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6.6 Surface integrals  (Page 8/27)

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Calculate ∬ S ( x 2 − z ) d S , where S is the surface with parameterization r ( u , v ) = ⟨ v , u 2 + v 2 , 1 ⟩ , 0 ≤ u ≤ 2 , 0 ≤ v ≤ 3 .

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Calculating the surface integral of a piece of a sphere

Calculate surface integral ∬ S f ( x , y , z ) d S , where f ( x , y , z ) = z 2 and S is the surface that consists of the piece of sphere x 2 + y 2 + z 2 = 4 that lies on or above plane z = 1 and the disk that is enclosed by intersection plane z = 1 and the given sphere ( [link] ).

A diagram in three dimensions of the upper half of a sphere. The center is at the origin, and the radius is 2. The top part above the plane z=1 is cut off and shaded; the rest is simply an outline of the hemisphere. The top section has center at (0,0,1) and radius of radical three.
Calculating a surface integral over surface S .

Notice that S is not smooth but is piecewise smooth; S can be written as the union of its base S 1 and its spherical top S 2 , and both S 1 and S 2 are smooth. Therefore, to calculate ∬ S z 2 d S , we write this integral as ∬ S 1 z 2 d S + ∬ S 2 z 2 d S and we calculate integrals ∬ S 1 z 2 d S and ∬ S 2 z 2 d S .

First, we calculate ∬ S 1 z 2 d S . To calculate this integral we need a parameterization of S 1 . This surface is a disk in plane z = 1 centered at ( 0 , 0 , 1 ) . To parameterize this disk, we need to know its radius. Since the disk is formed where plane z = 1 intersects sphere x 2 + y 2 + z 2 = 4 , we can substitute z = 1 into equation x 2 + y 2 + z 2 = 4 :

x 2 + y 2 + 1 = 4 ⇒ x 2 + y 2 = 3 .

Therefore, the radius of the disk is 3 and a parameterization of S 1 is r ( u , v ) = ⟨ u cos v , u sin v , 1 ⟩ , 0 ≤ u ≤ 3 , 0 ≤ v ≤ 2 π . The tangent vectors are t u = ⟨ cos v , sin v , 0 ⟩ and t v = ⟨ − u sin v , u co s v , 0 ⟩ , and thus

t u × t v = | i j k cos v sin v 0 − u sin v u cos v 0 | = ⟨ 0 , 0 , u cos 2 v + u sin 2 v ⟩ = ⟨ 0 , 0 , u ⟩ .

The magnitude of this vector is u . Therefore,

∬ S 1 z 2 d S = ∫ 0 3 ∫ 0 2 π f ( r ( u , v ) ) ‖ t u × t v ‖ d v d u = ∫ 0 3 ∫ 0 2 π u d v d u = 2 π ∫ 0 3 u d u = 2 π 3 .

Now we calculate ∬ S 2 d S . To calculate this integral, we need a parameterization of S 2 . The parameterization of full sphere x 2 + y 2 + z 2 = 4 is

r ( ϕ , θ ) = ⟨ 2 cos θ sin ϕ , 2 sin θ sin ϕ , 2 cos ϕ ⟩ , 0 ≤ θ ≤ 2 π , 0 ≤ ϕ ≤ π .

Since we are only taking the piece of the sphere on or above plane z = 1 , we have to restrict the domain of ϕ . To see how far this angle sweeps, notice that the angle can be located in a right triangle, as shown in [link] (the 3 comes from the fact that the base of S is a disk with radius 3 ) . Therefore, the tangent of ϕ is 3 , which implies that ϕ is π / 6 . We now have a parameterization of S 2 :

r ( ϕ , θ ) = ⟨ 2 cos θ sin ϕ , 2 sin θ sin ϕ , 2 cos ϕ ⟩ , 0 ≤ θ ≤ 2 π , 0 ≤ ϕ ≤ π / 3 .
A diagram of a plane within the three-dimensional coordinate system. Two points are marked on the z axis: (0,0,2) and (0,0,1). The distance from the origin to (0,0,1) is marked as 1, the horizontal distance between the point (0,0,1) and a point of the sphere is labeled radical three, and the angle between the origin and the point on the sphere is theta. There is a line drawn from the origin to the point on the sphere, and this forms a triangle.
The maximum value of ϕ has a tangent value of 3 .

The tangent vectors are

t ϕ = ⟨ 2 cos θ cos ϕ , 2 sin θ cos ϕ , −2 sin ϕ ⟩ and t θ = ⟨ −2 sin θ sin ϕ , u cos θ sin ϕ , 0 ⟩ ,

and thus

t ϕ × t θ = | i j k 2 cos θ cos ϕ 2 sin θ cos ϕ −2 sin ϕ −2 sin θ sin ϕ 2 cos θ sin ϕ 0 | = ⟨ 4 cos θ sin 2 ϕ , 4 sin θ sin 2 ϕ , 4 cos 2 θ cos ϕ sin ϕ + 4 sin 2 θ cos ϕ sin ϕ ⟩ = ⟨ 4 cos θ sin 2 ϕ , 4 sin θ sin 2 ϕ , 4 cos ϕ sin ϕ ⟩ .

The magnitude of this vector is

‖ t ϕ × t θ ‖ = 16 cos 2 θ sin 4 ϕ + 16 sin 2 θ sin 4 ϕ + 16 cos 2 ϕ sin 2 ϕ = 4 sin 4 ϕ + cos 2 ϕ sin 2 ϕ .

Therefore,

∬ S 2 z d S = ∫ 0 π / 6 ∫ 0 2 π f ( r ( ϕ , θ ) ) ‖ t ϕ × t θ ‖ d θ d ϕ = ∫ 0 π / 6 ∫ 0 2 π 16 cos 2 ϕ sin 4 ϕ + cos 2 ϕ sin 2 ϕ d θ d ϕ = 32 π ∫ 0 π / 6 cos 2 ϕ sin 4 ϕ + cos 2 ϕ sin 2 ϕ d ϕ = 32 π ∫ 0 π / 6 cos 2 ϕ sin ϕ sin 2 ϕ + cos 2 ϕ d ϕ = 32 π ∫ 0 π / 6 cos 2 ϕ sin ϕ d ϕ = 32 π [ − cos 3 ϕ 3 ] 0 π / 6 = 32 π [ 1 3 − 3 8 ] = 32 π 3 − 4 3 .

Since ∬ S z 2 d S = ∬ S 1 z 2 d S + ∬ S 2 z 2 d S , we have ∬ S z 2 d S = ( 2 π − 4 ) 3 + 32 π 3 .

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Calculate line integral ∬ S ( x − y ) d S , where S is cylinder x 2 + y 2 = 1 , 0 ≤ z ≤ 2 , including the circular top and bottom.

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Scalar surface integrals have several real-world applications. Recall that scalar line integrals can be used to compute the mass of a wire given its density function. In a similar fashion, we can use scalar surface integrals to compute the mass of a sheet given its density function. If a thin sheet of metal has the shape of surface S and the density of the sheet at point ( x , y , z ) is ρ ( x , y , z ) , then mass m of the sheet is m = ∬ S ρ ( x , y , z ) d S .

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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