From a Surface to Its Projection
A scalar surface integral adds up the values of a scalar field across a surface, with each small contribution weighted by the area of the surface piece. When the surface is a graph \(z=f(x,y)\), the integral can be computed over a region in the \(xy\)-plane. The key is to account for two things: the scalar field’s value at the point on the graph, and the amount by which surface area differs from projected area.
Let \(D\) be a region in the \(xy\)-plane and let \(f:D\to\mathbb{R}\) be continuously differentiable. Its graph is the surface \(S\) consisting of points \((x,y,f(x,y))\) for \((x,y)\in D\). If \(\Phi(x,y,z)\) is a continuous scalar field defined on \(S\), its value at the graph point over \((x,y)\) is \(\Phi(x,y,f(x,y))\). This substitution is essential: the field must be evaluated at points on the surface being integrated.
The formula separates the work into manageable pieces: substitute \(z=f(x,y)\) into the scalar field, compute the graph area factor, and integrate the resulting function over \(D\). The surface is not being replaced by its projection; the factor preserves the effect of the surface’s slope.
Why the Graph Formula Works
A graph has a natural parametrization: use \((x,y)\) itself as the pair of parameters. The point on the surface is \(\mathbf{r}(x,y)=(x,y,f(x,y))\). For a regular parametrized surface, the scalar surface integral is the integral of the field evaluated at the parametrized point, multiplied by the magnitude of the cross product of the two tangent vectors. Applying that definition gives the graph formula directly.
Set \(\mathbf{r}(x,y)=(x,y,f(x,y))\) for \((x,y)\in D\).
Differentiating with respect to each parameter gives \(\mathbf{r}_x=(1,0,f_x)\) and \(\mathbf{r}_y=(0,1,f_y)\).
The cross product is \(\mathbf{r}_x\times\mathbf{r}_y=(-f_x,-f_y,1)\), whose magnitude is \(\sqrt{f_x^2+f_y^2+1}\).
At parameter values \((x,y)\), the surface point has height \(f(x,y)\), so the field value is \(\Phi(x,y,f(x,y))\).
By the definition of a scalar integral on a parametrized patch, the integrand in parameter space is the field value times the tangent cross-product magnitude. Substituting the expressions above yields \[ \iint_S\Phi\,dS = \iint_D\Phi(\mathbf{r}(x,y))\lVert\mathbf{r}_x\times\mathbf{r}_y\rVert\,dA = \iint_D\Phi(x,y,f(x,y))\sqrt{1+f_x^2+f_y^2}\,dA. \] This also explains the geometric role of the square root: a small projected rectangle becomes a tilted surface piece, whose area is multiplied by the magnitude of the cross product of its tangent directions.
The factor is always at least \(1\), since \(1+f_x^2+f_y^2\geq 1\). It equals \(1\) where both partial derivatives vanish, and increases where the graph is steeper. The scalar field itself may be positive, negative, or variable; it is a separate factor from the area correction.
Worked Examples on Rectangular Regions
Worked Example: A Plane and Its Height Field
Let \(S\) be the graph \(z=x+y\) over \(D=[0,1]\times[0,2]\), and let \(\Phi(x,y,z)=z\). Find \(\iint_S\Phi\,dS\).
First, \(f_x=1\) and \(f_y=1\), so the graph area factor is constant: \[ \sqrt{1+f_x^2+f_y^2}=\sqrt{1+1+1}=\sqrt{3}. \] On the graph, \(z=x+y\), so the scalar field restricts to \(\Phi(x,y,f(x,y))=x+y\). Therefore, \[ \iint_S\Phi\,dS =\sqrt{3}\int_0^1\int_0^2(x+y)\,dy\,dx. \] Evaluate the inner integral: \[ \int_0^2(x+y)\,dy =\left[xy+\frac{y^2}{2}\right]_0^2 =2x+2. \] Then \[ \sqrt{3}\int_0^1(2x+2)\,dx =\sqrt{3}\left[x^2+2x\right]_0^1 =3\sqrt{3}. \] For a check, integrating \(x+y\) over the rectangle gives an \(x\)-contribution of \(2\int_0^1x\,dx=1\) and a \(y\)-contribution of \(\int_0^2y\,dy=2\), for a total of \(3\). Multiplying by the constant area factor \(\sqrt{3}\) confirms \(3\sqrt{3}\).
Worked Example: A Polynomial Field on a Sloping Plane
Let \(S\) be the graph \(z=x+2y\) over \(D=[0,1]\times[0,1]\), and let \(\Phi(x,y,z)=xz\). Compute the scalar surface integral.
The partial derivatives are \(f_x=1\) and \(f_y=2\), so \[ \sqrt{1+f_x^2+f_y^2} =\sqrt{1+1+4} =\sqrt{6}. \] Substitute the graph equation into the field: \[ \Phi(x,y,f(x,y))=x(x+2y)=x^2+2xy. \] Thus, \[ \iint_S\Phi\,dS =\sqrt{6}\int_0^1\int_0^1(x^2+2xy)\,dy\,dx. \] Integrating first with respect to \(y\) gives \[ \int_0^1(x^2+2xy)\,dy =\left[x^2y+xy^2\right]_0^1 =x^2+x. \] Then \[ \sqrt{6}\int_0^1(x^2+x)\,dx =\sqrt{6}\left[\frac{x^3}{3}+\frac{x^2}{2}\right]_0^1 =\sqrt{6}\left(\frac13+\frac12\right) =\frac{5\sqrt{6}}{6}. \] For a check, the antiderivative \(x^3/3+x^2/2\) differentiates to \(x^2+x\), and its endpoint difference is \(5/6\). The constant graph factor therefore gives the stated result.
Radial Graphs Over Disks
If the projection region is a disk and both the graph and the restricted field depend only on the distance \(r=\sqrt{x^2+y^2}\) from the origin, polar coordinates often make the double integral simpler. In polar coordinates, \(dA=r\,dr\,d\theta\). The graph factor must still be computed from \(f_x\) and \(f_y\); it is not replaced by the polar area factor \(r\). These are distinct multipliers: one adjusts surface area, and the other expresses projected area in polar coordinates.
Worked Example: Area of a Radial Graph
Find the area of the graph \(z=(x^2+y^2)/2\) over the unit disk \(D\). For area, the scalar field is \(\Phi=1\).
Here \(f_x=x\) and \(f_y=y\), so the graph factor is \[ \sqrt{1+x^2+y^2}=\sqrt{1+r^2}. \] The disk has \(0\leq r\leq 1\) and \(0\leq\theta\leq 2\pi\). Therefore, \[ \operatorname{Area}(S) =\int_0^{2\pi}\int_0^1\sqrt{1+r^2}\,r\,dr\,d\theta =2\pi\int_0^1r\sqrt{1+r^2}\,dr. \] Use \(u=1+r^2\), so \(du=2r\,dr\). Then \[ 2\pi\int_0^1r\sqrt{1+r^2}\,dr =\pi\int_1^2u^{1/2}\,du =\frac{2\pi}{3}\left[u^{3/2}\right]_1^2 =\frac{2\pi}{3}(2\sqrt{2}-1). \] As a check, the derivative of \(\frac{2}{3}(1+r^2)^{3/2}\) is \(2r\sqrt{1+r^2}\), so \(\frac13(1+r^2)^{3/2}\) is an antiderivative of \(r\sqrt{1+r^2}\). Its endpoint difference, multiplied by \(2\pi\), gives the same area.
Worked Example: Integrating the Height Over a Radial Graph
On the same graph \(z=(x^2+y^2)/2\) over the unit disk, find the integral of the height field \(\Phi(x,y,z)=z\).
On the graph, \(\Phi=r^2/2\), and the graph factor remains \(\sqrt{1+r^2}\). Hence \[ \iint_S z\,dS =\int_0^{2\pi}\int_0^1\frac{r^2}{2}\sqrt{1+r^2}\,r\,dr\,d\theta =\pi\int_0^1r^3\sqrt{1+r^2}\,dr. \] Set \(u=1+r^2\). Then \(r^2=u-1\) and \(r\,dr=du/2\), so \(r^3\,dr=(u-1)\,du/2\). The bounds change from \(r=0,1\) to \(u=1,2\). Thus, \[ \pi\int_0^1r^3\sqrt{1+r^2}\,dr =\frac{\pi}{2}\int_1^2(u-1)u^{1/2}\,du. \] An antiderivative is \[ \frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}, \] because its derivative is \(u^{3/2}-u^{1/2}=(u-1)u^{1/2}\). Evaluating, \[ \frac{\pi}{2}\left[\frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2}\right]_1^2 =\frac{\pi}{2}\left(\frac{4\sqrt{2}}{15}+\frac{4}{15}\right) =\frac{2\pi(\sqrt{2}+1)}{15}. \] The positive result is consistent with both the nonnegative height and the positive area factor. Differentiating the displayed antiderivative recovers the substituted integrand, and the endpoint difference confirms the evaluation.
A Common Error: Omitting the Area Factor
A frequent mistake is to restrict the field correctly but then integrate it over the projection as though projected area and surface area were the same. In the polynomial-plane example, this incorrect calculation would be \[ \int_0^1\int_0^1(x^2+2xy)\,dy\,dx=\frac{5}{6}. \] That is the integral over the square in the \(xy\)-plane, not the scalar surface integral. The graph has area factor \(\sqrt{6}\), so the correct result is \(\sqrt{6}\cdot 5/6=5\sqrt{6}/6\). The correction is to include the graph factor before evaluating the double integral.
A second check is to ensure the field is evaluated on the surface. For \(\Phi=xz\) and \(z=x+2y\), the projected integrand is \(x(x+2y)\), not \(xz\) with \(z\) left as an independent variable. The graph equation specifies the height at every point of the projected region.
Check Your Understanding
For each question, identify the restricted field and the graph area factor before attempting the integral.
- For \(z=3x-y\) over a rectangle, what is the graph area factor?
- If \(\Phi(x,y,z)=y+z\) and \(z=x^2+y^2\), what function must be integrated over the projection, before including the area factor?
- Why does polar-coordinate area contribute a factor \(r\) in addition to the graph area factor?
- For \(z=(x^2+y^2)/2\), write the area integral over a disk of radius \(a\) using polar coordinates.
- What error is made if the field is restricted to the graph but the graph area factor is omitted?