Parametrized Surfaces and Surface Integralspeople.ku.edu/~s334m797/Math127Fall2018/Week13post.pdfSurface Integrals of Vector Fields (I)Tangent Lines and Planes of Parametrized Surfaces

Section 16.4Parametrized Surfaces and Surface Integrals

(I) Parametrizing Surfaces

(II) Surface Area

(III) Scalar Surface Integrals

MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 1 / 41

In chapter 13 we parametrized curves using a single parameter. Moreparameters are not required because a curve is a one dimensional object.

In this section, we explore the parametrization of surfaces. Consider avector function ~r (u, v) of two variables

~r (u, v) = 〈f (u, v), g(u, v), h(u, v)〉with points (u, v) in the domain D which lie in a uv -plane. The set ofpoints satisfying the equation is the surface parametrized by ~r .


If a surface S is parametrized by ~r (u, v), then the surface can be sketchedusing the grid lines of the domain.

Grid lines along a parametrized surface are the image of the vertical andhorizontal lines in the domain, the uv -plane.

For a constant a, ~r (a, v) is a parametrization of the vertical grid lines.

For a constant b, ~r (u, b) is a parametrization of the horizontal grid lines.


Example 1: Find a parametrization of the plane containing the point Pand the vectors ~a and ~b.

Solution: Points in the plane can befound by starting at P and movingalong the vectors ~a then ~b.

~r (u, v) = P + u~a + v~b

Example 1b: The plane containing the points (3, 0, 0), (0, 2, 0), and(0, 0, 1) contains the vectors 〈−3, 0, 1〉 and 〈−3, 2, 0〉 and has a normalvector ~n = 〈2, 3, 6〉.

Scalar Equation:

2x + 3y + 6z = 6

Parametrization:

~r (u, v) = 〈−3u − 3v , 2v , u + 1〉


Example 2: Find a parametrization for the cylinder x2 + y2 = r2.

Solution: We use the cylindrical coordinates transformation.

~r (θ, z) = 〈r cos(θ), r sin(θ), z〉 [0, 2π)× [−∞,∞]


Example 3: Find a parametrization for the sphere x2 + y2 + z2 = r2.

Solution: We use the spherical coordinates transformation.

~r (θ, φ) = 〈r cos(θ) sin(φ), r sin(θ) sin(φ), r cos(φ)〉 [0, 2π)× [0, π]


Example 4: Identify the surface parametrized by

~r (u, v) = 〈u cos(v), u sin(v), u〉Solution: Eliminate u and v to obtain an equation in (x , y , z):

x2 + y2 =(u cos(v)

)2+(u sin(v)

)2= u2 = z2

The parametric surface is a cone!

Two-variable functions z = f (x , y) define asurface and can be easily parametrized as

~r (x , y) = 〈x , y , f (x , y)〉


Example 5: Let ~r (u, v) = 〈u cos(v), u sin(v), v〉.What are the horizontal grid lines, when v is constant?

What are the vertical grid lines, when u is constant?

Solution:


Example 6: Find a parametrization for the hyperboloid of one sheet

x2 + y2 − z2 = 1

Solution: Don’t forget that there are an infinite number ofparametrizations for any surface!

We will find two for this surface:

(I) Set z = u, then x2 + y2 = 1 + c2. We can then let

x =√

1 + u2 cos(v) y =√

1 + u2 sin(v)

(II) Let z = tan(u), then x2 + y2 = sec2(u). We can then let

x = tan(u) cos(v) y = tan(u) sin(v)


Surface Area of a Parametrized Surface

In section 13.3 we calculated the arclength of a parametrized curve andthen we used the resulting formula to define line integrals.

In a smiliar fashion, we will calculate the surface area of a surfaceparametrized by ~r (u, v) over the domain R.

We partition R into m × n subregions Rij . Each subregion corresponds toa portion of the surface Sij .

S = ∪Sij Sij = ~r (Dij), the Image of Dij under ~r


Similar to the Jacobian calculations of section 15.6, we approximate Sij asa rectangle of width ~ru(ui , vj) ∆u and length ~rv (ui , vj) ∆v .

Using the parallelogram property of the cross product:

Area(Sij) ≈ |~ru(ui , vj) ∆u × ~rv (ui , vj) ∆v | = |~ru(ui , vj) × ~rv (ui , vj) | ∆u ∆v

The area of S is approximatelym∑i=1

n∑j=1

|~ru(ui , vj) × ~rv (ui , vj) | ∆u ∆v .

Taking the limit as (m, n)→ (∞,∞) yields

Area(S) =

∫∫D|~ru(u, v) × ~rv (u, v) | dA


Area(S) =

∫∫D|~ru(u, v) × ~rv (u, v) | dA

Example 7: Find the surface area of the cylinder x2 + y2 = 4 betweenz = 0 and z = 5.

Solution: First, parameterize the cylinder as

~r (u, v) = 〈2 cos(u), 2 sin(u), v〉 [0, 2π]× [0, 5]

Second, calculate |~ru × ~rv |.

~ru(u, v) = 〈−2 sin(u), 2 cos(u), 0〉 ~rv (u, v) = 〈0, 0, 1〉

|~ru × ~rv | = |〈2 cos(u), 2 sin(u), 0〉| = 2

Area(S) =

∫ 2π

0

∫ 5

02 dv du = 20π


Surface Area of a 2-Variable Function

The surface area of the graph of z = f (x , y) defined on D has a naturalparametrization ~r (x , y) = 〈x , y , f (x , y)〉.

~rx(x , y) = 〈1, 0, fx(x , y)〉 ~ry (x , y) = 〈0, 1, fy (x , y)〉

Since ~rx × ~ry = 〈−fx ,−fy , 1〉, we have |~rx × ~ry | =√

1 + f 2x + f 2y .

The area of the surface is given by∫∫D

√1 + fx(x , y)2 + fy (x , y)2 dA


Example 8: Calculate the surface area of the sphere of radius a.

Solution: Half of the sphere is the graph z = f (x , y) =√a2 − x2 − y2

where x2 + y2 ≤ a2.

fx(x , y) =−x√

a2 − x2 − y2fy (x , y) =

−y√a2 − x2 − y2

Thus the area of the sphere is

2

∫∫D

√√√√1 +

(−x√

a2 − x2 − y2

)2

+

(−y√

a2 − x2 − y2

)2

dA

= 2

∫ 2π

0

∫ a

0

a√a2 − r2

r dr dθ = 4πa2


Example 9: A torus (surface of a donut) can be obtained by rotating thecircle on the yz-plane centered at (0, b, 0) with radius a (b > a > 0)about the z-axis. What is the surface area of the torus?

Solution: The torus has a parametrization

x(θ, α) = b cos(θ) + a cos(α) cos(θ)y(θ, α) = b sin(θ) + a cos(α) sin(θ)z(θ, α) = a sin(α)

where 0 ≤ θ ≤ 2π and 0 ≤ α ≤ 2π.

|~rθ × ~rα| = a(b + a cos(α))

The surface area is calculated∫ 2π

0

∫ 2π

0a(b + a cos(α)) dθ dα = 2πa

∫ 2π

0(b + a cos(α)) dα = 4π2ab


Surface Integrals - Scalar Surface Integral

Suppose that f (x , y , z) gives the density of some quantity per unit area ateach point on a surface S parametrized by ~r (u, v) over the domain R.

Then the total amount of the quantity is calculated as∫∫Sf (x , y , z) dS =

∫∫Rf (~r (u, v)) |~ru × ~rv | dA

The symbol dS is the surface element or the surface area differential.

dS = |~ru × ~rv | dAuv


Example 10: Suppose that the mass density of a surface is given byf (x , y , z) = x2. Compute the total mass of the sphere x2 + y2 + z2 = 1.

Solution: First, parametrize the sphere using spherical coordinates.

~r (θ, φ) = 〈sin(φ) cos(θ), sin(φ) sin(θ), cos(φ)〉 |~ru × ~rv | = sin(φ)

Total Mass:∫∫Sx2 dS =

∫ 2π

0

∫ π

0(sin(φ) cos(θ))2 sin(φ) dφ dθ

=

∫ 2π

0

∫ π

0sin(φ)3 cos(θ) dφ dθ

=cos3(φ)

3− cos(φ)

∣∣∣∣π0

sin(θ)

∣∣∣∣2π0

=4π

3


Example 1


IClicker 1

(A) 26√

2π

(B) 27√

2π

(C) 26π

(D) 26√

2


IClicker 2

(A) 54

√0

(B) 54

√29

(C) 14

√29

(D) 59MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 19 / 41

Section 16.5Surface Integrals of Vector Fields

(I) Tangent Lines and Planes of Parametrized Surfaces

(II) Oriented Surfaces

(III) Vector Surface Integrals

(IV) Flux, Fluid Flow, Electric and Magnetic Fields


Let S be parametrized by ~r (u, v) on the domain R.

If ~ru is nonzero at (a, b), then the tangent line to the horizontal grid lineat ~r (a, b) has a directional vector ~ru(a, b).

If ~rv is nonzero at (a, b), then the tangent line to the vertical grid line at~r (a, b) has a directional vector ~rv (a, b).

If ~ru × ~rv is nonzero at (a, b), then the tangent plane to S at ~r (a, b) has anormal vector (~ru × ~rv ) (a, b).


If it is possible to choose a unit normal vector ~n at every points so that ~nvaries continuously over the surface, then the surface is called an orientedsurface and the choice of ~n provides an orientation for the surface.

S is smooth if ~ru × ~rv is nonzero for all points in R.

If S is a smooth orientable surface which is parametrized by ~r (u, v), thenthe surface is automatically supplied with the orientation of the unitnormal vector

~n =~ru × ~rv|~ru × ~rv |


Not all surfaces are can be oriented! Discovered in 1858 by AugustMobius, the Mobius strip is a surface which cannot be oriented.

If an ant were to crawl along the stripstarting at a point P, it would eventuallyend up on the other side of the strip.

If the any continued to crawl in the samedirection, then it would end up back at thesame point P without ever crossing an edge.

The Mobius strip has only one side!

For r ∈ [−0.5, 0.5] and θ ∈ [0, 2π], theMobius strip can be parametrized as

x = 2 cos(θ) + r cos(θ/2)y = 2 sin(θ) + r cos(θ/2)

z = r sin(θ/2)


A closed surface is the boundary of a solid region.

By convention, the positive orientation of a closed surface is the one forwhich the normal vectors point outward from the solid region.

Example 1: The unit sphere parametrized using the spherical coordinatetransformation has positive orientation.

~r (φ, θ) = 〈sin(φ) cos(θ), sin(φ) sin(θ), cos(φ)〉

~rφ × ~rθ =⟨sin2(φ) cos(θ), sin2(φ) sin(θ), sin(φ) cos(φ)

⟩MATH 127 (Section 16.4) Parametrized Surfaces and Surface Integrals The University of Kansas 24 / 41

The normal component of ~F to S is ~F · ~n.

If ~F is a continuous vector field defined on an oriented surface S with unitnormal vector ~n, then the vector surface integral of ~F over S is∫∫

S~F · d ~S =

∫∫S~F · ~n dS


If ~F is a continuous vector field defined on an oriented surface S with unitnormal vector ~n, then the vector surface integral of ~F over S is∫∫

S~F · d ~S =

∫∫S~F · ~n dS

The integral is also called the flux of ~F across S.

Let S be parametrized by ~r (u, v) over R.∫∫S~F · d ~S =

∫∫S~F · ~n dS

=

∫∫R~F (~r (u, v)) ·

~ru × ~rv|~ru × ~rv |

|~ru × ~rv | dA

=

∫∫R~F (~r (u, v)) · (~ru × ~rv ) dA


Example 2: Find the flux of ~F (x , y , z) =⟨0, yz , z2

⟩outward through the

surface S cut from the cylinder y2 + z2 = 1, z ≥ 0, by the planes x = 0and x = 1.

Solution: The surface can be parametrized with an outward orientation as

~r (x , θ) = 〈x , sin(θ), cos(θ)〉 x ∈ [0, 1] θ ∈ [−π/2, π/2]

~rx × ~rθ = 〈1, 0, 0〉 × 〈0, cos(θ),− sin(θ)〉 = 〈0, sin(θ), cos(θ)〉

~F (~r (x , θ)) =⟨0, sin(θ) cos(θ), cos2(θ)

⟩∫∫S~F · d ~S =

∫∫R~F · (~rφ × ~rθ) dA

=

∫ π/2

−π/2

∫ 1

0sin2(θ) cos(θ) + cos3(θ) dx dθ

=

∫ π/2

−π/2sin2(θ) cos(θ) + cos3(θ) dθ = 2


Example 3: Find the flux of the vector field ~F (x , y , z) = 〈z , y , x〉 acrossthe unit sphere x2 + y2 + z2 = 1

Solution: The unit sphere can be parametrized as

~r (φ, θ) = 〈sin(φ) cos(θ), sin(φ) sin(θ), cos(φ)〉 φ ∈ [0, π] θ ∈ [0, 2π]

~rφ × ~rθ =⟨sin2(φ) cos(θ), sin2(φ) sin(θ), sin(φ) cos(φ)

⟩~F (~r (φ, θ)) = 〈cos(φ), sin(φ) sin(θ), sin(φ) cos(θ)〉∫∫

S~F · d ~S =

∫∫R~F · (~rφ × ~rθ) dA

=

∫ 2π

0

∫ π

02 sin2(φ) cos(φ) cos(θ) + sin3(φ) sin2(θ) dφ dθ

=4π

3


Fluid Flux

For a fluid with velocity vector field ~V , the flow rate across the surface S(volume per unit time) is ∫∫

S~V · d ~S


Example 4: A fluid has density ρ kg/m3 and flows with velocity~V (x , y , z) =

⟨z , y2, x2

⟩where x , y , z are measured in meters and the

components of ~V in meters per second. Find the rate of flow outwardthrough the cylinder x2 + y2 = 4, 0 ≤ z ≤ 1.

Solution: First, parametrize the surface:

~r (θ, z) = 〈2 cos(θ), 2 sin(θ), z〉 θ ∈ [0, 2π] z ∈ [0, 1]

~rθ × ~rz = 〈−2 sin(θ), 2 cos(θ), 0〉 × 〈0, 0, 1〉 = 〈2 cos(θ), 2 sin(θ), 0〉

Note that ~r has positive orientation as it points out from the cylinder.

ρ

∫∫S~V · d ~S = ρ

∫∫R

⟨z , 4 sin2(θ), 4 cos2(θ)

⟩· 〈2 cos(θ), 2 sin(θ), 0〉 dA

= ρ

∫ 2π

0

∫ 1

02z cos(θ) + 8 sin3(θ) dz dθ

= ρ

(sin(θ) +

1

3cos3(θ)− cos(θ)

)∣∣∣∣2π0

= 0kg

s


Example 5: Calculate the flux of ~F (x , y , z) = 〈0, y ,−z〉 through thesurface S consisting of the paraboloid y = x2 + z2, y ≤ 1 and the diskx2 + z2 ≤ 1, y = 1.

Solution: S is the closed surface consisting of S1 + S2.S1 is parametrized by

~r (x , z) =⟨x , x2 + z2, z

⟩on x2 + z2 ≤ 1.

~rx × ~rz = 〈2x ,−1, 2z〉

S2 is parametrized by

~s (r , θ) = 〈r sin(θ), 1, r cos(θ)〉

on r ∈ [0, 1], θ ∈ [0, 2π].

~sr × ~sθ = 〈0, r , 0〉The parametrization of S points outward from the surface.∫∫

S~F · d ~S =

∫∫S1

~F · d ~S +

∫∫S2

~F · d ~S

=

∫∫R1

−x2 − z2 − 2z2 dA +

∫∫R2

1− r2 cos2(θ) dA

= −π + 5π/3 = 2π/3


Types of Integral

Let f be a scalar function and ~F a vector field.

Scalar Line Integral along a curve C parametrized by ~r (t) on [a, b].∫Cf ds =

∫ b

af (~r (t))

∣∣~r ′(t)∣∣ dt

Vector Line Integral along an oriented curve C:∫C~F · d~r =

∫ b

a

~F (~r (t)) · ~r ′(t) dt

Scalar Surface Integral over a surface S parametrized by ~r (u, v) on R:∫∫Sf dS =

∫∫Rf (~r (u, v)) |~ru × ~rv | dA

Vector Surface Integral over an oriented surface S:∫∫S~F · d ~S =

∫∫R~F (~r (u, v)) · (~ru × ~rv ) dA


Documents

Parametrized Surfaces and Surface Integralspeople.ku.edu/~s334m797/Math127Fall2018/Week13post.pdfSurface Integrals of Vector Fields (I)Tangent Lines and Planes of Parametrized Surfaces