Revolving A Graph About The Y-Axis

Why the Graph’s Orientation Matters

A curve can be described using \(y\) as a function of \(x\), or using \(x\) as a function of \(y\). When the curve is revolved about the \(y\)-axis, writing it as \(x=g(y)\) makes the distance from the axis especially clear: at height \(y\), the radius is the horizontal distance \(|g(y)|\).

Surface area depends on both this radius and the length of the small piece of the curve being revolved. The radius determines the circumference of a thin band, while the curve’s arc length determines the band’s slanted width. Keeping these two roles distinct is the key to setting up the integral correctly.

Core formula: If the curve \(x=g(y)\), for \(a\le y\le b\), is continuously differentiable and generates the surface once, then its surface area about the \(y\)-axis is

S=2\pi\int_a^b |g(y)|\sqrt{1+[g'(y)]^2}\,dy.

The factor \(|g(y)|\) is the radius, and \(\sqrt{1+[g'(y)]^2}\,dy\) is the arc-length element.

Radius and Arc Length

At a fixed value of \(y\), the point on the profile has coordinates \((g(y),y)\). Its distance to the \(y\)-axis, whose points have \(x=0\), is \(|g(y)|\). The absolute value matters because a distance cannot be negative: a point on the left side of the axis and a point the same distance to its right produce the same radius when revolved.

For a small change \(dy\), the horizontal change along the graph is approximately \(dx=g'(y)\,dy\). The corresponding small arc length \(ds\) follows from the Pythagorean theorem:

ds=\sqrt{(dx)^2+(dy)^2} =\sqrt{[g'(y)]^2+1}\,dy.

This expression is the arc-length element for a graph expressed as \(x=g(y)\). It is not generally equal to \(dy\): that would ignore the horizontal change in the curve.

Why the Surface-Area Formula Works

Imagine dividing the profile into many short pieces. When each piece is revolved about the \(y\)-axis, it forms a narrow band. If the piece has arc length approximately \(\Delta s\) and its radius is approximately \(r=|g(y)|\), the band’s area is approximately its circumference times its slanted width, \(2\pi r\,\Delta s\).

The approximation can be made precise using the area of a frustum of a cone. If the two endpoint radii are \(r_1\) and \(r_2\), and the slant length is \(\ell\), its lateral area is \(\pi(r_1+r_2)\ell\). For a sufficiently short piece, \(r_1\) and \(r_2\) are both close to the radius \(r\), so this area is close to \(2\pi r\ell\). Adding the band areas over a partition and taking the limit gives the surface-area integral.

Find the radius.
At height \(y\), measure the distance from the profile to the axis: \(r(y)=|g(y)|\).

Find the arc-length element.
For \(x=g(y)\), use \(ds=\sqrt{1+[g'(y)]^2}\,dy\).

Multiply circumference by arc length.
The local band area is \(dS=2\pi r(y)\,ds\).

Integrate over the specified interval.
Substitute the radius and \(ds\), then integrate from \(y=a\) to \(y=b\).

Substitution of the radius and arc-length element yields the formula:

dS=2\pi |g(y)|\,ds =2\pi |g(y)|\sqrt{1+[g'(y)]^2}\,dy, \qquad S=\int_a^b dS.

Worked Examples

Worked Example: A Slanted Line Segment

Revolve the segment \(x=2+y\), for \(0\le y\le2\), about the \(y\)-axis. Find its surface area.

On this interval \(g(y)=2+y\), which is positive, so the radius is \(r(y)=2+y\). The derivative and arc-length factor are

g'(y)=1, \qquad \sqrt{1+[g'(y)]^2}=\sqrt{1+1}=\sqrt{2}.

Therefore, the surface-area integral is

\begin{aligned} S &=2\pi\int_0^2 (2+y)\sqrt{2}\,dy\\ &=2\pi\sqrt{2}\left[2y+\frac{y^2}{2}\right]_0^2\\ &=2\pi\sqrt{2}\left((4+2)-0\right)\\ &=12\pi\sqrt{2}. \end{aligned}

Check using the frustum formula. The endpoint radii are \(2\) and \(4\). The segment’s slant length is \(\sqrt{(2-0)^2+(4-2)^2}=\sqrt{8}=2\sqrt{2}\). Thus its lateral area is \(\pi(2+4)(2\sqrt{2})=12\pi\sqrt{2}\), agreeing with the integral.

Worked Example: A Vertical Line Produces a Cylinder

Revolve the segment \(x=3\), for \(-1\le y\le2\), about the \(y\)-axis.

Here \(g(y)=3\), so the radius is \(3\), and \(g'(y)=0\). Hence \(ds=\sqrt{1+0}\,dy=dy\). The height of the segment is \(2-(-1)=3\), giving

\begin{aligned} S &=2\pi\int_{-1}^{2}3\,dy\\ &=6\pi[y]_{-1}^{2}\\ &=6\pi(2-(-1))\\ &=18\pi. \end{aligned}

Check against the cylinder’s lateral area: circumference times height is \(2\pi(3)(3)=18\pi\). The integral agrees. The top and bottom circular ends are not included because the rotated segment generates only the cylinder’s curved side.

Worked Example: A Spherical Zone from a Circular Profile

Revolve the curve \(x=\sqrt{4-y^2}\), for \(0\le y\le1\), about the \(y\)-axis. This is a portion of the circle \(x^2+y^2=4\), so the resulting surface is part of a sphere of radius \(2\).

Throughout this interval the radius is \(g(y)=\sqrt{4-y^2}\). Differentiating gives

g'(y)=\frac{-y}{\sqrt{4-y^2}}, \qquad 1+[g'(y)]^2 =1+\frac{y^2}{4-y^2} =\frac{4}{4-y^2}.

Thus \(ds=\frac{2}{\sqrt{4-y^2}}\,dy\). The radius and arc-length factor cancel in the integrand:

\begin{aligned} S &=2\pi\int_0^1 \sqrt{4-y^2}\,\frac{2}{\sqrt{4-y^2}}\,dy\\ &=4\pi\int_0^1 1\,dy\\ &=4\pi[y]_0^1\\ &=4\pi. \end{aligned}

Check using the spherical-zone area formula \(2\pi Rh\), where \(R\) is the sphere’s radius and \(h\) is the zone’s vertical height. Here \(R=2\) and \(h=1-0=1\), so the area is \(2\pi(2)(1)=4\pi\), as obtained by integration.

The profile’s arc length over this interval is not \(1\). In fact,

\int_0^1 ds =\int_0^1\frac{2}{\sqrt{4-y^2}}\,dy =2\arcsin\left(\frac12\right) =\frac{\pi}{3}.

This is a spherical zone of radius \(2\) and vertical height \(1\), so its area is \(2\pi\cdot2\cdot1=4\pi\). Its area is not found by multiplying the sphere’s circumference by the profile’s arc length.

Worked Example: A Second Spherical Zone

Revolve \(x=\sqrt{9-y^2}\), for \(0\le y\le2\), about the \(y\)-axis. This profile lies on a sphere of radius \(3\).

The radius is \(\sqrt{9-y^2}\), and

g'(y)=\frac{-y}{\sqrt{9-y^2}}, \qquad 1+[g'(y)]^2 =1+\frac{y^2}{9-y^2} =\frac{9}{9-y^2}.

Therefore \(ds=\frac{3}{\sqrt{9-y^2}}\,dy\). Substituting into the surface-area integral gives

\begin{aligned} S &=2\pi\int_0^2 \sqrt{9-y^2}\,\frac{3}{\sqrt{9-y^2}}\,dy\\ &=6\pi\int_0^2 1\,dy\\ &=6\pi[y]_0^2\\ &=12\pi. \end{aligned}

Check: the sphere has radius \(3\), and the zone has vertical height \(2\), so its area is \(2\pi(3)(2)=12\pi\). This confirms the integral.

Conditions and Common Setup Errors

The formula applies directly when \(g\) is continuously differentiable on the interval, so that the arc-length factor is defined, and the chosen curve segment generates the surface being measured once. The integral’s limits must describe the actual vertical extent of that segment. If the curve is described by different formulas on different intervals, set up the integral separately on each smooth piece and add the resulting areas, provided the pieces together describe the intended surface without counting any part more than once.

The absolute value in \(|g(y)|\) is essential whenever the profile lies to the left of the \(y\)-axis. For example, if \(g(y)=-2\), the radius is \(2\), not \(-2\). A negative radius would incorrectly produce a negative surface-area contribution.

Common mistake: Using \(2\pi\int_a^b |g(y)|\,dy\) as the surface area assumes that the profile’s arc-length element is simply \(dy\). For a slanted curve, this omits the factor \(\sqrt{1+[g'(y)]^2}\). The correct setup is \(2\pi\int_a^b |g(y)|\sqrt{1+[g'(y)]^2}\,dy\).

For instance, applying the mistaken setup to \(x=2+y\) on \(0\le y\le2\) gives \(2\pi\int_0^2(2+y)\,dy=12\pi\). But the actual slanted width is \(\sqrt{2}\,dy\), not \(dy\). The correct area is \(12\pi\sqrt{2}\), as the frustum check confirms.

A related error is to confuse the profile’s arc length with the surface area. Arc length measures distance along the generating curve. Surface area measures the circumference of each rotational band multiplied by its slanted width and accumulated along the curve. The circular-profile example shows why both factors are needed.

Summary

For a profile \(x=g(y)\) revolved about the \(y\)-axis, use the distance \(|g(y)|\) as the radius. The arc-length element is \(ds=\sqrt{1+[g'(y)]^2}\,dy\), so the surface-area integral is \(2\pi\int_a^b|g(y)|\sqrt{1+[g'(y)]^2}\,dy\). Check that the interval and curve describe the intended surface, then verify the setup against familiar geometry when possible.

Check Your Understanding

Use the radius and arc-length element separately in each problem before evaluating the integral.

For \(x=g(y)\) revolved about the \(y\)-axis, what is the radius at height \(y\), and why might an absolute value be needed?
Derive the arc-length element \(ds\) for a graph \(x=g(y)\) from the Pythagorean theorem.
Set up, but do not evaluate, the surface-area integral for \(x=1+2y\) on \(0\le y\le1\).
Why does revolving the segment \(x=3\) on an interval of length \(h\) produce lateral area \(6\pi h\)?
For the profile \(x=\sqrt{R^2-y^2}\), explain why the radius and arc-length factors simplify in the surface-area integrand.