Line and surface integrals · Definition

Surface integral

multivariable calculusvector calculus

Before you start you need to know

It is the analogue of the scalar line integral for surfaces: it computes areas, masses of curved sheets and averages over a surface.

Definition

The surface integral of a function \(f\) over a surface \(S\), parametrized by \(\mathbf r\) over a region \(D\), adds up the values of \(f\) weighted by the area of each piece of \(S\).

\[\iint_S f\,dS=\iint_D f(\mathbf r(u,v))\,\lVert\mathbf r_u\times\mathbf r_v\rVert\,du\,dv\]

The factor \(\lVert\mathbf r_u\times\mathbf r_v\rVert\) converts the area \(dA\) of the parameter domain into the actual area \(dS\) of the surface piece: a \(du\times dv\) rectangle becomes a parallelogram with sides \(\mathbf r_u\,du\) and \(\mathbf r_v\,dv\), whose area is that norm times \(dA\). With \(f=1\) we get the surface area, \(\iint_D\lVert\mathbf r_u\times\mathbf r_v\rVert\,dA\).

For a graph \(z=g(x,y)\) the formula reduces to \(dS=\sqrt{1+g_x^2+g_y^2}\,dA\). In the \(xy\)-plane it equals \(1\) and we recover the double integral. For the cylinder of radius \(a\) it is \(a\,du\,dv\), and for the sphere of radius \(a\), \(a^2\sin\varphi\,d\varphi\,d\theta\); from the latter comes the area \(4\pi a^2\).

The result does not depend on the chosen parametrization, only on the surface and on \(f\). If \(f\) is a density, the integral is the mass of the sheet; divided by the area, it gives the average value of \(f\) over \(S\). In practice we choose the parametrization that makes the norm of the cross product simplest: polar for paraboloids and cones, spherical for spheres.

A typical mistake is forgetting the norm and writing just \(f\,dA\), which measures the projection, not the surface. Another is evaluating \(f(x,y,z)\) without substituting \(x,y,z\) by the components of \(\mathbf r\). This concept uses the parametric surface and the double integral, and it is the scalar surface integral; with the normal, flux will come later.

Think about it

Compute the area of the part of the plane \(z=2x+y\) over the square \(0\le x,y\le1\).

\(dS=\sqrt{1+4+1}\,dA=\sqrt6\,dA\), so the area is \(\sqrt6\).

Why does the area of a sphere of radius \(a\) grow like \(a^2\) and not like \(a\)?

Because \(\lVert\mathbf r_\varphi\times\mathbf r_\theta\rVert=a^2\sin\varphi\): each tangent direction is scaled by \(a\).

Exercises 10

  1. For \(\mathbf r(u,v)=\langle u,\,v,\,u+v\rangle\) with \(0\le u,v\le1\), the surface area is:
  2. For \(\mathbf r(u,v)=\langle u,\,v,\,u^2+v^2\rangle\), the area element \(dS\) is:
  3. The area of the cylinder \(x^2+y^2=1\), \(0\le z\le3\) (without caps) is:
  4. Let \(S\) be the cylinder \(x^2+y^2=1\), \(0\le z\le3\). The value of \(\iint_S z\,dS\) is:
  5. Let \(S\) be the unit sphere. With \(dS=\sin\varphi\,d\varphi\,d\theta\), the value of \(\iint_S z^2\,dS\) is:
  6. The area of a sphere of radius \(3\) is:
  7. If \(S\) is a flat region \(D\) in the \(xy\)-plane (parametrized by \(\mathbf r(x,y)=\langle x,y,0\rangle\)), then \(\iint_S f\,dS\) equals:
  8. The area of the part of the plane \(z=x+2y\) over the square \(0\le x,y\le1\) is:
  9. The area of the part of the paraboloid \(z=x^2+y^2\) with \(z\le1\) is:
  10. Let \(S\) be the cylinder \(x^2+y^2=1\), \(0\le z\le3\). The value of \(\iint_S z^2\,dS\) is:

References

  • James Stewart, Daniel K. Clegg, Saleem Watson, Calculus: Early Transcendentals, 9th edition, Cengage, 2021. §16.7
  • Joel Hass, Christopher Heil, Przemyslaw Bogacki, Maurice D. Weir, George B. Thomas Jr., Thomas' Calculus, 15th edition, Pearson, 2023. §16.6
  • Jerrold E. Marsden, Anthony Tromba, Vector Calculus, 6th edition, W. H. Freeman, 2012. §7.4