Scalar Integrals on Implicit Surfaces

From an Implicit Equation to a Scalar Integral

A surface is sometimes described by an equation such as $F(x,y,z)=0$, rather than by a parametrization or an explicit formula for one coordinate. To integrate a scalar function over such a surface, we can often solve the equation locally for one coordinate and then use the graph formula for surface integrals.

The important point is that two substitutions are needed. The scalar function must be evaluated at points on the surface, and the surface-area element must account for the surface’s slope. Omitting either step changes the integral.

Definition: If $\phi(x,y,z)$ is a scalar function defined on a surface patch $S$, its scalar surface integral is $\iint_S \phi\,dS$. Here $dS$ denotes surface area. The value of $\phi$ is integrated over the surface with respect to area; no orientation is required.

Writing an Implicit Patch as a Graph

Suppose the patch is contained in the level surface $F(x,y,z)=0$, where $F$ has continuous first partial derivatives. At a point where $F_z\ne 0$, the implicit function theorem guarantees that, in a sufficiently small neighborhood, the surface can be written as a graph $z=g(x,y)$. The graph satisfies

$$ F(x,y,g(x,y))=0. $$

For the graph over a region $D$ in the $xy$-plane, the surface-area element is $dS=\sqrt{1+g_x^2+g_y^2}\,dA$, where $dA$ is planar area. Thus, for a continuous scalar function $\phi$, the surface integral is

\iint_S \phi(x,y,z)\,dS = \iint_D \phi(x,y,g(x,y)) \sqrt{1+g_x^2+g_y^2}\,dA.

The domain $D$ must describe exactly the projection of the patch being integrated. If an implicit equation describes more than one sheet over the same region, each sheet must be treated separately.

Deriving the Area Factor from the Implicit Equation

The graph formula can be expressed using derivatives of $F$, which is useful when solving explicitly for $g$ would be cumbersome. Differentiate $F(x,y,g(x,y))=0$ with respect to $x$. The chain rule gives $F_x+F_zg_x=0$, so $g_x=-F_x/F_z$. Differentiating with respect to $y$ gives $g_y=-F_y/F_z$. In these expressions, the partial derivatives of $F$ are evaluated at $(x,y,g(x,y))$.

\sqrt{1+g_x^2+g_y^2} = \sqrt{1+\frac{F_x^2}{F_z^2}+\frac{F_y^2}{F_z^2}} = \frac{\sqrt{F_x^2+F_y^2+F_z^2}}{|F_z|}.

The absolute value is essential: the area factor is positive whether $F_z$ is positive or negative. Writing $\nabla F=(F_x,F_y,F_z)$, the resulting formula is

Theorem: If $F$ has continuous first partial derivatives, $F_z\ne 0$ on a patch that is a single graph $z=g(x,y)$, and $\phi$ is continuous there, then $$ \iint_S \phi\,dS = \iint_D \phi(x,y,g(x,y)) \frac{|\nabla F(x,y,g(x,y))|}{|F_z(x,y,g(x,y))|} \,dA. $$

This is the graph formula with the derivatives of $g$ replaced by derivatives of the implicit defining function. The condition $F_z\ne 0$ is not optional: it is what permits this local representation and keeps the displayed denominator nonzero. If $F_z=0$ but another partial derivative is nonzero, a different coordinate may instead be solved for.

Worked Examples

Worked Example: A Scalar Integral over an Implicit Plane

Let $S$ be the patch $z-2x+y=0$ over $D=[0,1]\times[0,2]$, and let $\phi(x,y,z)=1+x$. Find $\iint_S\phi\,dS$.

Set $F(x,y,z)=z-2x+y$. Then $F_x=-2$, $F_y=1$, and $F_z=1$, so $F_z\ne 0$. The equation gives $z=2x-y$, and the area factor is

\frac{\sqrt{F_x^2+F_y^2+F_z^2}}{|F_z|} = \frac{\sqrt{(-2)^2+1^2+1^2}}{1} = \sqrt{6}.

The integrand $1+x$ does not depend on $z$, so its substitution is unchanged. Therefore

\begin{aligned} \iint_S(1+x)\,dS &=\int_0^2\int_0^1(1+x)\sqrt{6}\,dx\,dy\\ &=\sqrt{6}\int_0^2\left[x+\frac{x^2}{2}\right]_0^1\,dy\\ &=\sqrt{6}\int_0^2\frac{3}{2}\,dy\\ &=3\sqrt{6}. \end{aligned}

As a check, the graph derivatives are $g_x=2$ and $g_y=-1$, giving $\sqrt{1+g_x^2+g_y^2}=\sqrt{1+4+1}=\sqrt{6}$, the same area factor. The planar integral of $1+x$ over $D$ is $3$, so the result $3\sqrt{6}$ agrees with the graph formula.

Worked Example: Height Integrated over a Spherical Cap

Let $S$ be the upper spherical cap $x^2+y^2+z^2=25$ with $z\ge 4$, and take $\phi(x,y,z)=z$. Find $\iint_S z\,dS$.

On this patch, $z=\sqrt{25-x^2-y^2}$. Its projection is the disk $D$ given by $x^2+y^2\le 9$: the boundary $z=4$ implies $x^2+y^2=25-16=9$. With $F=x^2+y^2+z^2-25$, we have $F_z=2z$, which is nonzero because $z\ge4$, and $|\nabla F|=2\sqrt{x^2+y^2+z^2}=10$. Hence the area factor is $10/(2z)=5/z$. Substituting the scalar field $z$ cancels that factor’s denominator:

z\,dS = z\frac{5}{z}\,dA = 5\,dA.

In polar coordinates, the disk has $0\le r\le3$ and $0\le\theta\le2\pi$, with $dA=r\,dr\,d\theta$. Thus

\begin{aligned} \iint_S z\,dS &=\int_0^{2\pi}\int_0^3 5r\,dr\,d\theta\\ &=\int_0^{2\pi}5\left[\frac{r^2}{2}\right]_0^3\,d\theta\\ &=\int_0^{2\pi}\frac{45}{2}\,d\theta\\ &=45\pi. \end{aligned}

For a check, the area of the projection disk is $\pi(3)^2=9\pi$, and the transformed integrand is the constant $5$. Their product is $5(9\pi)=45\pi$, as obtained by integration.

Worked Example: A Height Integral over a Paraboloid Patch

Let $S$ be the patch $z=x^2+y^2$ above the disk $x^2+y^2\le1$, and let $\phi(x,y,z)=z$. Find $\iint_S z\,dS$.

Use $F(x,y,z)=z-x^2-y^2$. Its derivatives are $F_x=-2x$, $F_y=-2y$, and $F_z=1$. Therefore the area factor is $\sqrt{1+4x^2+4y^2}$. The scalar field becomes $x^2+y^2$ on the graph. In polar coordinates, $x^2+y^2=r^2$ and $dA=r\,dr\,d\theta$, so

\begin{aligned} \iint_S z\,dS &=\int_0^{2\pi}\int_0^1 r^2\sqrt{1+4r^2}\,r\,dr\,d\theta\\ &=2\pi\int_0^1 r^3\sqrt{1+4r^2}\,dr. \end{aligned}

Set $u=1+4r^2$. Then $du=8r\,dr$, $r^2=(u-1)/4$, and $r^3\,dr=(u-1)\,du/32$. The bounds become $u=1$ at $r=0$ and $u=5$ at $r=1$. Hence

\begin{aligned} \iint_S z\,dS &=\frac{\pi}{16}\int_1^5 (u-1)\sqrt{u}\,du\\ &=\frac{\pi}{16}\left[\frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}\right]_1^5\\ &=\frac{\pi}{16}\left(\frac{20\sqrt{5}}{3}+\frac{4}{15}\right)\\ &=\frac{\pi(25\sqrt{5}+1)}{60}. \end{aligned}

To check the antiderivative, differentiating $\frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}$ gives $u^{3/2}-u^{1/2}=(u-1)\sqrt{u}$, the required integrand. The final value is positive, consistent with $z\ge0$ on the patch.

A Common Mistake: Substituting the Scalar but Not the Area

For the spherical cap example, one might substitute $z=\sqrt{25-r^2}$ into the scalar and then mistakenly integrate $z\,dA$, as if the curved patch had the same area element as its projection. This incorrect calculation gives

\begin{aligned} \int_0^{2\pi}\int_0^3 \sqrt{25-r^2}\,r\,dr\,d\theta &=2\pi\left[-\frac{(25-r^2)^{3/2}}{3}\right]_0^3\\ &=\frac{122\pi}{3}. \end{aligned}

This is not the surface integral: it uses $dA$ instead of $dS$. The correct area factor is $5/z$, so the transformed integrand is $z(5/z)=5$, and the correct integral is $45\pi$. In general, substituting into $\phi$ does not make the area factor disappear; it disappears only if the algebra actually cancels it.

When the Formula Applies

Before using the implicit graph formula, check that the chosen coordinate can be solved for throughout the patch and that the projection describes the intended surface exactly once. A nonzero $F_z$ ensures a local graph in $z$, but a large patch may need to be divided into smaller pieces if one graph does not cover it. The whole level surface might contain several sheets, even when each sheet is locally a graph.

If $F_z=0$ at a point, the formula with denominator $|F_z|$ cannot be used there. That does not by itself mean the surface is singular. For example, another partial derivative might be nonzero, allowing the surface to be written locally as $x=h(y,z)$ or $y=h(x,z)$. The appropriate area factor then uses the corresponding projection. If all three partial derivatives vanish, this graph argument provides no regularity guarantee at that point.

Key takeaway: For a regular implicit patch that is a graph $z=g(x,y)$, evaluate the scalar field on the graph and multiply by $\sqrt{1+g_x^2+g_y^2}$. Equivalently, use the factor $|\nabla F|/|F_z|$, provided $F_z\ne0$ on the patch.

Check Your Understanding

For each question, identify both the scalar substitution and the appropriate area factor before evaluating or simplifying.

For $F(x,y,z)=z-x-y=0$, what is the graph area factor over the $xy$-plane?
Why must the implicit graph formula use $|F_z|$, rather than $F_z$, in its denominator?
For the sphere $x^2+y^2+z^2=25$, where $z\ge4$, what is the projection region in the $xy$-plane?
What condition allows the implicit function theorem to represent a patch locally as $z=g(x,y)$?
If $F_z=0$ but $F_x\ne0$, which coordinate could be solved for locally instead?