Surface Area from an Implicit Equation

From an Equation to a Surface Area Integral

An equation such as $G(x,y,z)=0$ can describe a surface without expressing any one coordinate explicitly. To calculate area, we can often solve locally for a coordinate and use the graph area formula. The key is to check that this coordinate really can be solved for on the patch, and that the resulting projection counts the surface points correctly.

The same surface may be a graph over one coordinate plane in one region and fail to be a graph over that plane elsewhere. A zero partial derivative signals that the corresponding graph description may fail locally. It does not, by itself, mean that the surface has a singularity.

Definition: Let $G$ be continuously differentiable near a point of the level surface $G(x,y,z)=0$. The point is regular if $\nabla G\ne\mathbf{0}$, where $\nabla G=(G_x,G_y,G_z)$. A regular patch is a portion of this surface consisting of regular points.

Choosing a Coordinate to Solve For

The implicit function theorem says that if $G_z\ne0$ at a regular point, then in a sufficiently small neighborhood the surface can be written as $z=f(x,y)$. In this neighborhood, implicit differentiation gives $G(x,y,f(x,y))=0$, and hence

G_x+G_z f_x=0,\qquad G_y+G_z f_y=0.

Thus $f_x=-G_x/G_z$ and $f_y=-G_y/G_z$, with the derivatives of $G$ evaluated on the surface. The graph area factor is $\sqrt{1+f_x^2+f_y^2}$. Substitution gives the area factor directly in terms of the implicit equation.

Theorem: If $G_z\ne0$ on a patch that is a single graph over a region $D$ in the $xy$-plane, then its area is $$ A=\iint_D\sqrt{1+\left(\frac{G_x}{G_z}\right)^2+ \left(\frac{G_y}{G_z}\right)^2}\,dA =\iint_D\frac{\|\nabla G\|}{|G_z|}\,dA. $$ The derivatives are evaluated at the corresponding surface points. The denominator is an absolute value because area is nonnegative.

The same reasoning works for either of the other coordinates. The table gives the area factor when the indicated coordinate is solved for.

Graph description	Required condition	Area factor over the projection
$z=f(x,y)$	$G_z\ne0$	$\\|\nabla G\\|/\|G_z\|$
$y=f(x,z)$	$G_y\ne0$	$\\|\nabla G\\|/\|G_y\|$
$x=f(y,z)$	$G_x\ne0$	$\\|\nabla G\\|/\|G_x\|$

In each row, the double integral is over the corresponding coordinate-plane projection. The formula applies to a single graph sheet; if the projection covers multiple sheets, calculate the area of each sheet separately and add the results.

Why the Area Factor Has This Form

For the graph $z=f(x,y)$, a small region in the $xy$-plane is stretched on the surface according to the graph area factor $\sqrt{1+f_x^2+f_y^2}$. Implicit differentiation expresses those slopes in terms of $G$. In particular,

1+f_x^2+f_y^2 =1+\frac{G_x^2}{G_z^2}+\frac{G_y^2}{G_z^2} =\frac{G_x^2+G_y^2+G_z^2}{G_z^2}.

Taking the nonnegative square root yields $\|\nabla G\|/|G_z|$. This proves the formula when $G_z\ne0$. It also explains why the partial derivative in the denominator matters: as it approaches zero, the surface may become steep relative to that projection, and the graph area factor may grow without bound.

Worked Example: An Implicit Paraboloid

Find the area of the patch $G(x,y,z)=z-x^2-y^2=0$ lying above the disk $x^2+y^2\le1$.

Here $G_z=1$, so the patch is a graph $z=x^2+y^2$ over the whole disk $D$. The derivatives are $G_x=-2x$, $G_y=-2y$, and $G_z=1$. Therefore the area factor is

\sqrt{1+4x^2+4y^2}.

Use polar coordinates on the disk: $x=r\cos\theta$, $y=r\sin\theta$, and $dA=r\,dr\,d\theta$, with $0\le r\le1$ and $0\le\theta\le2\pi$. The area is

\begin{aligned} A &=\int_0^{2\pi}\int_0^1 r\sqrt{1+4r^2}\,dr\,d\theta\\ &=2\pi\int_0^1 r\sqrt{1+4r^2}\,dr. \end{aligned}

Set $u=1+4r^2$. Then $du=8r\,dr$; when $r=0$, $u=1$, and when $r=1$, $u=5$. Thus

\begin{aligned} A &=\frac{2\pi}{8}\int_1^5 u^{1/2}\,du\\ &=\frac{\pi}{4}\left[\frac{2}{3}u^{3/2}\right]_1^5\\ &=\frac{\pi}{6}(5\sqrt{5}-1). \end{aligned}

Check the integration by differentiating the antiderivative $(1+4r^2)^{3/2}/12$: its derivative is $r\sqrt{1+4r^2}$, the required integrand. The bounds give $(5\sqrt5-1)/12$ before multiplication by $2\pi$, confirming the area.

Projection Failure Is Not Necessarily Surface Failure

Solving for $z$ requires $G_z\ne0$, but it does not follow that the surface is singular when $G_z=0$. At a regular point, at least one of $G_x,G_y,G_z$ is nonzero. If $G_z=0$ but $G_y\ne0$, for example, the implicit function theorem permits solving locally for $y$ instead. A projection fails when its chosen coordinate cannot serve as a local graph coordinate; the surface itself may remain perfectly regular.

If all three partial derivatives vanish, the equation does not establish regularity at that point. The level set might genuinely be singular, or the particular function $G$ might be a poor defining equation for a smooth surface. In either case, the regular graph area formula cannot be applied there without further analysis.

Worked Example: A Spherical Cap and Its Equator

Find the area of the cap of the sphere $G(x,y,z)=x^2+y^2+z^2-25=0$ above the plane $z=4$.

On this cap, $z\ge4$, so $G_z=2z\ne0$ and the upper part is a graph over the disk $x^2+y^2\le9$. Since $G_x=2x$, $G_y=2y$, and $G_z=2z$, the area factor is

\frac{\|\nabla G\|}{|G_z|} =\frac{\sqrt{4x^2+4y^2+4z^2}}{2z} =\frac{5}{z} =\frac{5}{\sqrt{25-x^2-y^2}}.

In polar coordinates, the area is

\begin{aligned} A &=\int_0^{2\pi}\int_0^3 \frac{5r}{\sqrt{25-r^2}}\,dr\,d\theta\\ &=10\pi\int_0^3\frac{r}{\sqrt{25-r^2}}\,dr. \end{aligned}

An antiderivative of $r/\sqrt{25-r^2}$ is $-\sqrt{25-r^2}$, since its derivative is $r/\sqrt{25-r^2}$. Therefore

A=10\pi\left[-\sqrt{25-r^2}\right]_0^3 =10\pi(-4+5)=10\pi.

At the equator $z=0$, $G_z=0$, so this sphere cannot be written as a smooth graph $z=f(x,y)$ there. However, $\nabla G=(2x,2y,2z)$ has length $10$ everywhere on the sphere. In particular, the equator is regular. It is the projection onto the $xy$-plane that fails at the equator, not the surface.

When a Projection Counts More Than One Sheet

A projection need not distinguish surface points: two different points on the surface can have the same projected coordinates. For example, solving an equation for $y$ may produce both a positive and a negative branch. The area integral over one branch gives only that branch’s area. Count each branch included in the patch exactly once.

Worked Example: Two Graph Patches on a Cylinder

Consider the cylinder $G(x,y,z)=x^2+y^2-25=0$, restricted to $-3\le x\le3$ and $0\le z\le2$. Find the combined area of the patches with $y>0$ and $y<0$.

On either patch, solve for $y$: $y=\pm\sqrt{25-x^2}$. Since $|x|\le3$, the quantity $25-x^2$ is at least $16$, so both graphs are smooth throughout this domain. The derivatives are $G_x=2x$, $G_y=2y$, and $G_z=0$. With $y$ as the graph coordinate, the area factor over the $xz$-plane is

\frac{\|\nabla G\|}{|G_y|} =\frac{\sqrt{4x^2+4y^2}}{2|y|} =\frac{5}{\sqrt{25-x^2}}.

The projected region has $x$ from $-3$ to $3$ and $z$ from $0$ to $2$. The area of either branch is

\begin{aligned} A_{\text{branch}} &=\int_0^2\int_{-3}^3\frac{5}{\sqrt{25-x^2}}\,dx\,dz\\ &=2\left[5\arcsin\left(\frac{x}{5}\right)\right]_{-3}^3\\ &=10\left(\arcsin\frac35-\arcsin\left(-\frac35\right)\right)\\ &=20\arcsin\frac35. \end{aligned}

The antiderivative check is $\frac{d}{dx}\bigl(5\arcsin(x/5)\bigr)=5/\sqrt{25-x^2}$. The two patches have equal area, so their combined area is $40\arcsin(3/5)$. At $x=\pm5$, $G_y=0$ and solving for $y$ fails; those points lie outside the stated patches. The calculation therefore uses no invalid graph description.

A Common Mistake: Using a Formula Where Its Denominator Vanishes

It is tempting to apply the $z=f(x,y)$ area factor across an entire surface even when $G_z$ vanishes. For the sphere in the cap example, the factor $\|\nabla G\|/|G_z|$ becomes $5/|z|$. This expression is undefined at the equator, so it cannot directly describe a graph over a region containing the equator.

The correction is to restrict that formula to a region where $G_z\ne0$, as in the cap above, or to choose a different local graph coordinate near the equator. Also check whether the projection is one-to-one: if two branches project onto the same region, integrating the area factor for just one branch omits the other.

Practical Steps for an Implicit Surface

Check regularity.
Determine whether $\nabla G\ne\mathbf{0}$ on the patch under consideration.

Choose a graph coordinate.
For a $z=f(x,y)$ description, check that $G_z\ne0$; for the other choices, check the corresponding partial derivative.

Determine the projection and its sheets.
Specify the projected region and check whether each point in it corresponds to one surface point or several.

Integrate and account for every patch.
Use the matching area factor, and split the surface into suitable graph patches if a chosen projection fails.

Implicit equations are useful because they describe surfaces compactly, but the equation alone does not determine a single graph or a single projection. A valid area calculation combines the right graph condition, the corresponding projection, and a careful count of the surface sheets.

Check Your Understanding

Use the graph conditions and area factors to reason through these questions.

For $G(x,y,z)=0$, what condition permits a local graph $z=f(x,y)$, and what is its area factor?
At a regular point where $G_z=0$, what can you conclude about the surface, and what should you check next?
If solving for $y$ gives two branches over the same projected region, how should the area be counted?
For the sphere $x^2+y^2+z^2=25$, why does the $xy$-projection fail as a smooth $z$-graph at the equator even though the sphere is regular there?
Which partial derivative must be nonzero to solve locally for $x$, and which coordinate plane is the projection?

Graph description	Required condition	Area factor over the projection
\(z=f(x,y)\)	\(G_z\ne0\)	\(\\|\nabla G\\|/\|G_z\|\)
\(y=f(x,z)\)	\(G_y\ne0\)	\(\\|\nabla G\\|/\|G_y\|\)
\(x=f(y,z)\)	\(G_x\ne0\)	\(\\|\nabla G\\|/\|G_x\|\)

Surface Area from an Implicit Equation

What You'll Learn

From an Equation to a Surface Area Integral

Choosing a Coordinate to Solve For

Why the Area Factor Has This Form

Worked Example: An Implicit Paraboloid

Worked Example: An Implicit Paraboloid

Projection Failure Is Not Necessarily Surface Failure

Worked Example: A Spherical Cap and Its Equator

Worked Example: A Spherical Cap and Its Equator

When a Projection Counts More Than One Sheet

Worked Example: Two Graph Patches on a Cylinder

A Common Mistake: Using a Formula Where Its Denominator Vanishes

Practical Steps for an Implicit Surface

Check Your Understanding

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