Lesson 32: Surface integral - II (17.6)

Accessible transcription generated on 4/8/2026

Original Notes

Page 1

Lesson 32: Surface integral - II (17.6)

A technical diagram illustrating the mapping from a 2D parameter space to a 3D surface. On the left, a 2D Cartesian coordinate system features a horizontal u-axis and a vertical v-axis. A region, labeled 'R' and outlined in blue, contains a point designated as '(a, b)'. A red arrow curves from this point to the right, labeled with the vector-valued function \(\vec{r}(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle\). On the right, a 3D Cartesian coordinate system with x, y, and z axes displays a blue curved surface labeled 'S'. The red arrow terminates at a point on surface S labeled \(\vec{r}(a,b)\). This visualization demonstrates how points from a planar region R are mapped to a surface S in three-dimensional space via parametrization.
Visual Description: A technical diagram illustrating the mapping from a 2D parameter space to a 3D surface. On the left, a 2D Cartesian coordinate system features a horizontal u-axis and a vertical v-axis. A region, labeled 'R' and outlined in blue, contains a point designated as '(a, b)'. A red arrow curves from this point to the right, labeled with the vector-valued function \(\vec{r}(u,v) = \langle x(u,v), y(u,v), z(u,v) \rangle\). On the right, a 3D Cartesian coordinate system with x, y, and z axes displays a blue curved surface labeled 'S'. The red arrow terminates at a point on surface S labeled \(\vec{r}(a,b)\). This visualization demonstrates how points from a planar region R are mapped to a surface S in three-dimensional space via parametrization.

Definition: The surface integral of a scalar function \(f\) over a parametric surface \(S\) is defined and computed as follows:

\[\iint_S f \, dS = \iint_R f(\vec{r}(u,v)) |\vec{r}_u \times \vec{r}_v| \, dA\]

\((u,v \text{ Bounds})\)

The surface integral has various physical interpretations depending on the nature of the function \(f\):

If \(f = 1 \implies \iint_S 1 \, dS = \text{Surface Area}\)

If \(f = \text{density of Surface} \implies \iint_S f \, dS = \text{Mass of Surface}\)

Page 2

Example: Surface Area Calculation

Find the area of the surface parametrized by:

\[\vec{r}(u,v) = \langle \cos u, \sin u, 2v \rangle, \quad 0 \leq u \leq \frac{3\pi}{2}, \quad 0 \leq v \leq 1\]

To find the surface area (\(SA\)), we set up the surface integral:

\[SA = \iint_S 1 \, dS\]

Using the parametrization \(\vec{r}(u, v)\) and its bounds:

\[= \iint_R |\vec{r}_u \times \vec{r}_v| \, dA = \int_0^1 \int_0^{\frac{3\pi}{2}} |\vec{r}_u \times \vec{r}_v| \, du \, dv\]

First, calculate the partial derivatives:

\[\vec{r}_u = \langle -\sin u, \cos u, 0 \rangle\] \[\vec{r}_v = \langle 0, 0, 2 \rangle\]

Next, calculate the cross product:

\[\vec{r}_u \times \vec{r}_v = \langle 2\cos u, 2\sin u, 0 \rangle\]

Find the magnitude of the cross product:

\[|\vec{r}_u \times \vec{r}_v| = \sqrt{(2\cos u)^2 + (2\sin u)^2 + 0^2} = \sqrt{4(\cos^2 u + \sin^2 u)} = 2\]

Now, substitute the magnitude into the integral:

\[SA = \int_0^1 \int_0^{\frac{3\pi}{2}} 2 \, du \, dv = 2 \times \frac{3\pi}{2} \times 1 = 3\pi\]

Page 3

Surface Area of a Cylinder

Verify that the surface area of a cylinder of radius \(R\) and height \(H\) is \(2\pi RH\).
A hand-drawn 3D diagram of a cylinder oriented along the z-axis of a Cartesian coordinate system. The cylinder's base is centered at the origin (labeled 0) in the xy-plane, and its top reaches a height labeled H on the z-axis. Blue lines outline the cylinder's structure, with circles at the top and bottom and vertical segments for the walls. Diagonal shading lines are drawn across the side surface. A blue curved arrow on the xy-plane indicates the angular sweep around the z-axis starting from the x-axis.
Visual Description: A hand-drawn 3D diagram of a cylinder oriented along the z-axis of a Cartesian coordinate system. The cylinder's base is centered at the origin (labeled 0) in the xy-plane, and its top reaches a height labeled H on the z-axis. Blue lines outline the cylinder's structure, with circles at the top and bottom and vertical segments for the walls. Diagonal shading lines are drawn across the side surface. A blue curved arrow on the xy-plane indicates the angular sweep around the z-axis starting from the x-axis.
The cylinder is defined by the equations: \[x^2 + y^2 = R^2, \quad 0 \leq z \leq H\] Parametrization: We can parametrize the surface of the cylinder as: \[\vec{r}(u, v) = \langle R \cos u, R \sin u, v \rangle\] where the bounds for the parameters are: \[0 \leq u \leq 2\pi\] \[0 \leq v \leq H\] To find the surface area element \(dA\), we first calculate the partial derivatives: \[\vec{r}_u = \langle -R \sin u, R \cos u, 0 \rangle\] \[\vec{r}_v = \langle 0, 0, 1 \rangle\] Next, we find the cross product: \[\vec{r}_u \times \vec{r}_v = \langle R \cos u, R \sin u, 0 \rangle\] The magnitude of the cross product is: \[|\vec{r}_u \times \vec{r}_v| = \sqrt{(R \cos u)^2 + (R \sin u)^2 + 0^2} = \sqrt{R^2(\cos^2 u + \sin^2 u)} = R\] The surface area (S.A.) is then calculated as the surface integral: \[\text{S.A.} = \iint_S 1 \, dA = \int_0^H \int_0^{2\pi} |\vec{r}_u \times \vec{r}_v| \, du \, dv\] \[\text{S.A.} = \int_0^H \int_0^{2\pi} R \, du \, dv = 2\pi RH\]

Page 4

Example: Parametric Surface Representation

Example: \(\vec{r}(u,v) = \langle u \cos v, u \sin v, 3u \rangle, \quad 0 \leq u \leq 4, \quad 0 \leq v \leq 2\pi\)

What does this surface represent?

From the parametric equations, we have:

\[ \begin{cases} x = u \cos v \\ y = u \sin v \\ z = 3u \end{cases} \]

By squaring and adding the components for \(x\) and \(y\), we find:

\[ x^2 + y^2 = u^2 \]

From the third component, we can isolate \(u\):

\[ u = \frac{z}{3} \]

Substituting this into the relation for \(x^2 + y^2\):

\[ x^2 + y^2 = \frac{z^2}{9} \rightarrow \text{Cone} \]

Applying the given domain constraints:

\[ 0 \leq u = \frac{z}{3} \leq 4 \rightarrow \text{part of single cone from } z=0 \text{ to } z=12 \]
A technical diagram illustrating the mapping from a 2D parametric domain to a 3D surface. On the left is the uv-plane, showing a rectangular domain bounded by 0 <= u <= 4 and 0 <= v <= 2\pi. Within this rectangle, three vertical lines are color-coded: a purple line at u=0, a red line at an intermediate u value, and a blue/green line at u=4. On the right, a 3D coordinate system (x, y, z) shows an upward-opening cone corresponding to the mapping. The purple line at u=0 maps to the origin (0,0,0). The red vertical line in the uv-plane maps to a red horizontal circular cross-section on the cone, labeled z=3u. The blue/green line at u=4 maps to the top circular rim of the cone, which is labeled at height z=12. The diagram visualizes how lines of constant 'u' in the parameter space correspond to circles of constant height and radius on the cone's surface.
Visual Description: A technical diagram illustrating the mapping from a 2D parametric domain to a 3D surface. On the left is the uv-plane, showing a rectangular domain bounded by 0 <= u <= 4 and 0 <= v <= 2\pi. Within this rectangle, three vertical lines are color-coded: a purple line at u=0, a red line at an intermediate u value, and a blue/green line at u=4. On the right, a 3D coordinate system (x, y, z) shows an upward-opening cone corresponding to the mapping. The purple line at u=0 maps to the origin (0,0,0). The red vertical line in the uv-plane maps to a red horizontal circular cross-section on the cone, labeled z=3u. The blue/green line at u=4 maps to the top circular rim of the cone, which is labeled at height z=12. The diagram visualizes how lines of constant 'u' in the parameter space correspond to circles of constant height and radius on the cone's surface.

Page 5

What is its Surface Area

\[ S.A. = \iint_S 1 \, dS = \iint_{\substack{u, v \\ \text{Bounds}}} |r_u \times r_v| \, du \, dv \] \[ = \int_0^{2\pi} \int_0^4 |r_u \times r_v| \, du \, dv \]

\(\vec{r}(u,v) = \langle u \cos v, u \sin v, 3u \rangle,\)
\(\Rightarrow r_u = \langle \cos v, \sin v, 3 \rangle\)
\(r_v = \langle -u \sin v, u \cos v, 0 \rangle\)
\(r_u \times r_v = \langle -3u \cos v, -3u \sin v, u \rangle\)
\(|r_u \times r_v| = \sqrt{9u^2(\cos^2 v + \sin^2 v) + u^2} = \sqrt{10} u\)

\[ SA = \int_0^{2\pi} \int_0^4 \sqrt{10} \cdot u \, du \, dv = \sqrt{10} \times 2\pi \times \left. \frac{u^2}{2} \right|_0^4 = \sqrt{10} \times 16\pi \]

Page 6

Review of Parametrization and Normal Vectors

Last class: Consider the plane \(4x + 3y + z = 12\) in the 1st octant.

Parametrization:

\[\vec{r}(u, v) = \langle u, v, 12 - 4u - 3v \rangle\]

where the domain is defined by:

\[u \ge 0, v \ge 0, 12 - 4u - 3v \ge 0\]

The cross product of the partial derivatives is:

\[\vec{r}_u \times \vec{r}_v = \langle 4, 3, 1 \rangle\]

This result is the Normal vector for the plane.

In general:

The cross product \(\vec{r}_u \times \vec{r}_v\) evaluated at \(\vec{r}(a, b)\) represents the Normal vector to the tangent plane at the point \(\vec{r}(a, b)\).

Page 7

What does \( \vec{r}_u \times \vec{r}_v \) represent?

A mathematical diagram illustrating the transformation from a 2D parameter space to a 3D surface. On the left, an irregular blue region is shown in the uv-coordinate plane. A point (a, b) is marked, with a horizontal green line representing a constant v and a vertical purple line representing a constant u passing through it. An arrow labeled with the vector function r(u, v) = <x(u,v), y(u,v), z(u,v)> indicates a mapping from this region to a 3D coordinate system on the right. In the 3D space, a blue curved surface S is shown. The point (a, b) from the uv-plane maps to the point r(a, b) on the surface. Two curves intersect at this point: a green curve (the image of the line where v is constant) and a purple curve (the image of the line where u is constant). At the intersection point, two tangent vectors are drawn: a green vector r_u tangent to the green curve and a purple vector r_v tangent to the purple curve. Additionally, a vector N is shown pointing perpendicular to the surface at that point.
Visual Description: A mathematical diagram illustrating the transformation from a 2D parameter space to a 3D surface. On the left, an irregular blue region is shown in the uv-coordinate plane. A point (a, b) is marked, with a horizontal green line representing a constant v and a vertical purple line representing a constant u passing through it. An arrow labeled with the vector function r(u, v) = indicates a mapping from this region to a 3D coordinate system on the right. In the 3D space, a blue curved surface S is shown. The point (a, b) from the uv-plane maps to the point r(a, b) on the surface. Two curves intersect at this point: a green curve (the image of the line where v is constant) and a purple curve (the image of the line where u is constant). At the intersection point, two tangent vectors are drawn: a green vector r_u tangent to the green curve and a purple vector r_v tangent to the purple curve. Additionally, a vector N is shown pointing perpendicular to the surface at that point.

\( \vec{r}_u \to \) tangent vector of the curve we get by fixing \( v \)

\( \vec{r}_v \to \) tangent vector of the curve we get by fixing \( u \)

\( \vec{r}_u, \vec{r}_v \) lie on the tangent plane at \( \vec{r}(a, b) \).

\( \vec{r}_u \times \vec{r}_v \) is a Normal vector.

A geometric diagram depicting a tangent plane represented as a parallelogram. Two vectors, labeled r_u and r_v, originate from the same point and lie within the plane. A third vector, labeled r_u x r_v, originates from the same point but is oriented vertically, perpendicular to the plane surface. This visually demonstrates that the cross product of two tangent vectors yields a vector normal to the plane.
Visual Description: A geometric diagram depicting a tangent plane represented as a parallelogram. Two vectors, labeled r_u and r_v, originate from the same point and lie within the plane. A third vector, labeled r_u x r_v, originates from the same point but is oriented vertically, perpendicular to the plane surface. This visually demonstrates that the cross product of two tangent vectors yields a vector normal to the plane.

Page 8

Finding \(r_u \times r_v\) for Surface given by \(z = f(x, y)\)

Parametrization: \(\vec{r}(u, v) = \langle u, v, f(u, v) \rangle\)

The partial derivative vectors are calculated as follows:

\[\vec{r}_u = \langle 1, 0, f_u \rangle\] \[\vec{r}_v = \langle 0, 1, f_v \rangle\]
\[\vec{r}_u \times \vec{r}_v = \langle -f_u, -f_v, 1 \rangle\]

The magnitude of the cross product vector is given by:

\[|\vec{r}_u \times \vec{r}_v| = \sqrt{f_u^2 + f_v^2 + 1}\]

The Surface Area (S.A.) is calculated using the following double integral:

\[S.A. = \iint\limits_{S} 1 \, dS = \iint\limits_{u, v \text{ Bounds}} \sqrt{f_u^2 + f_v^2 + 1} \, dA\]

Page 9

Example: Surface Area of a Plane

Find the area of the part of the plane \(2x + y + z = 8\) in the 1st octant.
A two-part diagram illustrating the geometry of the surface area problem. The left part is a 2D graph in the uv-plane showing a shaded triangular region labeled R. This region R is bounded by the v-axis (where it has an intercept at (0,8)), the u-axis (where it has an intercept at (4,0)), and the diagonal line 2u + v = 8. The right part of the diagram is a 3D perspective sketch of the first octant showing a shaded triangular surface labeled S. This surface S represents the portion of the plane 2x + y + z = 8 that lies within the positive coordinate axes.
Visual Description: A two-part diagram illustrating the geometry of the surface area problem. The left part is a 2D graph in the uv-plane showing a shaded triangular region labeled R. This region R is bounded by the v-axis (where it has an intercept at (0,8)), the u-axis (where it has an intercept at (4,0)), and the diagonal line 2u + v = 8. The right part of the diagram is a 3D perspective sketch of the first octant showing a shaded triangular surface labeled S. This surface S represents the portion of the plane 2x + y + z = 8 that lies within the positive coordinate axes.
The surface area of \(S\) is given by the integral: \[\text{Surface Area of } S = \iint_S 1 \, dS = \iint_R |r_u \times r_v| \, dA\] From the equation of the plane, we can determine the cross product of the partial derivatives: \[r_u \times r_v = \langle 2, 1, 1 \rangle\] The magnitude is \(|r_u \times r_v| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{6}\). Substituting this into the integral: \[= \iint_R \sqrt{6} \, dA = \sqrt{6} \iint_R 1 \, dA\] The term \(\iint_R 1 \, dA\) represents the Area of R, where \(R\) is the triangular projection on the base plane. Based on the 2D diagram, this is a triangle with base 4 and height 8. \[= \sqrt{6} \times \left( \frac{1}{2} \times 4 \times 8 \right)\] \[= 16\sqrt{6}\]

Page 10

Example: Find area of part of \(z = 2 - x^2 - y^2\), above \(z = 1\).

A 3D coordinate plot illustrating a downward-opening paraboloid described by the equation \(z = 2 - x^2 - y^2\). The apex of the paraboloid is located on the z-axis at \(z=2\). The surface is truncated by a horizontal plane at \(z=1\). The specific portion of the surface being analyzed is labeled \(S\), which is the dome-like section situated between the levels \(z=1\) and \(z=2\). The x, y, and z axes are clearly shown for spatial orientation.
Visual Description: A 3D coordinate plot illustrating a downward-opening paraboloid described by the equation \(z = 2 - x^2 - y^2\). The apex of the paraboloid is located on the z-axis at \(z=2\). The surface is truncated by a horizontal plane at \(z=1\). The specific portion of the surface being analyzed is labeled \(S\), which is the dome-like section situated between the levels \(z=1\) and \(z=2\). The x, y, and z axes are clearly shown for spatial orientation.

Let the surface be defined by \(f(x, y) = 2 - x^2 - y^2\). We parameterize the surface as: \[ \vec{r}(u, v) = \langle u, v, f(u, v) \rangle \] The cross product of the partial derivatives is: \[ \vec{r}_u \times \vec{r}_v = \langle -f_u, -f_v, 1 \rangle = \langle 2u, 2v, 1 \rangle \] The magnitude of this cross product, representing the differential area element, is: \[ |\vec{r}_u \times \vec{r}_v| = \sqrt{4u^2 + 4v^2 + 1} \]

\(u, v\) Bounds:

A 2D coordinate graph showing the region of integration \(R\) in the \(uv\)-plane. The horizontal axis is labeled \(u\) and the vertical axis is labeled \(v\). The region \(R\) is a shaded unit circle centered at the origin, representing the projection of the 3D surface \(S\) onto the \(xy\)-plane. The circle is bounded by the inequality \(u^2 + v^2 \le 1\), and the interior is filled with blue diagonal hatch lines.
Visual Description: A 2D coordinate graph showing the region of integration \(R\) in the \(uv\)-plane. The horizontal axis is labeled \(u\) and the vertical axis is labeled \(v\). The region \(R\) is a shaded unit circle centered at the origin, representing the projection of the 3D surface \(S\) onto the \(xy\)-plane. The circle is bounded by the inequality \(u^2 + v^2 \le 1\), and the interior is filled with blue diagonal hatch lines.

To determine the limits for integration, we consider the condition that the surface must be above \(z=1\): \[ z = 2 - x^2 - y^2 \ge 1 \] Rearranging the terms, we find: \[ x^2 + y^2 \le 1 \] Substituting the parameters \(u=x\) and \(v=y\), we get the region \(R\) in the \(uv\)-plane: \[ u^2 + v^2 \le 1 \]

Page 11

Surface Area Calculation

\[ SA = \iint\limits_{\substack{u, v \\ \text{Bounds}}} |r_u \times r_v| \, dA = \iint\limits_{\substack{u, v \\ \text{Bounds}}} \sqrt{4u^2 + 4v^2 + 1} \, dA \]

Use polar to compute:

\[ u = r \cos \theta, \quad v = r \sin \theta \] \[ dA = r \, dr \, d\theta \] \[ 0 \leq r \leq 1 \] \[ 0 \leq \theta \leq 2\pi \] \[ = \int_{0}^{2\pi} \int_{0}^{1} \sqrt{4r^2 + 1} \cdot r \, dr \, d\theta \]
\[ \begin{aligned} 4r^2 + 1 &= t \\ 8r \, dr &= dt \\ r = 0 &\rightsquigarrow t = 1 \\ r = 1 &\rightsquigarrow t = 5 \end{aligned} \]
\[ = \int_{0}^{2\pi} \int_{1}^{5} \frac{1}{8} t^{1/2} \, dt \, d\theta \] \[ = 2\pi * \frac{1}{8} * \frac{2}{3} t^{3/2} \bigg|_{1}^{5} = \frac{\pi}{6} [5^{3/2} - 1] \]

Page 12

Worked Example: Surface Integral over a Spherical Cap

eg: Evaluate \(\iint_S xz \, ds\), where \(S\) is the part of a Sphere of Radius 2 (meaning \(\rho = 2\)), above \(z = 1\).

A 3D coordinate system diagram illustrating a spherical cap of radius rho = 2. The cap is the portion of the sphere that lies above the plane z = 1. A vector rho is drawn from the origin to the surface of the sphere. The polar angle phi is shown being measured from the positive z-axis, and the azimuthal angle theta is shown in the xy-plane starting from the positive x-axis. A dashed line indicates the circular boundary where the sphere intersects the plane z = 1.
Visual Description: A 3D coordinate system diagram illustrating a spherical cap of radius rho = 2. The cap is the portion of the sphere that lies above the plane z = 1. A vector rho is drawn from the origin to the surface of the sphere. The polar angle phi is shown being measured from the positive z-axis, and the azimuthal angle theta is shown in the xy-plane starting from the positive x-axis. A dashed line indicates the circular boundary where the sphere intersects the plane z = 1.

We use the standard spherical coordinate transformations:

\[ x = \rho \sin \phi \cos \theta \] \[ y = \rho \sin \phi \sin \theta \] \[ z = \rho \cos \phi \]

Parametrization!

\[ \vec{r}(\phi, \theta) = \langle 2 \sin \phi \cos \theta, 2 \sin \phi \sin \theta, 2 \cos \phi \rangle \]

To determine the limits of integration:

\[ \left. \begin{aligned} 0 & \leq \theta \leq 2\pi \\ 0 & \leq \phi \leq \text{hit } z=1 \end{aligned} \right\} \begin{aligned} 2 \cos \phi &= 1 \\ \cos \phi &= 1/2 \\ \Rightarrow \phi &= \pi/3 \end{aligned} \]

Setting up the integral:

\[ \iint_S xz \, ds = \int_0^{2\pi} \int_0^{\pi/3} \underbrace{(2 \sin \phi \cos \theta)}_{x} \underbrace{(2 \cos \phi)}_{z} |\vec{r}_\phi \times \vec{r}_\theta| \, d\phi \, d\theta \]

Page 13

Computing \(\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}\)

\[\mathbf{r}_{\phi} = \langle 2\cos\phi \cos\theta, 2\cos\phi \sin\theta, -2\sin\phi \rangle\] \[\mathbf{r}_{\theta} = \langle -2\sin\phi \sin\theta, 2\sin\phi \cos\theta, 0 \rangle\] \[\mathbf{r}_{\phi} \times \mathbf{r}_{\theta} = \langle 4\sin^2\phi \cos\theta, 4\sin^2\phi \sin\theta, 4\sin\phi \cos\phi [\cos^2\theta + \sin^2\theta] \rangle\] \[= \langle 4\sin^2\phi \cos\theta, 4\sin^2\phi \sin\theta, 4\sin\phi \cos\phi \rangle\] \[= 4\sin\phi \langle \sin\phi \cos\theta, \sin\phi \sin\theta, \cos\phi \rangle\] \[|\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}| = 4\sin\phi \sqrt{\sin^2\phi [\cos^2\theta + \sin^2\theta] + \cos^2\phi} = 4\sin\phi\]
In general, if \(\vec{r}(\phi, \theta) = \langle \rho\sin\phi \cos\theta, \rho\sin\phi \sin\theta, \rho\cos\phi \rangle\), then: \[|\mathbf{r}_{\phi} \times \mathbf{r}_{\theta}| = \rho^2 \sin\phi\]