Parameterized Surfaces and Surface Integrals

1 Problems

1. Consider the circle in the xz-plane with center at x = 5, z = 0 andradius 3. Let S be the surface obtained by rotating this circle aboutthe z-axis (this surface is called a torus).

(a) Find parametric equations to describe this surface.

(b) Set up an integral to compute the surface area of S.

(c) Set up an integral to find

S

z2 dA.

(d) Orient S with unit normal vectors pointing away from the inside

of the torus. Set up an integral to find

S

(x i+ y j) dA.

2. Let S be the surface defined by the equations x2 + y2 = 9, 0 z 4,oriented with unit normal pointing outwards. Evaluate the followingintegrals.

(a)

S

(x+ y + z) dA

(b)

S

(xz i+ x j+ z2 k

) dA3. Use geometric reasoning to compute the following surface integrals:

(a) The integral S

4 dA

where S is the boundary of the region x2 + y2 9, 4 z 11.

(b) The integral S

z dA

where S is the disc x2 + y2 4 on the plane z = 3.

(c) The integral S

F dA

where F(x, y, z) = (x+ 3y5, y + 10xz, z xy) and S is the upperhalf (z 0) of the sphere x2 + y2 + z2 = 1, oriented with unitnormal pointing outwards. (Hint: Use symmetry.)

2 Solutions

1. Consider the circle in the xz-plane with center at x = 5, z = 0 andradius 3. Let S be the surface obtained by rotating this circle aboutthe z-axis (this surface is called a torus).

(a) Find parametric equations to describe this surface.

(b) Set up an integral to compute the surface area of S.

(c) Set up an integral to find

S

z2 dA.

(d) Orient S with unit normal vectors pointing away from the inside

of the torus. Set up an integral to find

S

(x i+ y j) dA.

Answer:

(a) If we look at the surface in the rz-plane, we have a circle of radius3 centered at r = 5, z = 0. We can parameterize the circle as:

r = 5 + 3 cos t

z = 3 sin t0 t 2pi

By letting = u, we can parameterize the surface in cylindricalcoordinates:

r = 5 + 3 cos t

= u

z = 3 sin t

0 t 2pi0 u 2pi

Since x = r cos and y = r sin , we can parameterize the surfacein Cartesian coordinates:

x = (5 + 3 cos t) cosu

y = (5 + 3 cos t) sinu

z = 3 sin t

0 t 2pi0 u 2pi

(b) Using the above parametrization, we compute the tangent vectorsto the surface, Tt and Tu:

Tt =

(x

t,y

t,z

t

)= (3 sin t cosu, 3 sin t sinu, 3 cos t)

Tu =

(x

u,y

u,z

u

)= ( sinu(5 + 3 cos t), cosu(5 + 3 cos t), 0)

Next, we compute Tt Tu:

Tt Tu =

i j k3 sin t cosu 3 sin t sinu 3 cos t

sinu(5 + 3 cos t) cos u(5 + 3 cos t) 0

= 3 cos t cosu(5 + 3 cos t) i 3 cos t sinu(5 + 3 cos t) j

+ (5 + 3 cos t)(3 sin t cos2 u 3 sin t sinu) k

= (5 + 3 cos t) (3 cos t cosu i 3 cos t sin u j 3 sin tk)Next, we compute Tt Tu:Tt Tu = (5 + 3 cos t)

9 cos2 t cos2 u+ 9 cos2 t sin2 u+ 9 sin2 t

= (5 + 3 cos t)9 cos2 t+ 9 sin2 t

= (5 + 3 cos t)9 = 15 + 9 cos t

Thus:dA = Tt Tu dt du = (15 + 9 cos t) dt du

Thus:

Surface Area =

2pi0

2pi0

(15 + 9 cos t) dt du

(c) From part (a), we have:

dA = Tt Tu dt du = (15 + 9 cos t) dt duThus:

S

z2 dA =

2pi0

2pi0

9 sin2 t (15 + 9 cos t) dt du

(d) First we need to check that we have oriented the surface correctly.We can do this by checking the direction of the normal vectorTt Tu at a point on the surface.For simplicity, we choose the point where t = 0 and u = 0. Plug-ging t = 0 and u = 0 into the parametric equations, we get thepoint (8, 0, 0). Plugging t = 0 and u = 0 into the formula forTt Tu, we see that the normal vector at this point is (24, 0, 0).This vector points into the torus instead of out of the torus, so wewill need to negate the the normal vector.

We compute (x i+ y j) (Tt Tu):(x i+ y j) (Tt Tu) = 3 cos t cos2 u(5 + 3 cos t)2 + 3 cos t sin2 u(5 + 3 cos t)2

= 3 cos t(5 + 3 cos t)2

Thus:S

(x i+ y j) dA = 2pi0

2pi0

3 cos t(5 + 3 cos t)2 dt du

2. Let S be the surface defined by the equations x2 + y2 = 9, 0 z 4,oriented with unit normal pointing outwards. Evaluate the followingintegrals.

(a)

S

(x+ y + z) dA

(b)

S

(xz i+ x j+ z2 k

) dAAnswer:

(a) First, we need to parameterize the surface. If we choose the pa-rameters t = and u = z, then a parametrization is

x = 3 cos t

y = 3 sin t

z = u

0 t 2pi0 u 4

Then:

Tt = (3 sin t, 3 cos t, 0)Tu = (0, 0, 1)

Thus:

Tt Tu =

i j k3 sin t 3 cos t 0

0 0 1

= 3 cos t i+ 3 sin t j

Thus:

dA = Tt Tudt du =9 cos2 t+ 9 sin2 t dt du = 3 dt du

Now, we can evaluate the integral:S

(x+ y + z) dA =

40

2pi0

(3 cos t+ 3 sin t+ u) 3 dt du

= 3

40

2piu du = 6pi

[1

2u2]40

= 48pi

(b) First we need to check that we have oriented the surface correctly.The normal vector Tt Tu = 3 cos t i + 3 sin t j = (x, y, 0) pointsoutwards, and so the orientation is correct.

Next, we compute (xz i+ x j+ z2 k) (Tt Tu):(xz i+ x j+ z2 k

) (Tt Tu) = 9u cos2 t+ 9 cos t sin t

Now, we can evaluate the integral:S

(xz i+ x j+ z2 k

) dA = 40

2pi0

(9u cos2 t+ 9 cos t sin t) dt du

=

40

2pi0

(9

2u (1 + cos(2t)) +

9

2sin(2t)

)dt du

=

40

[9

2u

(t+

1

2sin(2t)

) 94cos(2t)

]2pi0

du

=

40

9piu du = 72pi

3. Use geometric reasoning to compute the following surface integrals:

(a) The integral S

4 dA

where S is the boundary of the region x2 + y2 9, 4 z 11.

(b) The integral S

z dA

where S is the disc x2 + y2 4 on the plane z = 3.

(c) The integral S

F dA

where F(x, y, z) = (x+ 3y5, y + 10xz, z xy) and S is the upperhalf (z 0) of the sphere x2 + y2 + z2 = 1, oriented with unitnormal pointing outwards. (Hint: Use symmetry.)

Answer:

(a) The region is a cylinder about the z-axis with radius 3 betweenz = 4 and z = 11. Since S is the boundary of this region, Sincludes the cylinder x2 + y2 = 9, 4 z 11, and the top andbottom circles. The integral

S

4 dA is 4 times the surface area

of S.

The surface area of the cylinder x2 + y2 = 9, 4 z 11 is2pirh = 2pi(3)(11 4) = 42pi. The surface area of the top circleis pir2 = pi(3)2 = 9pi, as is the surface area of the bottom circle.Thus, the surface area of S is 42pi + 9pi + 9pi = 60pi, so

S

4 dA = 4(60pi) = 240pi

(b) Since the surface is on the plane z = 3, the integral is

S

3 dA,

which equals 3 times the surface area of S. Since S is the discx2 + y2 4, the area of S is pir2 = 4pi. Thus:

S

z dA = 3(4pi) = 12pi

(c) We want to write the integral as a scalar surface integral

S

F N dAwhere N is the unit normal vector to the surface. Since S is partof a sphere centered at the origin, the vector x i+y j+z k is normalto S and points in the outward direction. So:

N =x i+ y j+ z kx2 + y2 + z2

On the surface S, x2 + y2 + z2 = 1, so

N = x i+ y j+ z k

Thus,

F N = (x+ 3y5, y + 10xz, z xy) (x, y, z)= x2 + y2 + z2 + 3xy5 + 10xyz xyz= 1 + 3xy5 + 9xyz

Thus, we want to evaluate the integralS

(1 + 3xy5 + 9xyz

)dA

This is now the tricky part, where we have to use symmetry. Sincethe surface S is symmetric over the yz-plane, the integral

S

(3xy5 + 9xyz

)dA

equals 0 (since the portion where x is negative cancels with theportion where x is positive). Thus:

S

F dA =

S

(1 + 3xy5 + 9xyz

)dA =

S

1 dA

This integral equals the surface area of S, which is half the surfacearea of a sphere of radius 1.

S

F dA =

S

dA =1

2(4pi) = 2pi