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Surface Area and Scalar Integrals · Tutorial 866 of 999

Surface Integrals by Projection

Relate surface area to projected area and use the projection direction to simplify surface and scalar-integral calculations.

Advanced 11 min read

What You'll Learn

  • State the projection formula using a unit normal and a unit projection direction.
  • Identify when a surface can be integrated over its projection.
  • Convert surface-area calculations into projected-area calculations.
  • Choose among projection directions by comparing their correction factors and projected regions.
  • Apply the projection formula to a scalar surface integral.
  • Recognize why a zero normal component makes the formula unusable in that direction.

Why Project a Surface?

A curved surface can be difficult to describe directly, even when its shadow on a coordinate plane has a simple shape. Projection lets us use that simpler planar region, while accounting for how much the surface is tilted relative to the plane.

The key adjustment depends on the angle between the surface and the projection plane. A surface nearly parallel to that plane has almost the same area as its projection. A surface that is nearly edge-on to the plane has much more area than its projection. The projection formula quantifies this difference.

The Projection Formula

Let \(S\) be a smooth surface patch, and let \(\mathbf{k}\) be a unit vector perpendicular to a plane \(P\). Orthogonal projection in the direction \(\mathbf{k}\) sends each point of \(S\) to the point directly below or above it in \(P\). Write \(D\) for the projected region, and \(dA\) for area in \(P\).

At a point on \(S\), let \(\mathbf{n}\) be either choice of unit normal. The sign of \(\mathbf{n}\) will not matter, because the formula uses an absolute value.

Theorem: Suppose the projection maps the surface patch one-to-one onto \(D\), and \(\mathbf{n}\cdot\mathbf{k}\ne 0\) throughout the patch. Then the surface-area element and projected-area element satisfy $$ dS=\frac{dA}{|\mathbf{n}\cdot\mathbf{k}|}. $$ Consequently, for a scalar function \(g\) that is integrable on \(S\), $$ \iint_S g\,dS = \iint_D \frac{g(\text{point on }S\text{ above the projected point})} {|\mathbf{n}\cdot\mathbf{k}|}\,dA. $$

In the scalar-integral formula, \(g\) is evaluated at the corresponding point on the surface, not just at the point in the projection plane. If a projected point corresponds to more than one surface point, the surface is not a single-valued patch over \(D\); the contributions from the different surface points must be handled separately.

The condition \(\mathbf{n}\cdot\mathbf{k}\ne 0\) is essential. If the normal is perpendicular to the projection direction, the surface is locally edge-on to the projection plane. Its projected area element is then zero, so division by that component cannot recover the surface area.

Why the Formula Works

A small piece of a smooth surface is approximated by a parallelogram. Its area is measured by the magnitude of the cross product of its two tangent sides. Projecting those sides into \(P\) gives a parallelogram whose area is the original area multiplied by the absolute normal component in the projection direction.

This relationship can be made precise with a local parametrization \(\mathbf{r}(u,v)\). The surface-area element is \(|\mathbf{r}_u\times\mathbf{r}_v|\,du\,dv\). The projected parallelogram has signed area \((\mathbf{r}_u\times\mathbf{r}_v)\cdot\mathbf{k}\,du\,dv\); its ordinary area uses the absolute value.

1
Write the surface-area element.
\(dS=|\mathbf{r}_u\times\mathbf{r}_v|\,du\,dv\).
2
Write the projected-area element.
\(dA=|(\mathbf{r}_u\times\mathbf{r}_v)\cdot\mathbf{k}|\,du\,dv\).
3
Use the unit normal.
\(\mathbf{n}=(\mathbf{r}_u\times\mathbf{r}_v)/|\mathbf{r}_u\times\mathbf{r}_v|\), up to sign, so \(dA=|\mathbf{n}\cdot\mathbf{k}|\,dS\).
4
Rearrange.
When \(|\mathbf{n}\cdot\mathbf{k}|>0\), division gives \(dS=dA/|\mathbf{n}\cdot\mathbf{k}|\).

This is a local area comparison. The one-to-one condition lets us add the local contributions over one projected region without counting any surface point more than once.

Projection of a Graph onto the \(xy\)-Plane

For a graph \(z=f(x,y)\), its unit normal can be chosen as \(( -f_x,-f_y,1)/\sqrt{1+f_x^2+f_y^2}\). Projection onto the \(xy\)-plane uses \(\mathbf{k}=(0,0,1)\), so the absolute normal component is \(1/\sqrt{1+f_x^2+f_y^2}\). The projection formula therefore gives the familiar graph area factor. This is one useful choice, but it is not the only possible projection direction.

Worked Example: Area of a Plane Patch from Its \(xy\)-Projection

Find the area of the patch \(z=2x+y\) above the rectangle \(0\le x\le1\), \(0\le y\le2\), using projection onto the \(xy\)-plane.

A normal vector is \((-2,-1,1)\), with length \(\sqrt{(-2)^2+(-1)^2+1^2}=\sqrt6\). Thus a unit normal is \(\mathbf{n}=(-2,-1,1)/\sqrt6\). For \(\mathbf{k}=(0,0,1)\),

$$ |\mathbf{n}\cdot\mathbf{k}|=\frac{1}{\sqrt6}. $$

The projected rectangle has area \((1-0)(2-0)=2\). The correction factor is the reciprocal of \(1/\sqrt6\), so

$$ \operatorname{Area}(S) =\iint_D\frac{1}{|\mathbf{n}\cdot\mathbf{k}|}\,dA =\iint_D\sqrt6\,dA =2\sqrt6. $$

As a check, a parametrization is \(\mathbf{r}(x,y)=(x,y,2x+y)\). Its tangent vectors are \((1,0,2)\) and \((0,1,1)\); their cross product is \((-2,-1,1)\), of length \(\sqrt6\). Integrating that constant area factor over a rectangle of area \(2\) also gives \(2\sqrt6\).

Choosing a Useful Projection Direction

The normal component controls the correction factor: a larger value of \(|\mathbf{n}\cdot\mathbf{k}|\) means a smaller factor between projected area and surface area. But the correction factor is only part of the calculation. A direction with a larger normal component may produce a more awkward projected region or a more complicated integrand. A good choice balances these features.

Worked Example: Two Projections of the Same Plane Patch

Consider the patch \(z=2x\), with \(0\le x\le1\) and \(0\le y\le3\). Compare projection onto the \(xy\)-plane with projection onto the \(yz\)-plane.

A unit normal is \(\mathbf{n}=(-2,0,1)/\sqrt5\).

Projection onto the \(xy\)-plane: Here \(\mathbf{k}=(0,0,1)\), so \(|\mathbf{n}\cdot\mathbf{k}|=1/\sqrt5\). The projected rectangle has area \(1\cdot3=3\). Therefore,

$$ \operatorname{Area}(S)=3\left(\frac{1}{1/\sqrt5}\right)=3\sqrt5. $$

Projection onto the \(yz\)-plane: Here \(\mathbf{k}=(1,0,0)\), so \(|\mathbf{n}\cdot\mathbf{k}|=2/\sqrt5\). Since \(z=2x\), the projected region is the rectangle \(0\le y\le3\), \(0\le z\le2\), with area \(3\cdot2=6\). Thus,

$$ \operatorname{Area}(S)=6\left(\frac{1}{2/\sqrt5}\right) =6\frac{\sqrt5}{2} =3\sqrt5. $$

The answers agree. The second projection has twice the projected area, but half the correction factor. It may be convenient if the \(yz\)-description or an integrand expressed in \(y,z\) is simpler. The direction \((0,1,0)\), by contrast, would not work: \(\mathbf{n}\cdot(0,1,0)=0\), and projection onto the \(xz\)-plane collapses the patch along its \(y\)-direction.

Using Projection for a Scalar Surface Integral

For a scalar surface integral, projection changes the area element, but it does not remove the scalar function. The function must be expressed at the surface point associated with each projected point. The resulting planar integral can be easier if the projected region and the combined integrand simplify.

Worked Example: Integrating Height over a Paraboloid Patch

Evaluate \(\iint_S z\,dS\) over the patch \(z=x^2+y^2\) above the unit disk \(D:\ x^2+y^2\le1\).

Use projection onto the \(xy\)-plane, with \(\mathbf{k}=(0,0,1)\). The graph derivatives are \(f_x=2x\) and \(f_y=2y\). A unit normal is \((-2x,-2y,1)/\sqrt{1+4x^2+4y^2}\), so

$$ |\mathbf{n}\cdot\mathbf{k}| =\frac{1}{\sqrt{1+4x^2+4y^2}}, \qquad dS=\sqrt{1+4x^2+4y^2}\,dA. $$

On the surface, \(z=x^2+y^2\). Therefore, the projected integral is

$$ \iint_S z\,dS = \iint_D (x^2+y^2)\sqrt{1+4x^2+4y^2}\,dA. $$

Use polar coordinates \(x=r\cos\theta\), \(y=r\sin\theta\), where \(0\le r\le1\), \(0\le\theta\le2\pi\), and \(dA=r\,dr\,d\theta\). Then

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

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

$$ \begin{aligned} \iint_S z\,dS &=\frac{\pi}{16}\int_1^5 (u-1)u^{1/2}\,du\\ &=\frac{\pi}{16} \left[ \frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2} \right]_1^5\\ &=\frac{\pi}{16} \left(\frac{20\sqrt5}{3}+\frac{4}{15}\right) =\frac{5\pi\sqrt5}{12}+\frac{\pi}{60}. \end{aligned} $$

For a check on the antiderivative, differentiating \(\frac25u^{5/2}-\frac23u^{3/2}\) gives \(u^{3/2}-u^{1/2}=(u-1)\sqrt{u}\), the integrand. The final value is positive, as expected because both \(z\) and \(dS\) are nonnegative on this patch.

A Common Mistake: Forgetting to Normalize the Normal

The formula uses a unit normal. Using an unnormalized normal changes the dot product and gives the wrong correction factor. The absolute value is also necessary: area cannot be negative, and reversing the normal must not change the result.

Worked Example: Correcting an Unnormalized-Normal Error

For the patch \(z=2x\) projected onto the \(yz\)-plane, the unnormalized normal \((-2,0,1)\) has dot product \(-2\) with \(\mathbf{k}=(1,0,0)\). An incorrect calculation might use \(dS=dA/|-2|=dA/2\). Since the projected area is \(6\), this would give area \(3\).

That result is wrong because \((-2,0,1)\) is not a unit normal. Its length is \(\sqrt5\), so the correct unit normal is \((-2,0,1)/\sqrt5\). Therefore, \(|\mathbf{n}\cdot\mathbf{k}|=2/\sqrt5\), and

$$ \operatorname{Area}(S)=\frac{6}{2/\sqrt5}=3\sqrt5. $$

This also agrees with the calculation using the \(xy\)-projection. Always normalize the normal before applying the projection formula.

When Projection Helps

Projection is useful when the surface has a simple shadow, when a coordinate choice makes the projected region easy to describe, or when the scalar function and area correction combine into a simpler integrand. Before using the formula, check three things:

  • The chosen direction is a unit vector perpendicular to the projection plane.
  • The surface is represented without overlap over the projected region, or separate surface contributions are accounted for.
  • The unit normal has nonzero dot product with the projection direction throughout the patch.

If the last condition fails at a point, that direction does not provide a valid finite area factor there. If it is close to failing, the correction factor can become large, which may make another projection direction more practical.

Check Your Understanding

Use the projection formula and the conditions stated above to answer these questions.

  1. Why must the normal in \(dS=dA/|\mathbf{n}\cdot\mathbf{k}|\) be a unit normal?
  2. A plane patch has unit normal \((1,2,2)/3\). What correction factor applies when projecting onto the plane perpendicular to \((0,0,1)\)?
  3. What geometric situation occurs when \(\mathbf{n}\cdot\mathbf{k}=0\), and why does the projection formula fail in that direction?
  4. For a scalar surface integral, where must the scalar function be evaluated after projection?
  5. In the plane-patch example, why did the \(xy\)- and \(yz\)-projections give the same surface area despite having different projected areas?

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