Choosing Coordinates Before Integrating
A surface integral can become much easier or harder depending on how the surface is parametrized. A useful choice does more than describe the surface correctly: it makes the parameter domain, surface-area factor, and transformed integrand fit together simply.
For a scalar function \(f\) on a surface \(S\), the integral is \(\iint_S f\,dS\). A parametrization \(\mathbf r(u,v)\) converts it to an integral over a parameter domain \(D\), with area factor \(\lVert\mathbf r_u\times\mathbf r_v\rVert\). Choosing parameters is therefore a way to organize the same surface integral—not a change to the quantity being integrated.
Area Factors to Compare
For a graph \(z=g(x,y)\) over a planar region \(D\), use \(\mathbf r(x,y)=(x,y,g(x,y))\). The surface-area factor is \(\sqrt{1+g_x^2+g_y^2}\), so \(dS=\sqrt{1+g_x^2+g_y^2}\,dx\,dy\). If the graph is radial, \(z=g(r)\), and its projection is naturally described in polar coordinates, this becomes \(dS=\sqrt{1+[g'(r)]^2}\,r\,dr\,d\theta\).
For the vertical cylinder of radius \(R\), a convenient parametrization is \(\mathbf r(\theta,z)=(R\cos\theta,R\sin\theta,z)\). Its area factor is \(R\), so \(dS=R\,d\theta\,dz\). More generally, a surface of revolution \((\rho(t)\cos\theta,\rho(t)\sin\theta,z(t))\) has area factor \(\rho(t)\sqrt{[\rho'(t)]^2+[z'(t)]^2}\).
For a sphere of radius \(R\), take polar angle \(\phi\) from the positive \(z\)-axis and azimuthal angle \(\theta\) around that axis. The parametrization \((R\sin\phi\cos\theta,R\sin\phi\sin\theta,R\cos\phi)\) gives \(dS=R^2\sin\phi\,d\phi\,d\theta\). These formulas are choices, not rules that require one coordinate system for every surface. Bounds and the transformed integrand decide which is most useful.
Why the Area Factors Appear
The factor \(\lVert\mathbf r_u\times\mathbf r_v\rVert\) measures the area of the small parallelogram spanned by the two tangent vectors \(\mathbf r_u\) and \(\mathbf r_v\). Thus a small rectangle of parameter area \(du\,dv\) corresponds, to first order, to a surface patch of area \(\lVert\mathbf r_u\times\mathbf r_v\rVert\,du\,dv\). This is why a surface integral uses that factor.
For a graph, \(\mathbf r_x=(1,0,g_x)\) and \(\mathbf r_y=(0,1,g_y)\), so \(\mathbf r_x\times\mathbf r_y=(-g_x,-g_y,1)\). Its length is \(\sqrt{1+g_x^2+g_y^2}\), yielding the graph factor. For the vertical cylinder, the tangent vectors are \((-R\sin\theta,R\cos\theta,0)\) and \((0,0,1)\); their cross product has length \(R\). The formulas reflect the local stretching of the parameter domain along the surface.
Worked Example: A Spherical Cap as a Graph or a Spherical Patch
Worked Example: A Spherical Cap as a Graph or a Spherical Patch
Let \(S\) be the part of the sphere \(x^2+y^2+z^2=4\) above the plane \(z=\sqrt{3}\). Evaluate \(\iint_S z\,dS\) using both a graph parametrization and spherical parameters.
Graph setup. The upper hemisphere is \(z=g(x,y)=\sqrt{4-x^2-y^2}\). The plane meets it where \(x^2+y^2=1\), so the projection is \(0\le r\le1\), \(0\le\theta\le2\pi\). Since \(g_x=-x/z\) and \(g_y=-y/z\), the graph area factor is \(\sqrt{1+r^2/z^2}=2/z\). Multiplying by the integrand \(z\) cancels the \(z\)-dependence:
In the last equality, \(\int_0^1 2r\,dr=1\), and the azimuthal interval has length \(2\pi\).
Spherical setup. The same cap has \(0\le\phi\le\pi/6\), because \(2\cos\phi\ge\sqrt{3}\), and \(0\le\theta\le2\pi\). Here \(z=2\cos\phi\) and \(dS=4\sin\phi\,d\phi\,d\theta\). Therefore,
Both calculations give \(2\pi\). The graph version has simple disk bounds and a cancellation in the integrand; spherical parameters describe the cap’s angular boundary directly. Either is efficient here, and agreement checks the setup.
Worked Example: A Conical Frustum in Cylindrical Parameters
Worked Example: A Conical Frustum in Cylindrical Parameters
Consider the conical frustum \(z=r\), \(1\le r\le2\), \(0\le\theta\le2\pi\), and integrate the scalar function \(f(x,y,z)=z\) over it. The frustum avoids the cone’s vertex, so the graph is smooth throughout this patch.
The rotational geometry and annular projection make cylindrical parameters natural. Use \(\mathbf r(r,\theta)=(r\cos\theta,r\sin\theta,r)\). Its tangent vectors are \((\cos\theta,\sin\theta,1)\) and \((-r\sin\theta,r\cos\theta,0)\). Their cross product is \((-r\cos\theta,-r\sin\theta,r)\), whose length is \(\sqrt{2}\,r\). Since \(z=r\),
As a check, the graph is \(z=\sqrt{x^2+y^2}\) on \(1\le r\le2\). Its gradient has squared length \(1\), so its area factor is \(\sqrt{2}\); polar \(dA\) contributes \(r\,dr\,d\theta\). The graph calculation is therefore \(\int_0^{2\pi}\int_1^2 r\sqrt{2}\,r\,dr\,d\theta\), exactly the same integral and value.
Worked Example: A Vertical Cylinder
Worked Example: A Vertical Cylinder
Let \(S\) be the part of \(x^2+y^2=4\) with \(0\le z\le3\), and let \(f(x,y,z)=x^2+y^2+z\). Find \(\iint_S f\,dS\).
The surface is already a vertical cylinder, so set \(\mathbf r(\theta,z)=(2\cos\theta,2\sin\theta,z)\), with \(0\le\theta\le2\pi\) and \(0\le z\le3\). The area factor is the radius \(2\), and the integrand becomes \(4+z\). Thus
For a check, integrating \(2(4+z)\) first in \(z\) gives \(2(12+9/2)=33\) for each fixed \(\theta\). Multiplying by the full azimuthal range \(2\pi\) again gives \(66\pi\). The cylindrical coordinates make both the bounds and the integrand independent of \(\theta\).
A Practical Comparison
Before calculating, compare the choices available. The best setup is often the one that simplifies more than one part of the integral at once:
| Surface or structure | Useful parameters | What to check |
|---|---|---|
| A single-valued height over a planar region | Graph parameters \((x,y)\) | Whether the projection is simple and the graph factor manageable |
| A radial graph or a rotational surface | Cylindrical parameters \((r,\theta)\) | Whether the bounds and integrand depend mainly on \(r\) |
| A spherical patch centered at the origin | Spherical angles \((\phi,\theta)\) | Whether angular bounds describe the patch directly |
| A vertical cylinder | Azimuth \(\theta\) and height \(z\) | Whether the scalar function simplifies on the fixed-radius surface |
A coordinate system can give a simple domain but an awkward area factor, or a simple factor but complicated bounds. Also consider the scalar function: for example, a function of \(x^2+y^2\) becomes a function of \(r\), while a function of \(z\) on a sphere becomes a function of \(\phi\). Evaluate the whole transformed integrand, not only the surface equation.
Common Pitfall: Losing the Radial Factor
In polar coordinates, planar area is \(dA=r\,dr\,d\theta\), not \(dr\,d\theta\). The extra \(r\) is also visible in the cone example’s cylindrical area factor \(\sqrt{2}\,r\). If that radial factor is omitted, the same frustum integral is incorrectly computed as
because \(\int_1^2r\,dr=3/2\). This is not the correct area element: the cross product has length \(\sqrt{2}\,r\), and the scalar function \(z=r\) supplies another factor of \(r\). Keeping both gives \(\int_0^{2\pi}\int_1^2\sqrt{2}\,r^2\,dr\,d\theta=14\pi\sqrt{2}/3\). When changing coordinates, identify the complete area factor before substituting the integrand.
Choosing Efficiently
A parametrization must cover the intended surface patch, and its parameter bounds must cover that patch without accidental omissions or repetitions that change the integral. Once those conditions are met, compare the resulting bounds, area factor, and scalar function. If two reasonable choices give substantially different-looking integrals, check the domain and the area factor carefully; if both setups are correct, they describe the same scalar surface integral.
The purpose is not to force every problem into one coordinate system. Use graph parameters for a convenient projection, cylindrical parameters for rotational or fixed-radius geometry, and spherical parameters when a sphere’s angular structure makes the patch and integrand simple. A short comparison before integrating can prevent unnecessary algebra and expose errors in the setup.
Check Your Understanding
For each question, focus on the surface, bounds, and area factor before attempting an integral.
- For the graph \(z=g(r)\) over a disk, what factor converts \(dr\,d\theta\) into surface area?
- A spherical cap on a sphere centered at the origin is bounded by a constant value of \(z\). Which angular bound describes its edge most directly?
- Why is the area factor for a vertical cylinder of radius \(R\) equal to \(R\)?
- On the conical frustum \(z=r\), \(1\le r\le2\), what is the area factor in \((r,\theta)\) parameters?
- When comparing two parametrizations, which three features should be assessed along with the surface’s geometric fit?