned} V &=\pi\int_0^3(9-6y+y^2)\,dy\\ &=\pi\left[9y-3y^2+\frac{y^3}{3}\right]_0^3\\ &=\pi\left[\left(27-27+9\right)-0\right]\\ &=9\pi. \end{aligned} $$

Check the antiderivative: differentiating \(9y-3y^2+y^3/3\) gives \(9-6y+y^2=(3-y)^2\). The volume is \(9\pi\) cubic units. Its radius decreases from \(3\) to \(0\), consistent with the triangle narrowing to the y-axis at \(y=3\).

Worked Example: A Radius That Changes Formula

A region is bounded by the y-axis and the two line segments \(x=y\) for \(0\le y\le1\), and \(x=2-y\) for \(1\le y\le2\). Find the volume when the region is rotated about the y-axis.

The region runs from \(y=0\) to \(y=2\). On the first interval the right boundary is \(x=y\), so the radius is \(r(y)=y\). On the second interval the boundary is \(x=2-y\), so the radius is \(r(y)=2-y\). Since the formula changes at \(y=1\), split the integral there:

$$ V=\pi\int_0^1 y^2\,dy+\pi\int_1^2(2-y)^2\,dy. $$

Evaluate each part:

$$ \begin{aligned} \pi\int_0^1 y^2\,dy &=\pi\left[\frac{y^3}{3}\right]_0^1 =\pi\left(\frac13-0\right) =\frac{\pi}{3},\\ \pi\int_1^2(2-y)^2\,dy &=\pi\left[-\frac{(2-y)^3}{3}\right]_1^2\\ &=\pi\left[\left(0\right)-\left(-\frac13\right)\right] =\frac{\pi}{3}. \end{aligned} $$

Adding the two volumes gives

$$ V=\frac{\pi}{3}+\frac{\pi}{3}=\frac{2\pi}{3}. $$

For the check, \(\frac{d}{dy}(y^3/3)=y^2\), and \(\frac{d}{dy}[-(2-y)^3/3]=(2-y)^2\). The two pieces have equal volumes because their radii each range from \(0\) to \(1\) over a height of \(1\).

A Common Radius Mistake

A radius is a distance from the axis, not the \(y\)-coordinate of a point on the boundary. In the parabola example, a common incorrect setup is

$$ V_{\text{wrong}}=\pi\int_0^4 y^2\,dy =\pi\left[\frac{y^3}{3}\right]_0^4 =\frac{64\pi}{3}. $$

This treats \(y\) as the radius. But at height \(y\), the horizontal distance from the y-axis to the curve \(y=x^2\) is \(x=\sqrt{y}\). The correct radius is \(\sqrt{y}\), so the correct integral and value are

$$ V=\pi\int_0^4(\sqrt{y})^2\,dy =\pi\int_0^4 y\,dy =8\pi. $$

A useful check is to state the radius in words before writing the integral: the radius is the horizontal distance from the y-axis to the boundary. Then square that distance to get the disk area.

Setting Up a Vertical-Axis Disk Integral

For a region that touches the y-axis and extends to the right, a reliable setup follows these steps:

  1. Find the lowest and highest \(y\)-values of the region.
  2. At a typical height \(y\), identify the horizontal boundary of the region.
  3. Write the radius as the horizontal distance from \(x=0\) to that boundary.
  4. Use \(A(y)=\pi[r(y)]^2\) for the disk area.
  5. Integrate with respect to \(y\), splitting the integral if the radius formula changes.

The essential choice is to use horizontal slices: they are perpendicular to the vertical axis, and their horizontal lengths become disk radii. Once the radius and the vertical bounds are correct, the volume integral follows directly.

Check Your Understanding

For each question, identify the horizontal distance that serves as the radius before evaluating or setting up an integral.

  1. A region is bounded by \(x=0\), \(y=x^2\), and \(y=9\). What are the \(y\)-bounds and the disk radius at height \(y\)?
  2. A region has horizontal slices from \(x=0\) to \(x=5-y\) for \(0\le y\le5\). Write the volume integral for rotation about the y-axis.
  3. For the region in question 2, evaluate the volume integral.
  4. A region’s radius is \(y\) for \(0\le y\le2\) and \(4-y\) for \(2\le y\le4\). Write a split disk-method integral.
  5. Why is the radius in the first example \(\sqrt{y}\), rather than \(y\)?

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