Surface Area of Graphs

Surface Area from a Graph

A surface given by $z=f(x,y)$ is often easier to describe than to parametrize with an unfamiliar coordinate system. If the surface lies over a planar region $D$, its area can be computed from the slopes of the graph above that region.

The key point is that a sloping piece of surface is generally larger than its projection onto the $xy$-plane. The partial derivatives $f_x$ and $f_y$ measure how the graph tilts in the two coordinate directions. Their values determine how much the area is stretched.

Definition: Let $D$ be a bounded planar region with piecewise smooth boundary, and let $f$ have continuous first partial derivatives on an open set containing $D$. The area of the graph $z=f(x,y)$ over $D$ is $$ A=\iint_D \sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\,dA. $$ Here $dA$ denotes area in the $xy$-plane. The integral must exist; under the stated regularity and boundedness conditions, its integrand is continuous on $D$.

The region $D$ is the graph’s projection onto the $xy$-plane, not a region measured along the surface. The integrand is at least $1$, so the graph’s area cannot be smaller than the area of its projection. Equality holds when both partial derivatives vanish throughout $D$, as for a horizontal graph.

Why the Area Factor Appears

A graph can be parametrized by its projected coordinates: $\mathbf r(x,y)=\langle x,y,f(x,y)\rangle$. A small rectangle in the parameter plane is mapped to a small surface patch. The two tangent directions to that patch are $\mathbf r_x=\langle 1,0,f_x\rangle$ and $\mathbf r_y=\langle 0,1,f_y\rangle$. The magnitude of their cross product gives the local area-stretch factor:

\mathbf r_x\times\mathbf r_y = \langle -f_x,-f_y,1\rangle, \qquad \left\lVert\mathbf r_x\times\mathbf r_y\right\rVert = \sqrt{f_x^2+f_y^2+1}.

Thus, a small projected area $dA$ corresponds locally to a surface area approximately $\sqrt{1+f_x^2+f_y^2}\,dA$. Adding these contributions across $D$ gives the stated integral. The factor also confirms that changing the sign of a slope does not change area: only the squares of the partial derivatives enter.

A Practical Setup

For each graph-area calculation, identify the projection region, compute both partial derivatives, and substitute them into the area factor. Then evaluate the double integral over the specified region. If the region or integrand has symmetry, choose coordinates that make the integration simpler, while keeping the area element appropriate to those coordinates.

Identify the graph and its domain.
Write the surface as $z=f(x,y)$ and determine its projection $D$ in the $xy$-plane.

Compute the slopes.
Find $f_x$ and $f_y$, then form $\sqrt{1+f_x^2+f_y^2}$.

Integrate over the projection.
Evaluate the resulting double integral using coordinates suited to $D$.

Check the result.
Confirm that the area is at least the projected area, and check any antiderivative or coordinate conversion used.

Worked Examples

Worked Example: A Plane over a Rectangle

Find the area of the plane $z=2x-y+3$ above the rectangle $D=[0,2]\times[0,1]$.

First compute the partial derivatives: $$ f_x=2,\qquad f_y=-1. $$ The area factor is constant: $$ \sqrt{1+f_x^2+f_y^2} =\sqrt{1+2^2+(-1)^2} =\sqrt{6}. $$ The rectangle has side lengths $2$ and $1$, so its projected area is $2$. Therefore, $$ A=\int_0^2\int_0^1\sqrt{6}\,dy\,dx =\int_0^2\sqrt{6}\,dx =2\sqrt{6}. $$ This is approximately $4.899$ square units to three decimal places. To check, the constant area factor $\sqrt 6$ multiplies the projected area $2$, giving $2\sqrt 6$; it is larger than $2$, as required for a non-horizontal plane.

Worked Example: A Radial Quadratic Graph over a Disk

Find the area of $z=(x^2+y^2)/2$ above the unit disk $D=\{(x,y):x^2+y^2\leq 1\}$.

The partial derivatives are $$ f_x=x,\qquad f_y=y. $$ Thus the area factor is $\sqrt{1+x^2+y^2}$. Since the region and integrand are radial, use polar coordinates, for which $x^2+y^2=r^2$ and $dA=r\,dr\,d\theta$: $$ A=\int_0^{2\pi}\int_0^1\sqrt{1+r^2}\,r\,dr\,d\theta. $$ For the inner integral, set $u=1+r^2$, so $du=2r\,dr$. When $r=0$, $u=1$; when $r=1$, $u=2$. Hence $$ \int_0^1 r\sqrt{1+r^2}\,dr =\frac12\int_1^2 u^{1/2}\,du =\frac13\left[u^{3/2}\right]_1^2 =\frac{2\sqrt{2}-1}{3}. $$ The angular integral contributes a factor of $2\pi$, so $$ A=\frac{2\pi}{3}(2\sqrt{2}-1). $$ As a check, differentiating $u^{3/2}/3$ gives $\sqrt u/2$, as required after the substitution. Also, the result exceeds the projected disk area $\pi$, because the graph is not flat; indeed, $2(2\sqrt{2}-1)/3>1$.

Worked Example: A Quadratic Graph over a Rectangle

Find the area of $z=x^2/2$ over $D=[-1,1]\times[0,1]$.

Here $f_x=x$ and $f_y=0$. The area factor is $\sqrt{1+x^2}$, independent of $y$, so $$ A=\int_{-1}^{1}\int_0^1\sqrt{1+x^2}\,dy\,dx =\int_{-1}^{1}\sqrt{1+x^2}\,dx =2\int_0^1\sqrt{1+x^2}\,dx. $$ An antiderivative is $$ F(x)=\frac12\left(x\sqrt{1+x^2}+\ln\left(x+\sqrt{1+x^2}\right)\right). $$ To verify it, differentiate the first term inside the parentheses: $$ \frac{d}{dx}\left(x\sqrt{1+x^2}\right) =\sqrt{1+x^2}+\frac{x^2}{\sqrt{1+x^2}}. $$ The derivative of the logarithm is $1/\sqrt{1+x^2}$. Combining the terms and multiplying by $1/2$ gives $$ F'(x)=\frac12\left( \sqrt{1+x^2}+\frac{x^2+1}{\sqrt{1+x^2}} \right) =\sqrt{1+x^2}. $$ Therefore, $$ A=2\bigl(F(1)-F(0)\bigr) =\sqrt{2}+\ln(1+\sqrt{2}). $$ Numerically, $\sqrt{2}\approx1.41421356$ and $\ln(1+\sqrt{2})\approx0.88137359$, so $A\approx2.29558715$, or $2.296$ square units to three decimal places. As a check, the antiderivative differentiates to the integrand, and the value is greater than the projected area $2$.

Common Pitfalls

A frequent error is to use $1+f_x^2+f_y^2$ as the area factor and omit the square root. For the plane in the first example, that incorrect calculation would give $A=6\cdot2=12$. But the required factor is $\sqrt{1+f_x^2+f_y^2}=\sqrt6$, so the correct area is $2\sqrt6$. The incorrect value also fails the direct area-stretch calculation: the tangent vectors span a parallelogram of area $\sqrt6$, not $6$.

Another error is to integrate over the wrong region or omit a coordinate area factor. In polar coordinates, $dA=r\,dr\,d\theta$, not $dr\,d\theta$; the extra $r$ accounts for the changing size of polar coordinate patches. Always state the projected region and the area element explicitly before evaluating the integral.

Key takeaway: For a continuously differentiable graph $z=f(x,y)$ over a suitable planar region $D$, compute its area by integrating $\sqrt{1+f_x^2+f_y^2}$ over $D$. The partial derivatives measure the tilt, and the square root converts projected area into surface area.

Why This Formula Is Useful

The graph formula turns a three-dimensional geometric measurement into a double integral over a planar region. It is especially effective when the surface is naturally described by a height function and its projection has simple boundaries. For a plane, the slopes are constant, so the area factor is constant. For a curved graph, the factor varies from point to point and the integral sums those varying local stretches.

It is also a useful consistency check to compare surface area with projected area. Since $1+f_x^2+f_y^2\geq1$, the integrand is at least $1$, and consequently $A\geq\iint_D1\,dA$. If a calculation gives less than the projected area, inspect the derivatives, the square root, and the integration limits.

Check Your Understanding

For each question, identify the graph’s projection and explain the role of the area factor.

For $z=3x+4y$, what is the constant area factor?
Why must the graph’s area be at least the area of its projection?
For $z=(x^2+y^2)/2$ over a disk of radius $R$, what expression replaces the upper radial limit in the polar integral?
What error results from using $1+f_x^2+f_y^2$ instead of its square root?
When converting a double integral over a disk to polar coordinates, what factor belongs in $dA$, and why?

What You'll Learn