MAT291: Lecture 21 - Scalar Integrals Over Surfaces

Overview

Parinaz Naseri develops scalar surface integrals by parametrizing a surface with two variables and deriving its area element from the cross product of its tangent vectors. Examples connect this method to gradients, Jacobians, spherical coordinates, and graph surfaces, then apply it to an upper hemisphere and a sphere patch bounded by a cylinder, obtaining area \(4\pi-8\) and weighted integral \(\iint_S z\,dS=2\pi\) for the patch.

Key takeaways

For a parametrized surface \(\mathbf r(u,v)\), the area element is \(dS=\lVert\mathbf r_u\times\mathbf r_v\rVert\,du\,dv\); this factor accounts for both tangent-vector lengths and their angle.
An implicit surface \(f(x,y,z)=c\) has normal \(\nabla f\), while a parametrized surface has normal \(\mathbf r_u\times\mathbf r_v\); the two can differ by scale and orientation.
For a graph \(z=g(x,y)\), use \(dS=\sqrt{1+g_x^2+g_y^2}\,dx\,dy\), then apply any additional change of variables to the projected domain.
On a sphere of radius \(R\), the spherical surface element is \(R^2\sin\phi\,d\phi\,d\theta\); for the upper hemisphere of radius 2, \(\iint_S z\,dS=8\pi\).
For the upper sphere \(x^2+y^2+z^2=4\) inside the cylinder \(x^2+y^2=2y\), polar bounds are \(0\leq\theta\leq\pi\), \(0\leq r\leq2\sin\theta\), yielding area \(4\pi-8\) and height-weighted integral \(2\pi\).

Chapters

0:00 Parametrize a Surface with Two Variables

A curve in 3D uses one parameter, such as \(\mathbf r(t)=(x(t),y(t),z(t))\), while a surface uses two, \(\mathbf r(u,v)=(x(u,v),y(u,v),z(u,v))\).
The surface describes the geometry of integration; the scalar function supplies a value at each point on that geometry.
Use a general \((u,v)\) parametrization when cylindrical or spherical coordinates do not conveniently describe the surface.

4:30 Build Surface Normals from Tangent Vectors

The partial derivatives \(\mathbf r_u\) and \(\mathbf r_v\) are tangent vectors formed by varying one parameter while holding the other fixed.
Their cross product \(\mathbf r_u\times\mathbf r_v\) is a normal vector; dividing by its magnitude gives a unit normal.
For an implicit surface \(f(x,y,z)=c\), the gradient \(\nabla f=(f_x,f_y,f_z)\) also gives a normal, potentially with opposite orientation.

8:45 Compare Gradient and Parametric Normals on a Cone

The cone \(z=\sqrt{x^2+y^2}\) has implicit equation \(x^2+y^2-z^2=0\) and parametrization \(\mathbf r(\rho,\theta)=(\rho\cos\theta,\rho\sin\theta,\rho)\).
The implicit gradient is \((2x,2y,-2z)\), which becomes \((2\rho\cos\theta,2\rho\sin\theta,-2\rho)\) on the cone.
Crossing \(\mathbf r_\rho\) with \(\mathbf r_\theta\) gives \((-\rho\cos\theta,-\rho\sin\theta,\rho)\); the gradient is \(-2\) times this vector, confirming the normals are parallel.

18:20 Derive the General Surface-Area Element

A tiny parameter rectangle maps approximately to a parallelogram whose sides follow \(\mathbf r_u\) and \(\mathbf r_v\).
Its area is \(\lVert\mathbf r_u\times\mathbf r_v\rVert\,du\,dv\), the surface-integral counterpart of the curve factor \(\lVert\mathbf r'(t)\rVert\,dt\).
The cross-product magnitude includes the angle between tangent directions; in orthogonal coordinates that angle is 90 degrees, so the sine factor is 1.
This area-scaling factor plays the role of a Jacobian determinant when changing from parameter coordinates to surface area.

23:40 Integrate Height over an Upper Hemisphere

For the upper hemisphere of radius 2, use \(\mathbf r(\phi,\theta)=(2\sin\phi\cos\theta,2\sin\phi\sin\theta,2\cos\phi)\), with \(0\leq\phi\leq\pi/2\) and \(0\leq\theta\leq2\pi\).
The spherical surface element is \(dS=4\sin\phi\,d\phi\,d\theta\), equivalently \(R^2\sin\phi\,d\phi\,d\theta\) for a sphere of radius \(R\).
Since \(z=2\cos\phi\), integrating \(z\,dS\) over the upper hemisphere gives \(8\pi\).
A spherical-coordinate surface has one coordinate fixed, here \(r=2\); its area factor matches the Jacobian-based change of variables.

33:30 Find the Area Element for a Graph Surface

For a graph \(z=g(x,y)\), parametrize with \(\mathbf r(x,y)=(x,y,g(x,y))\).
The tangent vectors are \(\mathbf r_x=(1,0,g_x)\) and \(\mathbf r_y=(0,1,g_y)\); their cross product is \((-g_x,-g_y,1)\).
Thus \(dS=\sqrt{1+g_x^2+g_y^2}\,dx\,dy\), converting projected area in the \(xy\)-plane to area on the surface.

36:05 Describe the Sphere Patch Inside a Cylinder

The example uses the upper half of the sphere \(x^2+y^2+z^2=4\), so \(z=\sqrt{4-x^2-y^2}\).
The bounding cylinder \(x^2+y^2=2y\) becomes \(r^2=2r\sin\theta\) in polar coordinates.
For the region inside the cylinder, the projected bounds are \(0\leq r\leq2\sin\theta\) and \(0\leq\theta\leq\pi\).

39:30 Convert the Sphere Patch Area Element to Polar Form

Applying the graph formula to \(z=\sqrt{4-x^2-y^2}\) gives \(dS=\frac{2}{\sqrt{4-x^2-y^2}}\,dx\,dy\).
The polar change of variables contributes \(dx\,dy=r\,dr\,d\theta\), so \(dS=\frac{2r}{\sqrt{4-r^2}}\,dr\,d\theta\).
The cylinder determines the integration bounds, while the sphere determines the surface-area scaling factor.

44:10 Compute the Sphere Patch Area

Integrating \(dS\) over \(0\leq r\leq2\sin\theta\), \(0\leq\theta\leq\pi\), gives the patch area \(4\pi-8\).
The radial antiderivative uses \(-2\sqrt{4-r^2}\), and substituting the upper bound introduces \(|\cos\theta|\).
The absolute value is necessary because \(\sqrt{\cos^2\theta}=|\cos\theta|\), especially across the full interval from 0 to \(\pi\).

46:58 Integrate Height over the Same Sphere Patch

For the weighted integral, the sphere height is \(z=\sqrt{4-r^2}\), using the same polar bounds as the area calculation.
Multiplying by \(dS=\frac{2r}{\sqrt{4-r^2}}\,dr\,d\theta\) cancels the square-root factors, leaving \(2r\,dr\,d\theta\).
The resulting integral evaluates to \(2\pi\); the worked example highlights that the scalar function changes while the surface geometry and bounds remain fixed.

Transcript excerpt

0:01 So, this week we talked about a scalar functions, integrals of a scalar functions over curves, right? Then we talked about vector functions over curves. Now, I know you're excited and you're like, "What about surfaces, right?" So, now we're going to start with a scalar functions over surfaces. We're going to do the same approach. First, we're going to assume we are in this general coordinate system or general domain. So, we are going to try to parametrically describe the surface that we're taking the integral on. Okay? Just like the curve problems or 1D

0:49 problems, remember you have a function, you have a curve. These are two different things. Here, it's the same thing. You have a function, you have a surface. The surface determines the geometry you're doing the integral on, and the functions determines what are the values at each point or what are the vectors at each point on this curve or on this surface. So, try to like distinguish between the two, okay? There is a big difference. We learned that a curve in general can be described in 3D by assigning three coordinates X, Y, Z to each point, right?

1:37 But you would only need one variable like T to describe the position at each distance T. Right? X T we define R on a curve as X T, Y T, Z T. Here though, for a surface, you would naturally need two dimensions. You're moving one dimension up. So, you will need two variables to describe each point on the surface >> [snorts] >> or each the the coordinates of each point on the surface with exactly two points, two variables. So, let's say this surface, the blue one, is defined in 3D. Just like the curve. So, every point

2:25 on this surface

This is the start of the transcript. Save this video free for the whole transcript, searchable and time-coded.

Related notes

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Parinaz Naseri.